Category: OP Malhotra Solutions

Parents can use Class 11 ISC Maths S Chand Solutions Chapter 16 The Straight Line Ex 16(h) to provide additional support to their children.

S Chand Class 11 ICSE Maths Solutions Chapter 16 The Straight Line Ex 16(h)

Question 1.
Find the equation of the straight line which passes through the point (4, 5) and is (i) parallel, (ii) perpendicular to the straight line 3x – 2y + 5 = 0.
Solution:
(i) given eqn. of straight line be
3x – 2y + 5 = 0 …(1)
∴ eqn. of line parallel to line (1) be given by
3x – 2y + k = 0 …(2)
eqn. (2) passes through the point (4, 5); we have
12 – 10 + k = 0 ⇒ k = – 2
putting k = – 2 in eqn. (2); we have
3x – 2y – 2 = 0
be the required eqn. of line.

(ii) eqn. of line ⊥ to line (1) be given by
2x + 3y + c = 0 …(3)
eqn. (3) passes through the point (4, 5); we have
8 + 15 + c = 0 ⇒ c = – 23
putting the value of c in eqn. (3);
2x + 3y – 23 = 0 be the required line.

Question 2.
Find the equation of the straight line which is such that
(i) it passes through the point (4, 3) and is parallel to the line 3x – 4y + 5 = 0.
(ii) it passes through the point (4, 3) and is perpendicular to the line 3x – 4y + 5 = 0.
Solution:
(i) eqn. of given line be
3x – 4y + 5 = 0 …(1)
The eqn. of line parallel to line (1) be given by
3x – 4y + k = 0 …(2)
Now eqn. (2) passes through the point (4, 3); we get
12 – 12 + k = 0 ⇒ k = 0
putting the value of k in eqn. (2); we get 3x – 4y = 0 be the required line.

(ii) The eqn. of line ⊥ to line 3x – 4y + 5 = 0 be given by
4x + 3y + k = 0 …(1)
Now eqn. (1) passes through the point (4, 3), we get
16 + 9 + k = 0 ⇒ k = – 25
putting the value of k in eqn. (1); we get
4x + 3y – 25 = 0 be the required line.

Question 3.
Find the equation of the straight line which passes through
(i) the origin and the point of intersection of the st. lines y – x + 7 = 0, y + 2x – 2 = 0 ;
(ii) the point (2, – 9) and the intersection of the lines 2x + 5y – 8 = 0 and 3x – 4y = 35 ;
(iii) the origin and the point of intersection of the lines ax + by + c = 0 and a’x + b’y + c = 0
Solution:
(i) The equations of given lines are ;
y – x + 7 = 0 …(1)
y + 2x – 2 = 0 …(2)
Thus the eqn. of line passes through the point of intersection of lines (1) and (2) be given by
y – x + 7 + k(y + 2x – 2) = 0 …(3)
Now eqn. (3) passes through origin (0, 0); we get
0 – 0 + 7 + k(0 + 0 – 2) = 0
⇒ 7 – 2k = 0 ⇒ k = \(\frac { 7 }{ 2 }\)
putting the value of k = \(\frac { 7 }{ 2 }\) in eqn. (3); we get
y – x + 7 + \(\frac { 7 }{ 2 }\)(y + 2x – 2) = 0
⇒ 12x + 9y = 0 ⇒ 4x + 3y = 0
which is the required line.

(ii) The eqns. of given lines are ;
2x + 5y – 8 = 0 …(1)
and 3x – 4y – 35 = 0 …(2)
Thus, the eqn. of line passes through the point of intersection of lines (1) and (2) be given by
2x + 5y – 8 + k(3x – 4y – 35) = 0 …(3)
eqn. (3) passes through the point (2, – 9).
4 – 45 – 8 + k(6 + 36 – 35) = 0
⇒ – 49 + 7k = 0
⇒ k = 7
putting the value of k in eqn. (3) ; we have
2x + 5y – 8 + 7(3x – 4y – 35) = 0
⇒ 23x – 23y – 253 = 0
⇒ x – y – 11 = 0
which is the required eqn.

(iii) The eqns. of given lines are
ax + by + c = 0 …(1)
and a’x + b’y + c’ = 0 …(2)
Thus, the eqn. of line passes through the point of intersection of lines (1) and (2) be given by
ax + by + c + k(a’x + b’y + c’) = 0 …(3)
Now eqn. (3) passes through origin (0, 0); we have
0 + 0 + c + k(0 + 0 + c’) = 0
⇒ c + kc’ = 0 ⇒ k = –\(\frac{c}{c^{\prime}}\)
putting the value of k in eqn. (3) ; we get
ax + by + c – \(\frac{c}{c^{\prime}}\) (a’x + b’y + c’) = 0
⇒ (ac’ – a’c) x + (bc’ – cb’) y = 0
which is the required eqn. of straight line.

Question 4.
Show that the equation
n(ax + by + c) = c(lx + my + n)
represents the line joining the origin to the point of intersection of
ax + by + c = 0 and lx + my + n = 0 .
Solution:
The equations of given lines are
ax + by + c = 0 …(1)
lx + my + n = 0 …(2)
Thus, the equation of line through the point of intersection of lines (1) and (2) be given by
ax + by + c + k(lx + my + n) = 0 …(3)
eqn. (3) passes through the point (0, 0); we get
0 + 0 + c + k(0 + 0 + n) = 0 ⇒ k = –\(\frac { c }{ n }\)
putting the value of k = –\(\frac { c }{ n }\) in eqn. (3) ; we get
ax + by + c –\(\frac { c }{ n }\)(lx + my + n) = 0
⇒ n(ax + by + c) – c (lx + my + n) = 0
which is the required line.

Question 5.
Find the equation of the line through the intersection of x – y = 1 and 2x – 3y + 1 = 0 and parallel to 3x + 4y = 12.
Solution:
The eqns. of given lines are
x – y – 1 = 0 …(1)
and 2x – 3y + 1 = 0 …(2)
Thus the eqn. of line passes through the point of intersection of (1) and (2) be given by
x – y – 1 + k(2x – 3y + 1) = 0
⇒ (1 + 2k) x + (- 1 – 3k) y – 1 + k = 0 …(3)
∴ slope of line (3) = –\(\frac{(1+2 k)}{-1-3 k}\) = \(\frac{1+2 k}{1+3 k}\)
It is given that line (3) is parallel to given line 3x + 4y – 12 = 0 whose slope be –\(\frac { 3 }{ 4 }\)
∴ \(\frac{1+2 k}{1+3 k}\) = –\(\frac { 3 }{ 4 }\) [∵ m₁ = m₂]
⇒ 4 + 8k = – 3 – 9k
⇒ 17k = – 7
⇒ k = \(\frac { -7 }{ 17 }\)
putting the value of k in eqn. (3); we have
\(\left(1-\frac{14}{17}\right) x\) + \(\left(-1+\frac{21}{17}\right) y\) y – 1 – \(\frac{7}{17}\) = 0
⇒ \(\frac { 3x }{ 17 }\) + \(\frac { 4y }{ 17 }\) – \(\frac { 24 }{ 17 }\) = 0
⇒ 3x + 4y – 24 = 0
which is the required r=eqn. of line

Question 6.
Find the equation of the line through the intersection of x + 2y + 3 = 0
and 3x + 4y + 7 = 0
and parallel to y – x = 8.
Solution:
The eqns. of given lines are
x + 2y + 3 = 0 …(1)
3x + 4y + 7 = 0 …(2)
Thus the eqn. of line through the intersection of lines (1) and (2) be given by
x + 2y + 3 + k(3x + 4y + 7) = 0
⇒ (1 + 3k) x + (2 + 4k) y + 3 + 7k = 0 …(3)
∴ slope of line (3) = –\(\frac{(1+3 k)}{2+4 k}\)
Also slope of given line y – x – 8 = 0 be 1 Since it is given that line (3) is parallel to y – x – 8 = 0 ∴ their slopes are equal.
∴ \(\frac{1+3 k}{2+4 k}\) = -1 ⇒ 1 + 3k = – 2 – 4k
⇒ 7k = – 3 ⇒ k = –\(\frac{3}{7}\)
putting the value of k in qn. (3) ; we get
\(\left(1-\frac{9}{7}\right) x\) + \(\left(2-\frac{12}{7}\right) y\) + 3 – 3 = 0
⇒ -2x + 2y = 0 ⇒ x – y = 0
which is the required line.

Question 7.
Find the equation of the line through the intersection of y + x = 0 and 2x – 3y + 7 = 0, and perpendicular to the line 2y – 3x – 5 = 0.
Solution:
equations of given lines are
y + x – 9 = 0 …(1)
and 2x – 3y + 7 = 0 …(2)
Thus, eqn. of line through the intersection of lines (1) and (2) be given by
(y + x – 9) + k (2x – 3y + 7) = 0
⇒ (1 + 2k) x + (1 – 3k) y – 9 + 7k = 0 …(3)
∴ slope of line (3) = \(-\frac{(1+2 k)}{1-3 k}\)
Also slope of given line 2y – 3x – 5 = 0 be \(\frac{-(-3)}{2}\) = \(\frac{3}{2}\)
Since it is given that line (3) is ⊥ to the line 2y – 3x – 5 = 0. Thus product of their slopes must be equal to -1 .
\(\left(\frac{-1-2 k}{1-3 k}\right)\) \(\left(\frac{3}{2}\right)\) = -1
⇒ \(\left(\frac{2 k+1}{3 k-1}\right)\) \(\frac{3}{2}\) = -1
⇒ 6k + 3 = – 6k + 2
⇒ 12k = – 1 ⇒ k = \(\frac{-1}{12}\)
putting the value of k in eqn. (3) ; we have

which is the required eqn. of line.

Practicing OP Malhotra Maths Class 11 Solutions Chapter 9 Complex Numbers Ex 9(f) is the ultimate need for students who intend to score good marks in examinations.

S Chand Class 11 ICSE Maths Solutions Chapter 9 Complex Numbers Ex 9(f)

Question 1.
Find the square root of the following complex numbers.
(i) 3 + 4i
(ii) – 8 + 6i
(iii) – 40 – 42i
(iv) i
(v) \(\left(\frac{2+3 i}{5-4 i}+\frac{2-3 i}{5-4 i}\right)\)
Solution:
(i) \(\sqrt{3+4 i}\) = x + iy; where x, y ∈ R
On squaring both sides ; we have
3 + 4i = (x + iy)²
⇒ 3 + 4i = x² – y² + 2ixy
On comparing real and imaginary parts on both sides ; we have
x² – y² = 3 …(1)
and 2xy = 4 …(2)
∴ x² + y² = \(\sqrt{\left(x^2-y^2\right)^2+(2 x y)^2}\)
= \(\sqrt{3^2+4^2}\) = 5 … (3)
On adding (1) and (3); we have
2x² = 8 ⇒ x² = 4 ⇒ x = ± 2
eqn. (3) – eqn. (1) gives ;
2y² = 2 ⇒ y = ± 1
Since xy be +ve
∴ x and y both are of same sign
Hence x = 2, y = 1 or x = – 2, y = – 1
∴ \(\sqrt{3+4 i}\) = 2 + i or – (2 + i)

(ii) \(\sqrt{- 8 + 6 i}\) = x + iy where x, y ∈ R
On squaring both sides ; we have
– 8 + 6i = (x + iy)² = x² – y² + 2ixy
On ωmparing real and imaginary parts on both sides, we get
x² – y² = – 8 … (1)
and 2xy = 6 … (2)
Now x² + y² = \(\sqrt{\left(x^2-y^2\right)^2+(2 x y)^2}\)
= \(\sqrt{(-8)^2+6^2}\)
= \(\sqrt{64+36}\) = 10 … (3)
On adding (1) and (3) ; we have
2x² = 2 ⇒ x = ± 1
eqn. (3) – eqn. (1) gives ;
2y² = 18 ⇒ y² = 9 ⇒ y = ± 3
Since xy be +ve ∴ both x and y are of same sign.
∴ x = 1; y = 3 or x = – 1, y = – 3
Thus \(\sqrt{-8+6 i}\) = 1 + 3i or – (1 + 3i)

(iii) \(\sqrt{-40-42 i}\) = x – iy where x, y ∈ R
On squaring both sides ; we have
– 40 – 42i = (x – iy)² = x² – y² – 2ixy
On comparing real and imaginary parts on both sides ; we have
– 40 = x² – y² …(1)
and 2xy = 42 …(2)
Now x² + y² = \(\sqrt{\left(x^2-y^2\right)^2+(2 x y)^2}\)
= \(\sqrt{(-40)^2+42^2}\)
= \(\sqrt{1600+1764}\)
= \(\sqrt{3364}\) = 58 … (3)
On adding (1) and (3) ; we have
2x² =18 ⇒ x = ± 3
eqn. (3) – eqn. (1) gives ;
2y² = 98 ⇒ y² = 49 ⇒ y = ± 7
Since xy be of +ve sign ∴ both x and y are of same sign
i.e. when x = 3, y = 7 or x = – 3, y = – 7
Thus, \(\sqrt{-40-42 i}\)
= 3 – 7i or – 3 + 7i
= ±(3 – 7i)

(iv) Let \(\sqrt{i}\) = x + iy;
On squaring; we have i = (x + iy)² = x² – y² + 2ixy
On comparing real and imaginary parts on both sides, we have
x² – y² = 0 …(1)
and 2xy = 1 …(2)
∴ x² + y² = \(\sqrt{\left(x^2-y^2\right)^2+(2 x y)^2}\)
= \(\sqrt{0^2+1^2}\) = 1 … (3)
On adding eqn. (1) and eqn. (3); we have

Question 2.
If ω is a cube root of unity, then
(i) ω + ω² = ….
(ii) 1 + ω = …
(iii) 1 + ω³ = ….
(iv) ω³ = ….
Solution:
Now ω be the cube root of unity
∴ ω = \(\frac{-1+\sqrt{3} i}{2}\) and ω² = \(\frac{-1-\sqrt{3} i}{2}\)

(i) 1 + ω² = \(\frac{-1+\sqrt{3} i}{2}+\frac{-1-\sqrt{3} i}{2}\)
= \(\frac { -2 }{ 2 }\) = – 1

(ii) 1 + ω = 1 + \(\frac{-1+\sqrt{3} i}{2}=\frac{2-1+\sqrt{3} i}{2}=\frac{1+\sqrt{3} i}{2}\) = – ω²

(iii) 1 + ω² = 1 + \(\frac{-1-\sqrt{3} i}{2}=\frac{2-1-\sqrt{3} i}{2}=\frac{1-\sqrt{3} i}{2}=-\left(\frac{-1+\sqrt{3} i}{2}\right)\) = – ω

(iv) ω³ = ω². ω = \(\left(\frac{-1-\sqrt{3} i}{2}\right)\left(\frac{-1+\sqrt{3} i}{2}\right)=\frac{(-1)^2-(\sqrt{3} i)^2}{4}=\frac{1+3}{4}\) = 1

Question 3.
If 1, ω, ω² are three cube roots of unity, prove that
(i) (1 + ω²)⁴ = ω
(ii) (1 + ω – ω²)³ = (1 – ω + ω²)³ = – 8
(iii) (1 – ω) (1 – ω²) = 3
(iv) \(\frac{1}{1+\omega}+\frac{1}{1+\omega^2}\) = 1
Solution:
(i) (1 + ω²)⁴ = (- ω)⁴= ω³ . ω = ω [∵ 1 + ω + ω² = 0 and ω³ = 1]

(ii) (1 + ω – ω²)² = (- ω² – ω²)³ = (- 2ω²)³ = – 8ω⁶ = – 8 (ω³)² = – 8
and (1 – ω + ω²)³ = (- ω – ω)³ = (- 2ω)³ = – 8ω³ = – 8 x 1 = – 8 [∵ 1 + ω + ω² = 0 and ω³ = 1]

(iii) (1 – ω) (1 – ω²) = 1 – ω – ω² + ω³ = 1 – (ω + ω²) + 1 [∵ ω³ = 1]
= 1 – (- 1) + 1 [∵ 1 + ω + ω² = 0]
= 1 + 1 + 1 = 3

(iv) \(\frac{1}{1+\omega}+\frac{1}{1+\omega^2}=\frac{1}{-\omega^2}+\frac{1}{-\omega}\) [∵ 1 + ω + ω² = 0]
= \(\frac{1+\omega}{-\omega^2}=\frac{-\omega^2}{-\omega^2}\) = 1

Question 4.
(i) (1 – ω – ω²)⁶ = 64
(ii) (1 + ω – ω²) (1 – ω + ω²) = 4
Solution:
(i) (1 – ω – ω²)⁶ = [1 – (ω + ω²)]⁶ = [1 – (- 1)]⁶ = 2⁶ = 64 [∵ 1 + ω + ω² = 0]

(ii) (1 + ω – ω²) (1 – ω + ω²) = (- ω² – ω²) (- ω – ω) [∵ 1 + ω + ω² = 0]
= (- 2ω²) (- 2ω)
= 4ω³
= 4 x 1 = 4

Question 5.
(3 + 5ω + 3ω²)⁶ = (3 + 5ω² + 3ω)⁶ = 64
Solution:
(3 + 5ω + 3ω²)⁶ = (3 + 5ω² + 3ω)⁶
= [3 (- ω) + 5ω]⁶ [∵ 1 + ω + ω² = 0]
= (2ω)⁶ = 64 (ω³)² = 64
and (3 + 5ω² + 3ω)⁶ = [3 (1 + ω) + 5ω²]⁶ = [3 (- ω²) + 5ω²]⁶ [∵ 1 + ω + ω² = 0]
= (2ω²)⁶ = 64ω¹² = 64 (ω³)⁴ = 64 [∵ ω³ = 1]

Question 6.
ω²⁸ + ω²⁹ + 1 = 0
Solution:
ω²⁸ + ω²⁹ + 1 = (ω³)⁹ . ω + (ω³)⁹ . ω² + 1
= 1⁹ . ω + 1⁹ . ω² + 1 [∵ ω³ = 1]
= ω + ω² + 1 = 0

Question 7.
Prove that \(\left(\frac{-1+i \sqrt{3}}{2}\right)^n+\left(\frac{-1-i \sqrt{3}}{2}\right)^n\) is equal to 2 if n be a multiple of 3 and is equal to – 1 if n be any other integer.
Or
If 1, ω, ω² are the cube roots of unity, prove that ωⁿ + ω²ⁿ = 2 or – 1 acωrding as n is a multiple of 3 or any other integer.
Solution:
We know that cube root of unity are 1, ω, ω²
where ω = \(\frac{-1+\sqrt{3} i}{2}\) and ω² = \(\frac{-1-\sqrt{3} i}{2}\)
∴ \(\left(\frac{-1+i \sqrt{3}}{2}\right)^n+\left(\frac{-1-i \sqrt{3}}{2}\right)^n=\omega^n+\left(\omega^2\right)^n=\omega^n+\omega^{2 n}\)

Case-I : When n be a multiple of 3
∴ n = 3k
∴ ωⁿ + ω²ⁿ = ω^3k + ω^6k = (ω³) + (ω³)^2k = 1^k + 1^2k =1 + 1 = 2

Case-II : When n be not a multiple of 3
∴ n = 3k + 1, 3k + 2

Subcase – I : When n = 3k + 1
ωⁿ + ω²ⁿ = ω^3k+1 + ω^2(3k+1) = ω^3k . ω + ω^6k . ω²
= 1 . ω + 1 . ω² = ω + ω² = – 1 [∵ ω + ω² +1 = 0 and ω^3k = (ω³)^k = 1]

Subcase-II : When n = 3k+2
∴ ωⁿ + ω²ⁿ = ω^3k+1 + ω^2(3k+2) = ω^3k. ω² + ω^6k + ω⁴ = (ω³)^k ω² + (ω³)^2k . ω³ . ω
= ω² + ω = – 1 [∵ (ω³)^k = 1^k = 1]
Thus, ωⁿ + ω²ⁿ = 2 or – 1
according as n is a multiple of 3 or any other integer.

Prove the following:

Question 8.
(1 – ω + ω²) (1 + ω – ω²) (1 – ω – ω²) = 8.
Solution:
(1 – ω + ω²) (1 + ω – ω²) (1 – ω – ω²) = (1 + ω² – ω) (1 + ω – ω²) (1 – (ω + ω²))
= (- ω – ω) (- ω² – ω²) (1 – (- 1)) [∵ 1 + ω + ω² = 0]
= (- 2ω) (- 2ω²)² = 8ω³ = 8 [∵ ω³ = 1]

Question 9.
(i) (1 + ω) (1 + ω²) (1 + ω⁴) (1 + ω⁸) …. to 2n factors = 1
(ii) (1 – ω + ω²) (1 – ω² + ω⁴) (1 – ω⁴ + ω⁸)…… to 2n factors = 2²ⁿ.
Solution:
(i) L.H.S = (1 + ω) (1 + ω²) (1 + ω⁴) (1 + ω⁸) …. to 2n factors
= (1 + ω) (1 + ω²) (1 + ω³ . ω) (1 + ω⁶ . ω²) … 2n factors
= (1 + ω) (1 + ω²) (1 + ω) (1 + ω²) …. 2n factors
= (1 + ω)ⁿ (1 + ω²)ⁿ = [(1 + ω) (1 + ω²)]ⁿ
= [1 + ω + ω² + ω³]ⁿ = [0 + 1]ⁿ = 1ⁿ = 1 [∵ 1 + ω + ω² = 0 and ω³ = 1]

(ii) (1 – ω + ω²) (1 – ω² + ω⁴) (1 – ω⁴ + ω⁸) to 2n factors
= (1 – ω + ω²) (1 – ω² + ω³ . ω) (1 – ω³ . ω + ω⁶ . ω²) to 2n factors
= (1 – ω + ω²) (1 – ω² + ω) (1 – ω + ω²)…. 2n factors [∵ ω³ = 1]
= [(1 – ω + ω²) …. n factors] [(1 – ω² + ω) …. n factors
= (- 2ω)ⁿ (- 2ω²)ⁿ [∵ 1 + ω + ω² = 0]
= 2ⁿ (ω . ω²)ⁿ (- 1)²ⁿ
= 2²ⁿ (ω³)ⁿ
= 2²ⁿ1ⁿ
= 2²ⁿ

Question 10.
\(\frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2}\) = ω
Solution:

Question 11.
\(\frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}+\frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2}\) = – 1
Solution:

Question 12.
If ω is a cube root of unity and n is a positive integer which is not a multiple of 3, then show that (1 + ωⁿ + ω²ⁿ) = 0.
Solution:
When n is not a multiple of 3
∴ n be of the form 3k + 1 or 3k + 2
Case – I. When n = 3k + 1
1 + ωⁿ + ω²ⁿ = 1 + ω^3k+1 + ω^6k+2 = 1 + (ω³)^k ω + (ω³)^2k . ω² = 1 + 1^k . ω + 1^2k ω²
= 1 + ω + ω² = 0

Case-II. When n = 3k+2
1 + ωⁿ + ω²ⁿ = 1 + ω^3k+2 + ω^2(3k+2)= 1 + ω^3k . ω² + ω^6k . ω⁴
= 1 + (ω³)^k . ω² + (ω³)^2k . ω³ω = 1 + ω² + ω = 0 [∵ ω³ = 1]

Question 13.
Show that (x + ωy + ω²z) (x + ω²y + ωz) = x² + y² + z² – yz – zx – xy.
Solution:
L.H.S = (x + ωy + ω²z) (x + ω²y + ωz)
= x² + (ω² + ω) xy + xz (ω + ω²) + ω³y² + yz (ω² + ω⁴) + ω³z²
= x² – xy + xz(- 1) + y² + yz (- 1) + z² [∵ 1 + ω + ω² = 1]
= x² + y² + z² – xy – yz – zx = R.H.S

Question 14.
Show that x³ + y³ = (x + y) (ωx + ω²y) (ω²x + ωy).
Solution:
R.H.S = (x + y) (ωx + ω²y) (ω²x + ω²y) = (x + y) (ω³x² + ω²xy + ω⁴xy + ω³y²)
= (x + y) [1 . x² + xy (ω² + ω³ . ω) + 1 . y²] [∵ ω³ = 1]
= (x + y) (x² + xy (ω² + ω) + y²) = (x + y) (x² – xy + y²) [∵ 1 + ω + ω² = 1]
= x³ + y³ = L.H.S

Question 15.
If 1, ω, ω² are cube roots of unity, prove that 1, ω² are vertices of an equilateral triangle.
Solution:
Since 1, ω and ω² are the cube root of unity.

Thus, 1, ω and ω² are the vertices of an equilateral triangle.

Well-structured OP Malhotra Maths Class 11 Solutions Chapter 9 Complex Numbers Ex 9(e) facilitate a deeper understanding of mathematical principles.

S Chand Class 11 ICSE Maths Solutions Chapter 9 Complex Numbers Ex 9(e)

Question 1.
Illustrate in the complex plane, the set of points satisfying the following conditions. Explain your answer:
(i) |z| ≤ 3
(ii) arg (z – 2) = \(\frac { π }{ 3 }\)
(iii) | i – 1 – 2z | > 9
Solution:
(i) Let z = x + iy
∴ | z | ≤ 3 ⇒ | x + iy ≤ 3
⇒ x² + y² ≤ 9
which represents the set of points in interior and on the circle with centre (0, 0) and radius 3.

(ii) arg (z – 2) = \(\frac { π }{ 3 }\)
⇒ arg (x + iy – 2) = \(\frac { π }{ 3 }\)
⇒ tan^-1\(\left(\frac{y}{x-2}\right)=\frac{\pi}{3}\)
⇒ \(\frac{y}{x-2}=\tan \frac{\pi}{3}\) = \(\sqrt{3}\) ⇒ y = \(\sqrt{3}\)(x – 2)
which represents the set of points on line which intersects x-axis at (2, 0) and making an angle of 60° with +ve direction of x-axis.

(iii) Given | i – 1 – 2z | > 9
| i – 1 – 2 (x + iy) | > 9, where z = x + iy

Question 2.
Illustrate and explain the region of the Argand’s plane represented by the inequality |z + i| ≥ |z + 2|.
Solution:
Let z = x + iy
given | z + i | ≥ | z + 2 |
⇒ | x + iy + i | ≥ | x + iy + 2 |
⇒ |x + i (y + 1) | ≥ |x + 2 + iy |
⇒ \(\sqrt{x^2+(y+1)^2}\) ≥ \(\sqrt{(x+2)^2+y^2}\)
On squaring both sides ; we have
x² + (y + 1)² ≥ (x + 2)² + y²
⇒ x² + y² + 2y + 1 ≥ x² + 4x + 4 + y²
⇒ 2y + 1 ≥ 4x + 4
⇒ y ≥ 2x + \(\frac { 3 }{ 2 }\)
which represents the set of points in the region lies above the line PQ intersects coordinate axes at (0, \(\frac { 3 }{ 2 }\)) and (- \(\frac { 3 }{ 4 }\), 0)

Question 3.
Illustrate and explain the set of points z in the Argand diagram, which represents | z – z₁ | ≤ 3 where z1 = 3 – 2i
Solution:
Given | z – z₁| ≤ 3 ;
where z₁ = 3 – 2i
⇒ | z – (3 – 2i) | ≤ 3
⇒ | x + iy – 3 + 2i | ≤ 3
⇒ | (x – 3) + i (y + 2) | ≤ 3
⇒ (x – 3)² + (y + 2)² ≤ 9
which represents the set of points in the interior and on the circle with centre (3, – 2) and radius 3.

Question 4.
If z = x + yi and ω = \(\frac{(1-z i)}{z-i}\) then
| ω | = 1 implies that in the complex plane
(a) z lies on the imaginary axis
(b) z lies on the real axis
(c) z lies on the unit circle
(d) None of these
Solution:
Given z = x + iy and w = \(\frac{1-z i}{z-i}\)
and |w| = 1 ⇒ \(\left|\frac{1-z i}{z-i}\right|\) = 1
⇒ |1 – zi | = |z – i| [∵ \(\left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|}\) ]
⇒ | 1 – (x + iy) i | = | x + iy – i |
⇒ |1 – ix + y| = |x + i(y – 1)|
⇒ 1(1 + y) – ix| = | x + | 0 – 1) |
⇒ \(\sqrt{(1+y)^2+(-x)^2}=\sqrt{x^2+(y-1)^2}\)
On squaring both sides ; we have
(1 + y)² + x² = x² + (y – 1)²
⇒ y² + 2y + 1 – x² = x²+y² – 2y + 1
⇒ 4y = 0
⇒ y = 0
which is the eqn. of x-axis. z lies on real axis i.e. x-axis

Question 5.
Find the locus of a complex number z such that arg \(\left(\frac{z-2}{z+2}\right)\) = \(\frac { π }{ 3 }\).
Solution:

Question 6.
If the amplitude of z – 2 – 3i is \(\frac { π }{ 4 }\), then find the locus of z = x +yi.
Solution:
Given z = x + iy
and amp (z – 2 – 3i) = \(\frac { π }{ 4 }\)
⇒ amp (x + iy – 2 – 3 z) = \(\frac { π }{ 4 }\)
⇒ amp [(x – 2) + i (y – 3)] = \(\frac { π }{ 4 }\)
⇒ tan^-1\(\left(\frac{y-3}{x-2}\right)=\frac{\pi}{4}\)
⇒ \(\frac{y-3}{x-2}=\tan \frac{\pi}{4}=1\)
⇒ y – 3 = x – 2
⇒ x – y + 1 = 0
Hence locus of z represents a straight line intersecting coordinate axes at (- 1, 0) and (0, 1).

Question 7.
Find the locus of z if
ω = \(\frac{z}{z-\frac{1}{3} i}\) = 1.
Solution:

Question 8.
A variable complex number z is such that the amplitude of \(\frac{z-1}{z+1}\) is always equal to \(\frac { π }{ 4 }\).
Illustrate the locus of z in the Argand plane.
Solution:

Question 9.
Find the radius and centre of the circle z\(\bar { z }\)+ (1 – i) z + (1 + i) \(\bar { z }\) – 7 =0.
Solution:
Given eqn. of circle be z\(\bar { z }\)+(1 – i)z + (1 + i)\(\bar { z }\) – 7 = 0 …(1)
where z = x + iy and \(\bar { z }\) = x – iy
∴ eqn. (1) becomes ;
⇒ (x + iy) (x – iy) + (1 – i) (x + iy) + (i + i) (x – iy) – 7 = 0
⇒ x² + y² + x + iy – ix + y + x – iy + ix + y – 7 = 0
⇒ x² + y² + 2x + 2y – 7 = 0
⇒ (x² + 2x) + (y² + 2y) – 7 = 0
⇒ (x² + 2x + 1) + (y² + 2y + 1) – 9 = 0
⇒ (x + 1)² + (y + 1)² = 9
which represents a circle with centre (- 1, – 1) and radius 3.

Question 10.
What is the region represented by the inequality 3 < | z – 2 – 3i|< 4 in the Argand plane.
Solution:
Given 3 < |z – 2 – 3i| < 4;

where z = x + iy
⇒ 3 < | x + iy – 2 – 3i| < 4
⇒ 3 < | (x – 2) + | (y – 3) | < 4
⇒ 3 < \(\sqrt{(x-2)^2+(y-3)^2}\) < 4
⇒ 9 < (x – 2)² + (y – 3)² < 4
which represents the set of points lies in the region between two concentric circles with centre (2, 3) and radius 2 and 3.

Students can track their progress and improvement through regular use of Class 11 ISC Maths S Chand Solutions Chapter 16 The Straight Line Ex 16(g).

S Chand Class 11 ICSE Maths Solutions Chapter 16 The Straight Line Ex 16(g)

Question 1.
Find the equations of the lines bisecting the angles between the following pairs of straight lines writing first at the bisector of the angle in which the origin lines :
(i) 3x – 4y + 10 = 0, 5x – 12y – 10 = 0;
(ii) 12x – 5y + 3 = 0, 4x + 3y – 2 = 0.
Solution:
(i) Writing given lines with positive constant terms, we get
3x – 4y + 10 = 0 …(1)
-5x + 12y + 10 = 0 …(2)
The equations of the angular bisectors of lines (1) and (2) are given by

Thus the eqn. of bisector in which origin lies (Taking positive sign) is given by
\(\frac{3 x-4 y+10}{5}\) = \(\frac{-5 x+12 y+10}{13}\)
⇒ 39x – 52y + 130 = – 25x + 60y + 50
⇒ 64x – 112y + 80 = 0
⇒ 4x – 7y + 5 = 0
The eqn. of other bisector be given by
\(\frac{3 x-4 y+10}{5}\) = \(-\frac{(-5 x+12 y+10)}{13}\)
⇒ 39x – 52y + 130 = 25x – 60y – 50
⇒ 14x + 8y + 180 = 0
⇒ 7x + 4y + 90 = 0

(ii) Writing the given eqns. with positive constant terms we have
12x – 5y + 3 = 0 …(1)
-4x – 3y + 2 = 0 …(2)
The equations of angular bisectors of lines (1) and (2) be given by

The bisector of the angle in which origin lies is given by
\(\frac{12 x-5 y+3}{13}\) = \(\frac{-4 x-3 y+2}{5}\)
⇒ 60x – 25y + 15 = – 52x – 39y + 26
⇒ 112x + 14y – 11 = 0
and the eqn. of other bisector be given by
\(\frac{12 x-5 y+3}{13}\) = –\(\left(\frac{-4 x-3 y+2}{5}\right)\)
⇒ 60x – 25y + 15 = + 52x + 39y – 26
⇒ 8x – 64y + 41 = 0

Question 2.
Find the equations of the bisectors of the angles between 4x + 3y – 4 = 0 and 12x + 5y – 3 = 0. Show that these bisectors are at right angles to each other.
Solution:
Eqns. of given lines are
4x + 3y – 4 = 0 …(1)
12x + 5y – 3 = 0 …(2)
The eqn’s of angular bisector of lines (1) and (2) be given by

Taking +ve sign ;
\(\frac{4 x+3 y-4}{5}\) = \(\frac{12 x+5 y-3}{13}\)
⇒ 52x + 39y – 52 = 60x + 25y – 15
⇒ 8x – 14y + 37 = 0 …(3)
Taking -ve sign ;
\(\frac{4 x+3 y-4}{5}\) = –\(\left(\frac{12 x+5 y-3}{13}\right)\)
⇒ 52x + 39x – 52 = – 60x – 25y + 15
⇒ 112x + 64x – 67 = 0 …(1)
slope of line (3) = m₁ = \(\frac{-8}{-14}\) = \(\frac{4}{7}\)
and slope of bisector (4) = m₂ = –\(\frac{112}{64}\) = –\(\frac{7}{4}\)
Here m₁m₂ = \(\frac{4}{7}\)\(\left(-\frac{7}{4}\right)\) = -1
Thus both bisector are at right angle to each other.

Question 3.
Find the locus of a point which moves so that the perpendiculars drawn from it to the two straight lines 3x + 4y = 5, 12x – 5y = 13 are equal.
Solution:
Given lines are
3x + 4y – 5 = 0 …(1)
and 12x – 5y – 13 = 0 …(2)
Let P(x, y) be any point whose locus is to be find out.
according to given condition we have ⊥ distance from P(x, y) to line (1)
= ⊥ distance from P(x, y) to line (2)

Taking +ve sign ;
13(3x + 4y – 5) = 5(12x – 5y – 13)
⇒ 21x – 77y = 0 ⇒ 3x – 11y = 0
Taking -ve sign ;
13(3x + 4y – 5) = – 5(12x – 5y – 13)
⇒ 99x + 27y – 130 = 0

Question 4.
Find the equations of the lines bisecting the angles between the lines 4x – 3y + 12 = 0 and 12x + 5y = 20. Find, without using the tables, the tangent of an angle between one of these bisectors and one of the origin lines.
Solution:
Eqns. of given lines are
4x – 3y + 12 = 0 …(1)
12x + 5y – 20 = 0 …(2)
The eqns. of angular bisectors of lines (1) and (2) are given by

Taking +ve sign ;
13 (4x – 3y + 12) = 5(12x + 5y – 20)
⇒ 8x + 64y – 256 = 0
⇒ x + 8y – 32 = 0 …(3)
Taking -ve sign ;
13(4x – 3y + 12) = – 5(12x + 5y – 20)
⇒ 112x – 14y + 56 = 0
⇒ 8x – y + 4 = 0 …(4)
Let us take the bisector (3) and line (1)
slope of bisector (3) = –\(\frac { 1 }{ 8 }\) = m₁
slope of line (1) = m₂ = \(\frac{-4}{-3}\) = \(\frac{4}{3}\)
Let θ be the angle between then

Let us take the bisector (4) and line (2).
∴ slope of bisector (4) = m₃ = \(\frac{-8}{-1}\) = 8
and slope of line (2) = m₄ =-\(\frac{12}{5}\)

Question 5.
Find the bisector of the acute angle between the lines :
(i) 3x + 4y = 11 and 12x – 5y = 2;
(ii) 5x = 12y + 24 and 12x = 5y + 10.
Solution:
(i) Writing the given equations with positive constant terms, we have
– 3x – 4y + 11 = 0 …(1)
– 12x + 5y + 2 = 0 …(2)
The eqns. of the angular bisector of lines (1) and (2) be given by

Taking +ve sign :
\(\frac{-3 x-4 y+11}{5}\) = \(\frac{-12 x+5 y+2}{13}\)
⇒ -39x – 52y + 143 = – 60x + 25y + 10
⇒ 21x – 77x + 133 = 0
⇒ 3x – 11x + 19 = 0 …(3)
Taking -ve sign ;
\(\frac{-3 x-4 y+11}{5}\) = \(-\left(\frac{-12 x+5 y+2}{13}\right)\)
⇒ – 39x – 52y + 143 = 60x – 25y – 10
⇒ 99x + 27y – 153 = 0
⇒ 11x + 3y – 17 = 0 …(3)
Out of (3) and (4), that bisector will be the bisector of acute angle which makes an angle θ < 45° with one of lines (1) and (2).
Consider the line (1) and bisector (4)
slope of line = m₁ = –\(\frac{3}{4}\)
slope of bisector (4) = m₂ = –\(\frac{11}{3}\)

Thus the bisector (4) will be bisector of the acute angle between lines (1) and (2).

(ii) Writing the given eqns. with positive constant terms
we have, – 5x + 12y + 24 = 0 …(1)
and – 12x + 5y + 10 = 0 …(2)
The eqns. of angular bisectors of lines (1) and (2) are given by

Taking +ve sign ;
– 5x + 12y + 24 = – 12x + 5y + 10
⇒ 7x + 7y + 14 = 0
⇒ x + y + 2 = 0 …(3)
Taking – ve sign ;
– 5x + 12y + 24 = + 12x – 5y – 10
⇒ 17x – 17y – 34 = 0
⇒ x – y – 2 = 0 …(3)
Out of bisector (3) and (4) that bisector will the bisector of the acute angle which makes an angle θ < 45° with any one of the lines (1) and (2).
Let us take the line (1) and bisector (4).
m₁ = slope of line (1) = –\(\frac{(-5)}{12}\) = \(\frac{5}{12}\)
m₂ = slope of bisector (4) = \(\frac{-1}{-1}\) = 1
If θ be the angle between line (1) and bisector (4)

Thus line (4) gives the bisector of acute angle between lines (1) and (2).

Question 6.
Prove that the perpendiculars drawn from any point of the line 2x + 11y = 5 to the lines 24x + 7y = 20 and 4x – 3y = 2 are equal in length.
Solution:
Equations of given lines are
2x + 11y – 5 = 0 …(1)
24x + 7y – 20 = 0 …(2)
4x – 3y – 2 = 0 …(3)
For any point on line (1) ; we put x = 0 in eqn. (1) ; we have
11y = 5 ⇒ y = \(\frac{5}{11}\)
∴ any point on line (1) be \(\left(0, \frac{5}{11}\right)\).
length of ⊥ from \(\left(0, \frac{5}{11}\right)\) to line (2)

Question 7.
A triangle is formed by the lines whose equations are
AB : x + y – 5 = 0,
BC : x + 7y – 7 = 0,
and CA : 7x + y + 14 = 0 .
Find (i) the bisector of the interior angle at B, and (ii) the bisector of the exterior angle at C.
Solution:
The bisector of the interior angle at B be given by

Taking +ve sign ;
\(\frac{x+7 y-7}{5}\) = x + y – 5
⇒ 4x – 2y – 18 = 0
⇒ 2x – y- 9 = 0 …(3)
Taking -ve sign ;
\(\frac{x+7 y-7}{5}\) = \(-\left(\frac{x+y-5}{1}\right)\)
⇒ x + 7y – 7 = – 5x – 5y + 25
⇒ 6x + 12y – 32 = 0
⇒ 3x + 6y – 16 = 0 …(4)
On solving AB and AC simultaneously; we have
x = –\(\frac { 19 }{ 6 }\) and y = \(\frac { 49 }{ 6 }\)
∴ Coordinates of A are \(\left(-\frac{19}{6}, \frac{49}{6}\right) \text {. }\)
On solving the eqns. of BC and AC; we have
x = \(\frac{-105}{48}\) = \(\frac{-35}{16}\) and y = \(\frac{147}{16 \times 7}\) = \(\frac{21}{16}\)
Out of lines (3) and (4), the line will be the required internal bisector of angle B for which points A and C are lies on its opposite sides.
Substituting the coordinates of points B and C in L.H.S of eqn. (4).

Thus points A and C are lies on opposite side of line (4).
∴ eqn. (4) will be the required internal bisector of angle B.
The bisector of interior angle at B are given by
\(\frac{7 x+y+14}{\sqrt{7^2+1^2}}\) = \(\pm \frac{x+7 y-7}{\sqrt{1^2+7^2}}\)
⇒ 7x + y + 14 = ± (x + 7y – 7)
i.e. 6x – 6y + 21 = 0
⇒ 2x – 2y + 7 = 0 …(5)
and 7x + y + 14 = – x – 7y + 7
⇒ 8x + 8y + 7 = 0 …(6)
On solving the eqns. of AB and BC be given by y = \(\frac { 1 }{ 3 }\) and x = \(\frac { 14 }{ 3 }\)
∴ Coordinates of B are \(\left(\frac{14}{3}, \frac{1}{3}\right) \text {. }\)
For point A \(\left(\frac{-19}{6}, \frac{49}{6}\right)\);
2 × \(\left(\frac{-19}{6}\right)\) – 2\(\left(\frac{49}{6}\right)\) + 7 = \(\frac{-38-98+42}{6}\) < 0 For point B\(\left(\frac{14}{3}, \frac{1}{3}\right)\); 2\(\left(\frac{14}{3}\right)\) – \(\frac{2}{3}\) + 7 > 0
Thus points A and B lies on opposite sides of eqn. (5).
∴ eqn. (5) will become the internal bisector of angle C. Thus eqn. (6) gives the bisector of exterior angle at C.

Question 8.
Find the centre of the inscribed circle of the triangle the equations of whose sides are y – 15 = 0, 12y + 5x = 0 and 4y – 3x = 0.
Solution:
Given eqns. of sides of triangle are
y – 15 = 0 …(1)
5x + 12y = 0 …(2)
and 4y – 3x = 0 …(3)
lines (1) and (2) intersects at C(- 36, 15)
lines (2) and (3) intersects at A(0, 0)
and lines (1) and (3) intersects at B (20, 15)

The eqns. of bisectors of the angle opposite to y – 15 = 0 are given by
\(\frac{4 y-3 x}{\sqrt{4^2+(-3)^2}}\) = \(\pm \frac{5 x+12 y}{\sqrt{5^2+12^2}}\)
⇒ \(\frac{4 y-3 x}{5}\) = \(\pm \frac{5 x+12 y}{13}\)
⇒ 52y – 39x = 25x + 60y
⇒ 64x + 8y = 0
⇒ 8x + y = 0 …(4)
and 52y – 39x = – 25x – 80y
⇒ 14x – 112y = 0
⇒ x – 8y = 0 …(5)
For point B(20, 15);
8 × 20 + 15 = 175 > 0
For point C(- 36, 15);
8 × (- 36) + 15 < 0
Thus the points B and C are lies on opposite sides of eqn. (4) and hence eqn. (4) will be the internal bisector of angle A. The internal bisectors of angle B are given by
\(\frac{4 y-3 x}{\sqrt{4^2+(-3)^2}}\) = \(\pm \frac{y-15}{1}\)
⇒ \(\frac{4 y-3 x}{5}\) = ± y – 15
i.e. 3x + y – 75 = 0 …(6)
and 3x – 9y + 75 = 0
⇒ x – 3y + 25 = 0 …(7)
For point A(0, 0);
L.H.S of eqn. (7) = 0 + 25 = 25 > 0
For point C(- 36, 15);
L.H.S = – 36 – 45 + 25 = – 56 < 0
Thus points A and C are lies on opposite sides of eqn. (7) and hence will be the internal bisector of angle B.
The internal bisector of angle C are given by
\(\frac{5 x+12 y}{\sqrt{5^2+12^2}}\) = ± (y – 15)
⇒ 5x – y + 195 = 0 …(8)
and 5x + 25y – 195 = 0
⇒ x + 5y – 39 = 0 …(9)
For point A(0, 0);
L.H.S of eqn. (9) = 0 + 0 – 39 = – 39 < 0 For point B (20, 15); L.H.S = 20 + 75 – 39 = 56 > 0
Thus points A and B are lies on opposite sides of eqn. (9) and it will be the internal bisector of angle C.
Clearly the centre of the inscribed circle of the triangle will be the Incentre i.e. the point of concurrence of all three internal bisectors of angles of Δ.
On solving eqn. (4) and (7) ; we have
x = – 1 and y = 8
The point (- 1, 8) also lies on eqn. (9).
Hence all the three internal bisectors intersects at point (- 1, 8) and will be the required centre of inscribed circle.

Question 9.
The co-ordinates of A, B, C are respectively (- 4, 0),(0, 2) and (- 3, 2). Find (i) the equation of the straight line which bisects the angle CAB internally; (ii) the co-ordinates of the point where this straight line meets the straight line joining C to the middle point of AB.
Solution:
eqn. of line AB be given by
y – 0 = \(\frac{2-0}{0+4}\) (x + 4)
⇒ y = \(\frac { 1 }{ 2 }\) (x + 4)
⇒ x – 2y + 4 = 0 …(1)

and eqn. of line AC be given by
y – 0 = \(\frac{2-0}{-3+4}\) (x + 4)
⇒ y = 2x + 8 ⇒ 2x – y + 8 = 0 ..(2)
The eqns. of bisectors of angle A be given by
\(\frac{x-2 y+4}{\sqrt{1^2+(-2)^2}}\) = \(\pm \frac{2 x-y+8}{\sqrt{2^2+(-1)^2}}\)
Taking +ve sign ;
x – 2y + 4 = 2x – y + 8
⇒ x + y + 4 = 0 …(3)
Taking -ve sign ;
x – 2y + 4 = – 2x + y – 8
⇒ 3x – 3y + 12 = 0
⇒ x – y + 4 = 0 …(4)
putting the coordinates of B and C in L.H.S of eqn. (4); we have
For point B(0, 2) ; 0 – 2 + 4 = 2 > 0
For point C(- 3, 2); – 3 – 2 + 4 = – 1 < 0
Thus points B(0, 2) and C(- 3, 2) are lies on opposite sides of eqn. (4).
Thus eqn. (4) will be the eqn. of internal bisector of angle A i.e. angle CAB.

(ii) Middle point of AB be given by
\(\mathrm{D}\left(\frac{-4+0}{2}, \frac{0+2}{2}\right)\) i.e. D(-2, 1)
Thus using two point form, eqn. of line CD be given by
y – 2 = \(\frac{1-2}{-2+3}\)(x + 3)
⇒ y – 2 = -(x + 3)
⇒ x + y + 1 = 0 …(5)
On solving eqn. (4) and eqn. (5); we have
x = –\(\frac { 5 }{ 2 }\) and y = \(\frac { 3 }{ 2 }\)
Thus required point of intersection be \(\left(-\frac{5}{2}, \frac{3}{2}\right)\).

Well-structured Class 11 ISC Maths S Chand Solutions Chapter 16 The Straight Line Ex 16(f) facilitate a deeper understanding of mathematical principles.

S Chand Class 11 ICSE Maths Solutions Chapter 16 The Straight Line Ex 16(f)

Question 1.
Find the distance of the point P from the line AB in the following cases :
(i) P(4, 2), AB is 5x – 12y – 9 = 0,
(ii) P(0, 0), AB is h(x + h) + k (y + k) = 0.
Solution:
(i) ⊥ distance of point P(4, 2) from given line AB

(ii) ⊥ distance of point P(0, 0) from given line AB

Question 2.
Calculate the length of the perpendicular from (7, 0) to the straight line 5x + 12y – 9 = 0 and show that it is twice the length of the perpendicular from (2, 1).
Solution:
Given eqn. of straight line be
5x +12y – 9 = 0
required length of ⊥ from P(7, 0) to given

Clearly the length of ⊥ from P(7, 0) to given line is twice the length of ⊥ from Q (2, 1) to same given line.

Question 3.
The point A(0, 0), B(1, 7), C(5, 1) are the vertices of a triangle. Find the length of the perpendicular from A to BC and hence the area of the △ABC.
Solution:
eqn. of line BC is given by using two point form, be

Question 4.
Find the lengths of altitudes of the triangle whose sides are given by
3x – 4y = 5, 4x + 3y = 5 and x + y = 1.
Solution:
Given eqns. of sides are
3x – 4y = 5 …(1)
4x + 3y = 5 …(2)
and x + y = 1 …(3)
First of all, we find all the vertices of triangle whose sides are along given lines (1), (2) and (3).
We solve three eqns. in pairs to get the vertices of triangle.
On solving eqn. (1) and (2) simultaneously, we have
x = \(\frac { 7 }{ 5 }\); y =\(\frac { -1 }{ 5 }\)
On solving eqn. (2) and (3) simultaneously ; we have
y = – 1 and x = 2
On solving eqn. (1) and (3) simultaneously ; we have
x = \(\frac { 9 }{ 7 }\) and y = \(\frac { -2 }{ 7 }\)
∴ length of altitude from A\(\left(\frac{7}{5}, \frac{-1}{5}\right)\) to BC

Question 5.
If P is the perpendicular distance of the origin from the line whose intercepts on the axes are a and b, show that
\(\frac{1}{p^2}\) = \(\frac{1}{a^2}\) = \(\frac{1}{b^2}\)
Solution:
eqn. of line whose intercepts on the axes are a and b be given by
\(\frac{x}{a}\) + \(\frac{y}{b}\) = 1 …(1)
p = length of ⊥ from (0, 0) to given eqn. (1)

Question 6.
Find the perpendicular distance between the lines
(i) 3x + 4y + 5 = 0, 3x + 4y + 17 = 0.
(ii) 9x + 40y – 20 = 0, 9x + 40y + 21 = 0.
(iii) y = mx + c, y = mx + d.
Solution:
(i) Given eqns. of lines are
3x + 4y + 5 = 0 …(1)
and 3x + 4y +17 =0 …(2)
Slope of both lines =-\(\frac{3}{4}\)
∴ lines (1) and (2) are parallel.
For any point on line (1), we put x = 0 in eqn. (1); we have y = –\(\frac{5}{4}\)
Thus any point on line (1) be \(\left(0,-\frac{5}{4}\right)\)
∴ required distance between parallel lines = ⊥ distance from \(\left(0,-\frac{5}{4}\right)\) to line (2)

(ii) Given lines are;
9x + 40y – 20 = 0 …(1)
9x + 40y – 21 = 0 …(2)
Slopes of lines (1) and (2) = –\(\frac{9}{40}\)
∴ both lines are parallel.
For any point on line (1); we put x = 0 in eqn. (1) ;
we have y = \(\frac{1}{2}\)
∴ any point on line (1) be \(\left(0, \frac{1}{2}\right) \text {. }\)
Thus required distance between parallel lines
= ⊥ distance of point \(\left(0, \frac{1}{2}\right)\) to line (2)

(iii) Equations of given lines are
y = mx + c …(1)
y = mx + d …(2)
Clearly both lines are parallel.
For any point on line (1) ;
we put x = 0 in eqn. (1); we have y = c
∴ any point on line (1) be (0, c).
∴ required distance between parallel lines = distance of point (0, c) from line (2)

Question 7.
Find the equations of two straight lines which are parallel to the straight line x + 7y + 2 = 0, and a unit distance from the point (2, – 1).
Solution:
eqn. of given straight line be
x + 7y + 2 = 0 …(1)
∴ eqn. of line parallel to line (1) be given by
x + 7y + k = 0 …(2)
given ⊥ distance from the point (2, – 1) to line (2) = 1

putting the value of k in eqn. (2); we have x + 7y + 5(1 ± √ 2) = 0 be the required eqn.’s of lines.

Question 8.
Find the equations of the two straight lines drawn through the point (0, 1) on which the perpendiculars dropped from the point (2, 2) are each of unit length.
Solution:
eqn. of straight line passing through the point (0, 1) and is having slope m be given by
y – 1 = m(x – 0) [using one point form]
⇒ mx – y + 1 = 0 …(1)
It is given that, length of ⊥ drawn from point (2, 2) to line (1) be of unit length.

On squaring both sides; we have
(2m – 1)² = m² + 1
⇒ 4m² – 4m + 1 = m² + 1
⇒ 3m² – 4m = 0
⇒ m(3m – 4) = 0 ⇒ m = 0, \(\frac { 4 }{ 3 }\)
putting m = 0 in eqn. (1); we have
– y + 1 = 0 ⇒ y = 1
putting m = \(\frac { 4 }{ 3 }\) in eqn. (1); we have
\(\frac { 4x }{ 3 }\) – y + 1 = 0
⇒ 4x – 3y + 3 = 0

Question 9.
A straight line is parallel to the lines 3x – y – 3 = 0 and 3x – y + 5 = 0, and lies between them. Find its equation if its distances from these lines are in the ratio 3 : 5.
Solution:
Given eqns. of lines are
3x – y – 3 = 0 …(1)
3x – y + 5 = 0 …(2)
Let the eqn. of line which is mid-parallel to lines (1) and (2) be given by
3x – y + k = 0 …(3)
To find any point on line (3); we put x = 0 in eqn. (3) ; we have, y = k
∴ any point on line (3) be (0, k).
Also it is given that, distances of line (3) from given lines (1) and (2) are in the ratio 3 : 5

On squaring both sides; we have
\(\frac{(k+3)^2}{(5-k)^2}\) = \(\frac{9}{25}\)
⇒ 25 (k² + 6k + 9) = 9 (k² – 10k + 25)
⇒ 16 k² + 240k = 0
⇒ 16 k (k + 15) = 0
⇒ k = 0, – 15
∴ from (3); 3x – y = 0
or 3x – y – 15 = 0
are the required eqns. of lines.

Question 10.
Find the equation of the locus of a point P which is equidistant from the st. line 3x – 4y + 2 = 0 and the origin.
Solution:
eqn. of given line be
3x – 4y + 2 = 0 …(1)
Let P(x, y) be any point whose locus is to be find out.
s.t distance of P(x, y) from O(0, 0) = length of ⊥ from P(x, y) to line (1)

on squaring both sides, we have
⇒ 25 (x² + y²) = 9x² + 16y² + 4 – 24xy – 16y + 12x
⇒ 16x² + 9y² + 24xy + 16y – 12x – 4 = 0
which is the required eqn. of locus.

Question 11.
A point P is such that the sum of the squares of its distances from the two axes of co-ordinates is equal to the square of its distance from the line x – y = 1. Find the equation of the locus of P.
Solution:
Eqn. of coordinate axes be given by x = 0 and y = 0
Let P(x, y) be any point whose locus is to be find out.
According to given condition, we have

Question 12.
Show that the equation to the parallel line mid-way between the parallel lines ax + by + c₁ = 0 and ax + by + c₂ = 0 is ax + by + \(\frac{c_1+c_2}{2}\) = 0.
Solution:
Eqns. of given lines re
ax + by + c₁ = 0 …(1)
and ax + by + c₂ = 0 …(2)
Let the eqn. of line which is mid-parallel to both given lines be
ax + by + c = 0 …(3)
For any point on line (3); we put x = 0 in eqn. (3); we have y =-\(\frac { c }{ b }\)
Thus any point on line (3) be \(\left(0,-\frac{c}{b}\right)\)
Since line (3) is mid-parallel between line (1) and line (2).
∴ ⊥ distance of point \(\left(0,-\frac{c}{b}\right)\) from line (1)
= ⊥ distance of point \(\left(0,-\frac{c}{b}\right)\) from line (2)

Question 13.
Prove that the line 12x – 5y – 3 = 0 is midparallel to the lines 12x – 5y + 7 = 0 and 12x – 5y – 13 =0 .
Solution:
Given eqns. of lines are
12x – 5y – 3 = 0 …(1)
12x – 5y + 7 = 0 …(2)
and 12x – 5y – 13 = 0 …(3)
Now line (1) is mid-parallel to lines (2) and (3)
If ⊥ distance of any point on line (1) from line (2) is equal to ⊥ distance of any point on line (1) from line (3).
For any point on line (1) ∴ we put x = 0 in eqn. (1) we have y = –\(\frac { 3 }{ 5 }\)
∴ Any point on line (1) be \(\left(0,-\frac{3}{5}\right)\)
⊥ distance of \(\left(0,-\frac{3}{5}\right)\) from line (2)

Students can track their progress and improvement through regular use of OP Malhotra Maths Class 11 Solutions Chapter 9 Complex Numbers Ex 9(d).

S Chand Class 11 ICSE Maths Solutions Chapter 9 Complex Numbers Ex 9(d)

Find the modulus and amplitude of the following complex numbers and hence express them into polar form.

Question 1.
\(\sqrt{3}\) + i
Solution:
Let z = \(\sqrt{3}\) + i = x + iy
Here x = \(\sqrt{3}\) > 0 and y = 1 > 0
∴ point (x, y) i.e. (-\(\sqrt{3}\), 1)
lies in first quadrant.

Question 2.
– \(\sqrt{3}\) + i
Solution:
Let z = – \(\sqrt{3}\) + i …(1)
put – \(\sqrt{3}\) = r cos θ …(2)
1 = r sin θ …(3)
On squaring and adding (1) and (2);
3 + 1 = r² (cos² θ + sin² θ)
⇒ r² = 4
⇒ r = 2 (r > 0)
On dividing (3) by (2); we have tan θ
tan θ = – \(\frac{1}{\sqrt{3}}\) = – tan \(\frac { π }{ 6 }\) = tan (π – \(\frac { π }{ 6 }\))
[∵ cos θ < 0 and sin θ > 0
∴ θ lies in 2nd quad]
⇒ θ = \(\frac { 5π }{ 6 }\)
⇒ arg (z) = \(\frac { 5π }{ 6 }\)
∴ from eqn. (1) ; we have
z = r [cos θ + i sin θ]
= 2 [cos \(\frac { 5π }{ 6 }\) + i sin \(\frac { 5π }{ 6 }\) ]

Question 3.
– 2 + 2\(\sqrt{3}\) i
Solution:
Let z = – 2 + 2\(\sqrt{3}\) i … (1)
put – 2 = r cos θ …(2)
2\(\sqrt{3}\) = r sin θ …(3)
since cos θ < 0 and sin θ > 0 ∴ θ lies in 2nd quadrant.
On squaring and adding (2) and (3); we have

which is the required polar form.

Question 4.
– 1 – i
Solution:
Let z = – 1 – i = x + iy
Here x = – 1 < 0 and y = – 1 < 0
∴ point (x, y) i.e. (- 1, – 1) lies in IIIrd quadrant.
∴ tan θ = \(\left|\frac{{Im}(z)}{{Re}(z)}\right|=\left|\frac{y}{x}\right|=\left|\frac{-1}{-1}\right|\) = 1
⇒ α = \(\frac { π }{ 4 }\)
∴ arg (z) = – (π – α) = – (π – \(\frac { π }{ 4 }\)) = – \(\frac { 3π }{ 4 }\)
and r = \(\sqrt{x^2+y^2}=\sqrt{(-1)^2+(-1)^2}=\sqrt{2}\)
Thus required polar form of z is given by
z = r [cos (arg z) + z sin (arg z)]
⇒ z = \(\sqrt{2}\left[\cos \left(\frac{-3 \pi}{4}\right)+i \sin \left(\frac{-3 \pi}{4}\right)\right]\)

Question 5.
– 2i
Solution:
Let z = – 2i × x + iy
∴ |z| = |- 2i| = 2
where x – 0 ; y = – 2
∴ point (x, y) i.e. (0, – 2) lies in IVth quadrant.
∴ tan θ = \(\left|\frac{{Im}(z)}{{Re}(z)}\right|=\left|\frac{-2}{0}\right|\) → ∞ ⇒ α = \(\frac { π }{ 2 }\)
∴ θ = arg (z) = – α = – \(\frac { π }{ 2 }\)
Thus polar form of z is given by z = | z | (cos θ + z sin θ)
= 2 \(\left[\cos \left(-\frac{\pi}{2}\right)+i \sin \left(-\frac{\pi}{2}\right)\right]\)

Question 6.
– 1 – \(\sqrt{3}\) i
Solution:
Let z = – 1 – \(\sqrt{3}\) i …(1)
put – 1 = r cos θ …(2)
– \(\sqrt{3}\) = r sin θ …(3)
Here cos θ, sin θ < 0
∴ θ lies in 3rd quadrant. On squaring and adding eqn. (2) and eqn. (3); we have
r² (cos² θ + sin² θ) = (- 1)² + (- \(\sqrt{3}\))² = 1 + 3 = 4
r² = 4 ⇒ r = 2 (∵ r > 0)
On dividing eqn. (3) by eqn. (2); we have
tan θ = \(\frac{-\sqrt{3}}{1}\) = \(\sqrt{3}\)
= \(\tan \left(\frac{\pi}{3}\right)=\tan \left[-\left(\pi-\frac{\pi}{3}\right)\right]\)
⇒ θ = – \(\frac { 2π }{ 3 }\)
∴ from (1) ; z = r [cos θ + i sin θ]
i.e. z = 2\(\left[\cos \left(-\frac{2 \pi}{3}\right)+i \sin \left(-\frac{2 \pi}{3}\right)\right]\)

Question 7.
– 2
Solution:
Let z = – 2 = x + iy
Here x = – 2 and y = 0
∴ (- 2, 0) lies on negative x-axis.
∴ | z | = | – 2 | = 2
∴ arg (z) = π ;
∴ z = 2 [cos π + i sin π] be the required polar form.

Question 8.
\(\frac{(1+i)^{13}}{(1-i)^7}\)
Solution:
Let

Further x = – 8 ; y = 0 i.e. point (- 8, 0) lies on negative real axis.
∴ arg (z) = π
∴ z = 8 [cos π + i sin π] be the required polar form

Question 9.
(3 + i) (4 + i)
Solution:
Let z = (3+ i) (4 + i)
= 12 + 3i + 4i – 1 = 11 + 7i …(1)
∴ | z | = \(\sqrt{11^2+7^2}\)
= \(\sqrt{121+49}=\sqrt{170}\)
put 11 = r cos θ …(2)
7 = r sin θ …(3)
On dividing (3) by (2); we have
tan θ = \(\frac { 7 }{ 11 }\)
⇒ θ = tan^-1\(\frac { 7 }{ 11 }\) = tan^-1(0.636)
⇒ θ = 32°29′
[since cos θ, sin θ > 0
∴ θ lies in first quadrant]
∴ arg (z) = θ = 32° 29′
∴ from (1); z = r [cos θ + i sin θ]
⇒ z = \(\sqrt{170}\) [cos (32° 29′) + i sin (32° 29′)]
[On squaring and adding eqn. (1) and (2); r = \(\sqrt{170}\) ]

Question 10.
\(\frac{(1+i)(2+i)}{(3+i)}\)
Solution:
Let z = \(\frac{(1+i)(2+i)}{3+i}=\frac{2+i+2 i-1}{3+i}\) = \(\frac{1+3 i}{3+i} \times \frac{3-i}{3-i}\)
⇒ z = \(\frac{3-i+9 i+3}{3^2-i^2}\)
⇒ z = \(\frac{6+8 i}{9+1}=\frac{3+4 i}{5}=\frac{3}{5}+\frac{4}{5} i\) … (1)
∴ | z | = \(\frac{\sqrt{3^2+4^2}}{5}=\frac{5}{5}\) = 1
Put \(\frac { 3 }{ 5 }\) = r cos θ … (2)
and \(\frac { 4 }{ 5 }\) = r sin θ … (2)
On squaring and adding eqn. (2) and (3);
(\(\frac { 3 }{ 5 }\))² = r² (cos² θ + sin² θ)
⇒ r² = \(\frac{9+16}{25}\) 1 (∵ r > 0)
On dividing eqn. (3) by eqn. (2); we have
tan θ = \(\frac{4}{5} \times \frac{5}{3}=\frac{4}{3}\) = 1.3333
⇒ θ = tan^-1 (1.3333) = 53° 8′ [∵ θ lies in first quadrant]
∴ from (1); z = r [cos θ + i sin θ]
⇒ z = 1 [cos (53° 8′) + i sin (53° 8′)]
which is the required polar form.

Question 11.
\(\frac{5-i}{2-3 i}\)
Solution:

Question 12.
\(\frac{(3+4 i)(4+5 i)}{(4+3 i)(6+7 i)}\)
Solution:
Let z = \(\frac{(3+4 i)(4+5 i)}{(4+3 i)(6+7 i)}\)

⇒ α = 18° 12′ and arg (z) = a = 18° 12′
∴ required polar form of z be given by z = | z | [cos α + i sin α]
⇒ z = \(\sqrt{\frac{41}{85}}\) [cos (18° 12′)+ i sin (18° 12′)]

Question 13.
Change the following complex numbers into polar form.
(i) – 4 + 4\(\sqrt{3}\) i
(ii) \(\frac{1+3 i}{1-2 i}\)
(iii) \(\frac{1+2 i}{1-(1-i)^2}\)
(iv) \(\frac{(3+4 i)(4+5 i)}{(4+3 i)(6+7 i)}\)
Solution:
(i) Let z = – 4 + 4\(\sqrt{3}\)i …(1)
put – 4 = r cos θ …(2)
4\(\sqrt{3}\) = r sin θ …(3)
On squaring and adding eqn. (2) and eqn. (3); we have
r² = (- 4)² + (4\(\sqrt{3}\))² = 16 + 48 = 64
⇒ r = 8 (∵ r > 0)
since cos θ < 0 and sin θ > 0 ∴ θ lies in 2nd quadrant.
On dividing eqn. (3) by eqn. (2); we have
tan θ = – \(\sqrt{3}\) = – tan\(\frac { π }{ 3 }\) = tan(π – \(\frac { π }{ 3 }\))
= tan\(\frac { 2π }{ 3 }\)
∴ θ = \(\frac { 2π }{ 3 }\)
∴ from (1); z = r [cos θ + i sin θ]
⇒ z = 8\(\left[\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right]\)
which is the required polar form

(ii) Let z = \(\frac{1+3 i}{1-2 i} \times \frac{1+2 i}{1+2 i}\)
⇒ z = \(\frac{1+2 i+3 i-6}{1^2-(2 i)^2}=\frac{5 i-5}{1+4}\) = i – 1
⇒ z = – 1 + i …(1)
put – 1 = r cos θ …(2)
and 1 = r sin θ …(3)
since cos θ < 0 and sin θ > 0
∴ θ lies in 2nd quadrant.
On squaring and adding eqn. (1) and (2); we have
r² (cos² θ + sin² θ) = (- 1)² + (1)²
⇒ r² = 2 ⇒ r = \(\sqrt{2}\) (∵ r > 0)
On dividing eqn. (3) by eqn. (2); we have
tan θ = – 1 = – tan \(\frac { π }{ 4 }\) = tan (π – \(\frac { π }{ 4 }\))
⇒ θ = π – \(\frac { π }{ 4 }\) = \(\frac { 3π }{ 4 }\)
∴ from (1); z = r [cos θ + 2 sin θ]
⇒ z = \(\sqrt{2}\left[\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right]\)
which is the required polar form.

(iii) Let z = \(\frac{1+2 i}{1-(1-i)^2}=\frac{1+2 i}{1-(1-1-2 i)}\)
⇒ z = \(\frac{1+2 i}{1+2 i}\) = x + iy
Here x = 1 ; y = 0
∴ point (x, y) i.e. (1,0)
lies on positive real axes.
∴ z = 1 = 1 + i0 = cos 0° + 2 sin 0°
which is the required polar form.

(iv) Let z = \(\frac{1+7 i}{(2-i)^2}=\frac{1+7 i}{4-1-4 i}\) = \(\frac{1+7 i}{3-4 i} \times \frac{3+4 i}{3+4 i}\)
⇒ z = \(\frac{3+4 i+21 i-28}{3^2-(4 i)^2}=\frac{-25+25 i}{9+16}\)
⇒ z = – 1 + i … (1)
put – 1 = r cos θ …(2)
and 1 = r sin θ … (3)
since cos θ < 0 and sin θ > 0 ∴ lies in 2nd quadrant.
On squaring and adding eqn. (1) and (2); we have
r² (cos² θ + sin² θ) = (- l)² + (l)²
⇒ r² = 2 ⇒ r = \(\sqrt{2}\) (∵ r > 0)
On dividing eqn. (3) by eqn. (2); we have
tan θ = – 1 = – tan \(\tan \frac{\pi}{4}=\tan \left(\pi-\frac{\pi}{4}\right)\)
⇒ θ = π – \(\frac { π }{ 4 }\) = \(\frac { π }{ 4 }\)
∴ from (1) ; z = r [cos θ + i sin θ]
⇒ z = \(\sqrt{2}\left[\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right]\)
which is the required polar form.

Question 14.
Given the complex number z = \(\frac{-1+\sqrt{3} i}{2}\) and w = \(\frac{-1-\sqrt{3} i}{2}\) (where i = \(\sqrt{-1}\))
(i) Prove that each of these complex numbers is the square of the other.
(ii) Calculate the modulus and argument of w and z.
(iii) Calculate the modulus and argument of \(\frac { w }{ z }\).
(iv) Represent z and w accurately on the complex plane.
Solution:
Given z = \(\frac{-1+\sqrt{3} i}{2}\) and w = \(\frac{-1-\sqrt{3} i}{2}\)

Thus each of the complex numbers is the square of the other.

(ii) z = – \(\frac{1}{2}+\frac{\sqrt{3}}{2}\) = x + iy

(iv) Clearly the complex number z is represented by point P\(\frac{-1-\sqrt{3} i}{2}\) and the complex number is represented by point Q\(\frac{-1-\sqrt{3} i}{2}\) in complex plane.

Students often turn to Class 11 ISC Maths S Chand Solutions Chapter 16 The Straight Line Ex 16(e) to clarify doubts and improve problem-solving skills.

S Chand Class 11 ICSE Maths Solutions Chapter 16 The Straight Line Ex 16(e)

Question 1.
Find the co-ordinates of the point of intersection of the straight lines
(i) 3x – 5y + 5 = 0, 2x + 3y – 22 = 0
(ii) 2x – 3y – 7 = 0, 3x – 4y – 13 = 0.
Solution:
(i) Given eqns. of lines are
3x – 5y + 5 = 0 …(1)
2x + 3y – 22 = 0 …(2)
To find the point of intersection of both lines we solve eqn. (1) and (2) simultaneously.
3 × eqn. (1) + 5 × eqn. (2); we have
19x – 95 = 0 ⇒ x = 5
∴ from (1); 15 – 5y + 5 = 0 ⇒ y = 4
∴ (5, 4) be the required point of intersection of both lines.

(ii) Given eqns. of lines are
2x – 3y – 7 = 0 …(1)
and 3x – 4y – 13 = 0 …(2)
For point of intersection of lines (1) and (2), we solve eqn. (1) and (2) simultaneously.
4 × eqn. (1) – 3 × eqn. (2); we have
8x – 9x – 28 + 39 = 0 ⇒ x = 11
putting x = 11 in eqn. (1); we have
22 – 3y – 7 = 0 ⇒ y = \(\frac { 15 }{ 3 }\) = 5
Thus (11, 5) be the required point of intersection of both lines.

Question 2.
Find the area of the triangle formed by the lines y + x – 6 = 0, 3y – x + 2 = 0 and 3y = 5x + 2.
Solution:
Given lines are
y + x – 6 = 0 …(1)
3y – x + 2 = 0 …(2)
and 3y = 5x + 2 …(3)
To find the point of intersection of lines (1) and (2) we have to solve (1) and (2) simultaneously.
on adding (1) and (2); we have
4y – 4 = 0 ⇒ y = 1
∴ from (1); 1 + x – 6 = 0 ⇒ x = 5
Thus lines (1) and (2) intersects at A (5, 1).
To find point of intersection of lines (2) and (3)
we solve (2) and (3) simultaneously.
5x + 2 – x + 2 = 0 ⇒ x = – 1
∴ from (2); 3y + 1 + 2 = 0 ⇒ y = – 1
Thus lines (2) and (3) intersection at B (- 1, – 1)
To find point of intersection of lines (1) and (3)
we solve eqn. (1) and (3) simultaneously
3 (6 – x) = 5x + 2
⇒ 18 – 3x = 5x + 2
⇒ 8x = 16 ⇒ x = 2
∴ from (1) ; y = 4
Thus lines (1) and (3) intersects at C(2, 4).
∴ required area of △ABC
= \(\frac { 1 }{ 2 }\) | (- 5 + 1) + (- 4 + 2) + (2 – 20) |
= \(\frac { 1 }{ 2 }\) | – 4 – 2 – 18 | = 12 sq. units

Question 3.
Find the orthocentre of the triangle whose angular points are (0, 0), (2, – 1), (- 1, 3).
Solution:
Let AL, BM and CN are the altitudes of the △ABC with vertices A(0, 0) ; B(2, – 1) and C(- 1, 3).

slope of BC = \(\frac{3+1}{-1-2}\) = \(-\frac{4}{3}\)
since AL ⊥ BC
∴ slope of AL = \(\frac{-1}{\text { slope of } B C}\) = \(\frac{-1}{\frac{-4}{3}}\) = \(\frac{3}{4}\)
slope of CA = \(\frac{3-0}{-1-0}\) = – 3
since BM ⊥ AC
∴ slope of BM = \(\frac{-1}{\text { slope of } A C}\) = \(\frac{-1}{-3}\) = \(\frac{1}{3}\)
slope of AB = \(\frac{-1-0}{2-0}\) = –\(\frac{1}{2}\)
since CN ⊥ AB
∴ slope of CN = \(\frac{-1}{\text { slope of } A B}\) = \(\frac{\frac{-1}{-1}}{2}\) = 2
Equation of line AL, passing through (0, 0) and having slope \(\frac{3}{4}\) is given by
y – 0 = \(\frac{3}{4}\)(x – 0) ⇒ y = \(\frac{3}{4}\)x …(1)
Equation of line BM, i.e. line passing
through B(2, – 1) and having slope \(\frac{1}{3}\) is given by
y + 1 = \(\frac{1}{3}\)(x – 2)
⇒ x – 3y – 5 = 0 …(2)
Eqn. of line CN i.e. line passing through C (- 1, 3)
and having slope 2 is given by
y – 3 = 2(x + 1)
⇒ 2x – y + 5 = 0 …(3)
For point of intersection of lines AL and BM,
we solve (1) and (2) simultaneously
x – \(\frac{9x}{4}\) – 5 = 0 ⇒ 4x – 9x – 20 = 0
⇒ -5x – 20 = 0 ⇒ x = – 4
∴ from (1) ; y = – 3
Thus (- 4, -3 ) be the point of intersection of lines (1) and (2).
The point (- 4, – 3) lies on eqn. (3) if 2(- 4) – (- 3) + 5 = 0
if 0 = 0, which is true.
Hence AL, BM and CN intersects at a point (- 4, – 3) and be the required orthocentre.

Question 4.
The vertices of a triangle are A (0, 5), B (- 1, – 2) and C(11, 7). Write down the equations of BC and the perpendicular from A to BC and hence find the co-ordinates of the foot of the perpendicular.
Solution:
Using two point form, eqn. of line joining B(- 1, – 2) and C (11, 7) be given by

y + 2 = \(\frac{7+2}{11+1}\)(x + 1)
⇒ y + 2 = \(\frac{9}{2}\)(x + 1)
⇒ y + 2 = \(\frac{3}{4}\)(x + 1)
⇒ 3x – 4y – 5 = 0 …(1)
∴ slope of line BC = –\(\frac{\text { coeff. of } x}{\text { coeff. of } y}\) = \(\frac{-3}{-4}\) = \(\frac{3}{4}\)

Let AD be the ⊥ drawn from A to BC
∴ slope of AD = \(\frac{-1}{\text { slope of } B C}\) = \(\frac{-1}{3 / 4}\) = \(\frac{-4}{3}\)
Thus required eqn. of line through the point A(0, 5) and ⊥ to BC be given by
y – 5 = –\(\frac{4}{3}\)(x – 0) ⇒ y – 5 = –\(\frac{4x}{3}\)
⇒ 4x + 3y – 15 = 0 …(2)
Let D be the foot of ⊥ which is the point of intersection of lines BC and AD.
eqn. (1) × 3 + 4 × eqn. (2); we have
9x – 15 + 16x – 60 = 0
⇒ 25x = 75 ⇒ x = 3
∴ from (1) ; 3 × 3 – 4y – 5 = 0
⇒ 4y = 4
⇒ y = 1
Thus coordinates of required foot of ⊥ are (3, 1).

Question 5.
Find the equation of the straight line passing through the point of intersection of the two lines x + 2y + 3 = 0 and 3x + 4y + 7 = 0 and parallel to the straight line y – x = 8.
Solution:
Given eqns. of lines are
x + 2y + 3 = 0 …(1)
and 3x + 4y + 7 = 0 …(2)
To find the point of intersection of lines (1) and (2); we solve (1) and (2) simultaneously eqn. (2) – 2 eqn. (1); we have
x + 1 = 0 ⇒ x = – 1
∴ from (1);
– 1 + 2y + 3 = 0 ⇒ y = – 1
Thus point of intersection of lines (1) and (2) be (- 1, – 1)
given eqn. of line be y – x = 8 …(3)
∴ slope of line (3) = \(\frac{-(-1)}{1}\) = 1
Thus slope of line || to line (3) be 1 .
Hence required eqn. of line through the point (- 1, – 1) and having slope 1 be given by y + 1 = 1(x + 1) ⇒ x – y = 0

Question 6.
Find the equation of the line through the intersection of y + x = 9 and 2x – 3y + 7 = 0, and perpendicular to the line 2y – 3x – 5 = 0.
Solution:
Given eqns. of lines are
y + x = 9 …(1)
and 2x – 3y + 7 = 0 …(2)
To find the point of intersection of (1) and (2), we solve (1) and (2) simultaneously
3 × eqn. (1) + eqn. (2) gives;
5x = 27 – 7 ⇒ 5x = 20 ⇒ x = 4
putting the value of x = 4 in eqn. (1) ; we have
y + 4 = 9 ⇒ y = 5
Thus (4, 5) be the point of intersection of given lines.
Also, eqn. of given line be
2y – 3x – 5 = 0 …(3)
∴ slope of line (3) = \(-\frac{\text { coeff. of } x}{\text { coeff. of } y}\) = \(\frac{-(-3)}{2}\) = \(\frac{3}{2}\)
Thus slope of line ⊥ to line (3) = –\(\frac{1}{\frac{3}{2}}\) = –\(\frac{2}{3}\)
Hence the eqn. of line through the point (4, 5) and having slope –\(\frac{2}{3}\) is given by
y – 5 = –\(\frac{2}{3}\)(x – 4) | y – y₁ = m(x – x₁)
⇒ 2x + 3y – 23 = 0

Question 7.
Prove that the three lines 5x + 3y – 7 = 0, 3x – 4y = 10, and x + 2y = 0 meet in a point.
Solution:
Equations of given lines are
5x + 3y – 7 = 0 …(1)
3x – 4y – 10 = 0 …(2)
and x + 2y = 0 …(3)
all the given lines meet at a point if point of intersection of any two lines say (1) and (2) lies on line (3).
For the point of intersection of line (1) and (2) we solve (1) and (2) simultaneously.
4 × eqn. (1) + 3 × eqn. (2); we have
29x – 58 = 0 ⇒ x = 2
putting x = 2 in eqn. (1); we have
10 + 3y – 7 = 0 ⇒ 3y = – 3 ⇒ y = – 1
Thus (2, – 1) be the point of intersection of lines (1) and (2).
Now point (2, – 1) lies on eqn. (3)
∴ 2 – 2 = 0 ⇒ 0 = 0, which is true.
Thus point of intersection of lines (1) and (2) lies on line (3). Therefore all the given lines meet in a point.

Question 8.
For what value of m are the three lines y = x + 1, y = 2(x + 1) and y = mx + 3 concurrent?
Solution:
Given eqns. of lines are
y = x + 1 …(1)
y = 2(x + 1) …(2)
y = mx + 3 …(3)
Three lines are concurrent if point of intersection of both lines (1) and (2) lies on line (3).
For line of intersection of eqn. (1) and (2) we solve eqn. (1) and (2) simultaneously.
From (1) and (2); we have
y = 2y ⇒ y = 0
∴ from (1); x = – 1
∴ (- 1, 0) be the point of intersection of lines (1) and (2). Since given lines are concurrent.
∴ (- 1, 0) lies on eqn. (3).
⇒ 0 = – m + 3 ⇒ m = 3

Question 9.
The co-ordinates of A, B and C are (3, 1), (1, 5) and (4, 2) respectively. P is the midpoint of BC and (i) Q lies on AC and is such that CQ : QA = 3 : 1, R lies on AB and is such that AR : RB = 1 : 3. Find the equation of the lines AP, BQ and CR and prove that lines are concurrent.
Solution:
The coordinates of points A, B and C are (3, 1), (1, 5) and (4, 2). Since P be the midpoint of BC using mid-point formula, we have
Coordinates of P are \(\left(\frac{1+4}{2}, \frac{5+2}{2}\right)\)
i.e. \(\left(\frac{5}{2}, \frac{7}{2}\right) \text {. }\)
Since Q lies on AC and is such that
CQ : QA = 3 : 1
∴ Q divides CA in the ratio 3 : 1
Then by section formula, we have

Coordinates of Q are
\(\left(\frac{3 \times 3+4 \times 1}{3+1}, \frac{3 \times 1+1 \times 2}{3+1}\right)\) i.e. \(\left(\frac{13}{4}, \frac{5}{4}\right)\)
Also R lies on AB such that AR : RB = 1 : 3
∴ R lies on AB in the ratio 1 : 3

Then by section formula coordinates of R are given by \(\left(\frac{1 \times 1+3 \times 3}{1+3}, \frac{1 \times 5+3 \times 1}{1+3}\right)\)
i.e. \(\left(\frac{10}{4}, \frac{8}{4}\right)\) i.e. \(\left(\frac{5}{2}, 2\right)\)
using two-point form, eqn. of line AP be given by

On solving eqn. (2) and (3); we have
y = 2 and 5x + 6 – 20 = 0 ⇒ x = \(\frac { 14 }{ 5 }\)
Thus \(\left(\frac{14}{5}, 2\right)\) be the point of intersection of lines (1) and (2), it lies on line (1)
if 5 × \(\frac { 14 }{ 5 }\) + 2 – 16 = 0 if 0 = 0, which is true. Hence all the three lines pass through the point \(\left(\frac{14}{5}, 2\right)\). Thus the lines AP, BQ and CR are concurrent.

Question 10.
The sides of a triangle are OA, OB, AB and have equations 2x – y = 0, 3x + y = 0, x – 3y + 10 = 0, respectively. Find the equation of the three medians of the triangle and verify that they are concurrent.
Solution:
Given eqns. of side OA, OB and ABof △OAB are
2x – y = 0 …(1)
3x + y = 0 …(2)
x – 3y + 10 = 0 …(3)

Cearly the lines (1) and (2) intersect at O(0, 0)
The coordinates of A can be found out by finding the point of intersection of line (1) and (3).
On solving eqn. (1) and (3) simultaneously we have, x – 3(2x) + 10 = 0 ⇒ x = 2
∴ from (1); y = 2x = 2 × 2 = 4
Thus coordinates of A are (2, 4)
Clearly lines (2) and (3) intersects at point B and its coordinates can be found out by solving eqn. (2) and (3) simultaneously.
∴ x – 3(-3x) + 10 = 0 ⇒ x = 1
∴ from (2); y = 3
Thus coordinates of B are (- 1, 3).
Let OL, AM and BN are the three medians of △ABC, where L, M and N are the midpoints of sides AB, OB and OA of △OAB respectively.
∴ Coordinates of L are \(\left(\frac{2-1}{2}, \frac{4+3}{2}\right)\)
i.e. \(\left(\frac{1}{2}, \frac{7}{2}\right)\)
Coordinates of M are \(\left(\frac{-1+0}{2}, \frac{3+0}{2}\right)\)
i.e. \(\left(-\frac{1}{2}, \frac{3}{2}\right)\)
and Coordinates of N are \(\left(\frac{2+0}{2}, \frac{4+0}{2}\right)\) i.e. (1, 2)
using two point form, eqn. of line OL be given by

Equation of median BN be given by
y – 3 = \(\frac{(2-3)}{1+1}\) (x + 1)
⇒ y – 3 = –\(\frac { 1 }{ 2 }\)(x + 1) ⇒ x + 2y – 5 = 0
Solving eqn. (4) and eqn. (5) ; we have
x – 7x + 2 = 0 ⇒ 6x = 2 ⇒ x = \(\frac { 1 }{ 3 }\)
∴ from (4); y = \(\frac { 7 }{ 3 }\)
Thus point of intersection of lines (4) and (5) be \(\left(\frac{1}{3}, \frac{7}{3}\right)\).
Now point \(\left(\frac{1}{3}, \frac{7}{3}\right)\) lies on eqn. (6)
if \(\frac { 1 }{ 3 }\) + \(\frac { 14 }{ 3 }\) – 5 = 0 if 0 = 0, which is true. Hence the three medians pass through the point \(\left(\frac{1}{3}, \frac{7}{3}\right)\) and hence all the medians are concurrent.

Question 11.
Show that the lines lx + my + n = 0, mx + ny + l = 0 and nx + ly + m = 0 are concurrent if l + m + n = 0.
Solution:
Given equations of lines are ;
lx + my + n = 0 …(1)
mx + ny + l = 0 …(2)
and nx + ly + m = 0 …(3)
Clearly all the three lines are concurrent if point of intersection of any two lines say line (1) and line (2) lies on line (3).
Clearly the line (1) and (2) pass through the point (1, 1) if l + m + n= 0
Also point (1, 1) lies on eqn. (3) if
l + m + n = 0
Aliter : The given lines are concurrent if the point of intersection of lines (1) and (2) lies on line (3).
For point of intersection of line (1) and (2) we solve (1) and (2) simultaneously

if 3lmn – n³ – l³ – m³ = 0
if l³ + m³ + n³ – 3lmn = 0
if (l + m + n) (l² + m² + n² – lm – mn – nl) = 0
if \(\frac { 1 }{ 2 }\) (l + m + n) [(l – m)² + (m – n)² + (n – l)²] = 0
since l ≠ m ≠ n otherwise given lines coincident.
∴ (l – m)² + (m – n)² + (n – l)² > 0
⇒ l + m + n = 0

Question 12.
Prove that the lines
(b – c) x + (c – a) y + (a – b) = 0,
(c – a) x + (a – b) y + (b – c) = 0
and (a – b) x + (b – c) y + (c – a) = 0 are concurrent.
Solution:
Given eqns. of lines are
(b – c) x + (c – a) y + (a – b) = 0 …(1)
(c – a) x + (a – b) y + (b – c) = 0 …(2)
and (a – b) x + (b – c) y + (c – a) = 0
Clearly the point (1, 1) lies on all three lines.
[∵ (b – c) × 1 + (c – a) × 1 + a – b = 0]
Thus all the three lines pass through the point (1, 1).
So all given lines intersect at one point (1, 1). Hence given three lines are concurrent.

Question 13.
Prove that the medians of a triangle are concurrent.
Solution:
Let one vertex of △OAB at origin and side OA along x-axis.
Let the coordinates of point O, A and B are (0, 0),(a, 0) and (m, n).
Let BL, OM and AN are the medians of △OAB,
where L, M and N are the mid-points of sides OA, AB and OB of △OAB respectively.

Thus coordinates of L are \(\left(\frac{a}{2}, 0\right)\)
Cooordinates of M are \(\left(\frac{a+m}{2}, \frac{n}{2}\right)\)
and coordinates of N are \(\left(\frac{m}{2}, \frac{n}{2}\right)\).
Thus eqn. of median BL be given by

For point of intersection of lines (1) and (2) ; we solve (1) and (2) simultaneously.
From (1) and (2);
2nx + \(\frac{(a-2 m) n x}{a+m}\) = na
⇒ 2n (a + m) x +(a – 2m) nx = na (a + m)
⇒ x[2na + 2mn + an – 2mn]=na (a + m)
⇒ 3anx = na (a + m) ⇒ x = \(\frac{a+m}{3}\)
∴from (2) y = \(\frac{n}{a+m}\) × \(\frac{a+m}{3}\) = \(\frac{n}{3}\)
Thus point of intersection of lines (1) and (2) be \(\left(\frac{a+m}{3}, \frac{n}{3}\right)\).
Thus, eqn. of AN be given by

Thus all the three medians BL, OM and AN pass through the point \(\left(\frac{a+m}{3}, \frac{n}{3}\right)\) and hence concurrent.

Practicing Class 11 ISC Maths S Chand Solutions Chapter 16 The Straight Line Ex 16(d) is the ultimate need for students who intend to score good marks in examinations.

S Chand Class 11 ICSE Maths Solutions Chapter 16 The Straight Line Ex 16(d)

Question 1.
Write down the slopes of the following lines :
(i) 2x + 3y + 1 = 0
(ii) 7x – 5y + 8 = 0
(iii) – 6y – 11x = 0
(iv) xx₁ + yy₁ = a²
(v) 3x + 4y – 2 (x – x₁) – 5 (y + y₁) + 2 = 0
Solution:
(i) Given eqn. of line be
2x + 3y + 1 = 0
∴ slope of given line = –\(\frac{\text { coeff. of } x}{\text { coeff. of } y}\) = –\(\frac{2}{3}\)

(ii) Given eqn. of line be 7x – 5y + 8 = 0
∴ slope of given line =-\(\frac{\text { coeff. of } x}{\text { cceff. of } y}\)
= \(\frac{-7}{-5}\) = \(\frac{7}{5}\)

(iii) Given eqn. of line be
– 6y – 11x = 0 ⇒ 6y + 11x = 0
∴ slope of given line = –\(\frac{\text { coeff. of } x}{\text { cceff. of } y}\) = –\(\frac{11}{6}\)

(iv) Given eqn. of line be xx₁ + yy₁ = a²
∴ slope of given line = –\(\frac{\text { coeff. of } x}{\text { cceff. of } y}\) = –\(\frac{x_1}{y_1}\)

(v) Given eqn. of line be
3x + 4y – 2 (x + x₁) – 5(y + y₁) + 2 = 0
⇒ x – y – 2x₁ – 5y₁ + 2 = 0
∴ slope of given line be = –\(\frac{\text { coeff. of } x}{\text { cceff. of } y}\) = \(\frac{-1}{-1}\) = 1

Question 2.
Find the value of k such that the line (k – 2) x+ (k + 3) y – 5 = 0 is
(i) parallel to the line 2x – y + 7 = 0
(ii) perpendicular to it.
Solution:
Given eqn. of line be
(x – 2)x + (k + 3) y – 5 = 0 …(1)
∴ slope of line (1) = m₁ = –\(\frac{(k-2)}{k+3}\)
and slope of line 2x – y + 7 = 0 be
m₂ = \(\frac{-2}{-1}\) = 2

(i) Since both lines are parallel ∴ m₁ = m₂
⇒ –\(\left(\frac{k-2}{k+3}\right)\) = 2
⇒ k – 2 = – 2(k + 3)
⇒ k – 2 = – 2k – 6
⇒ 3k = -6 + 2 = -4
⇒ k = –\(\frac{4}{3}\)

(ii) Since both given lines are perpendicular
∴ m₁m₂ = – 1
⇒ –\(\left(\frac{k-2}{k+3}\right) 2\) = -1
⇒ 2(k – 2) = k + 3
⇒ 2k – 4 = k + 3 ⇒ k = 7

Question 3.
Prove that the lines
(i) 3x + 4y – 7 = 0 and 28x – 21y + 50 = 0 are mutually perpendicular ;
(ii) px + qy – r = 0 and – 4px – 4qy + 5s = 0 are parallel.
Solution:
(i) Given eqns. of lines are
3x + 4y – 7 = 0 …(1)
and 28x – 21 y + 50 = 0 …(2)
slope of line (1) = m₁ = –\(\frac{3}{4}\)
slope of line (2) = m₂ = \(\frac{-28}{-21}\) = \(\frac{28}{21}\)
Here m₁m₂ = \(\left(-\frac{3}{4}\right)\)\(\left(\frac{28}{21}\right)\) = 1
Thus both lines (1) and (2) are mutually perpendicular.
(ii) eqns. of given lines are
px + qy – r = 0 …(1)
and -4px – 4py + 5s = 0 …(2)
slope of line (1) = m₁ = –\(\frac{\text { coeff. of } x}{\text { coeff. of } y}\)
= –\(\frac{p}{q}\)
slope of line (2) = m₂ = \(\frac{-4 p}{-4 q}\) = \(\frac{p}{q}\)
Here m₁ = m₂
Thus both gives lines are parallel.

Question 4.
Find the slope of the line which is perpendicular to the line 7x + 11y – 2 = 0.
Solution;
eqn. of given line be 7x + 11y – 2 = 0 …(1)
∴ slope of line (1) = m = –\(\frac{\text { coeff. of } x}{\text { coeff. of } y}\)
= –\(\frac{7}{11}\)
Thus slope of line ⊥ to line (1)
= \(\frac{-1}{\text { slope of line (1) }}\) = \(\frac{-1}{-\frac{7}{11}}\) = \(\frac{11}{7}\)

Question 5.
Determine the angle between the lines whose equations are
(i) 3x + y – 7 = 0 and x + 2y + 9 = 0,
(ii) 2x – y + 3 = 0 and x + y – 2 = 0.
Solution:
(i) Given eqns. of lines are
3x + y – 7 = 0 …(1)
and x + 2y + 9 = 0 …(2)
∴ slope of line (1) = m₁ = –\(\frac{\text { coeff. of } x}{\text { coeff. of } y}\)
= –\(\frac{3}{1}\) = – 3
and slope of line (2) = m₂ = –\(\frac{1}{2}\)
Let θ be the acute angle between the given lines

⇒ tan θ = 1 ⇒ θ = 45°

(ii) eqns. of given lines are ;
2x – y + 3 = 0 …(1)
and x + y – 2 = 0 …(2)
Slope of line (1) = m₁ = –\(\frac{\text { coeff. of } x}{\text { coeff. of } y}\)
= –\(\frac{(+2)}{-1}\) = 2
slope of line (2) = m₂ = \(\frac{(-1)}{1}\) = – 1
Let θ be the acute angle between the given lines

Question 6.
Use tables to find the acute angle between the lines 2y + x = 0 and \(\frac { x }{ 1 }\) + \(\frac { y }{ 2 }\) = 2.
Solution:
Given equations of lines are
2y + x = 0 …(1)
and \(\frac { x }{ 1 }\) + \(\frac { y }{ 2 }\) = 2 …(2)
∴ slope of line (1) = m₁ = –\(\frac{\text { coeff. of } x}{\text { coeff. of } y}\) = –\(\frac { 1 }{ 2 }\)
slope of line (1) = m₁ = –\(\frac{\text { coeff. of } x}{\text { coeff. of } y}\)
= –\(\frac { 1 }{ 2 }\)
∴ slope of line (2) = m₂ = \(\frac{-1}{\frac{1}{2}}\) = – 2
Let θ be the acute angle between the given lines (1) and (2).

Question 7.
Reduce the following equations to the normal form and find the values of p and α
(i) √3x – y + 2 = 0.
(ii) 3x + 4y + 10 = 0 (Use tables).
Solution:
(i) eqn. of given line be √3x – y + 2 = 0
⇒ √3x – y = – 2
⇒ -√3x + y = 2
⇒ –\(\frac{\sqrt{3}}{2} x\) + \(\frac { y }{ 2 }\) = 1 …(1)
On comparing eqn. (1) with
x cos α + y sin α = p
we have, cos α = \(\frac{-\sqrt{3}}{2}\) …(2)
and sin α = \(\frac { 1 }{ 2 }\) …(3)
and p = 1
On dividing eqn. (3) by eqn. (2); we have
tan α = –\(\frac{1}{\sqrt{3}}\) = – tan\(\frac{\pi}{6}\) = tan\(\left(\pi-\frac{\pi}{6}\right)\)
[Here α lies in 2nd quadrant ∴ sin α > 0 and cos α < 0]
α = \(\frac{5 \pi}{6}\) or 150° and p = 1

(ii) given eqn. of line be,
3x + 4y + 10 = 0
⇒ 3x + 4y = – 10
⇒ -3x – 4y = 10

Comparing eqn. (1) with
x cos α + y sin α = p
cos α = –\(\frac { 3 }{ 5 }\) …(2)
and sin α = –\(\frac { 4 }{ 5 }\) …(3)
and p = 2
since sin α, cos α < 0
∴ α lies in 3rd quadrant.
On dividing eqn. (3) by eqn. (2); we have
tan α = \(\frac{\frac{-4}{5}}{\frac{-3}{5}}\) = \(\frac { 4 }{ 3 }\)
= tan\(\left(180^{\circ}+\tan ^{-1} \frac{4}{3}\right)\)
α = 180° + 53°81′ = 233°8′
and p = 2

Question 8.
Put the equation 12y = 5x + 65 in the form x cos θ + y sin θ = p and indicate clearly in a rough diagram the position of the straight line and the meaning of the constant θ and p.
Solution:
Given eqn. of line be
12y = 5x + 65
⇒ – 5x + 12y = 65
On dividing eqn. (1) throughout by
\(\sqrt{(-5)^2+12^2}\) i.e. 13; we have
\(\frac{-5}{13}\)x + \(\frac{12}{13}\)y = \(\frac{65}{13}\) = 5 …(2)
On comparing eqn. (1) with
x cos θ + y sin θ = p
we have, cos θ = \(\frac { -5 }{ 13 }\) …(3)
and sin θ = \(\frac { 12 }{ 13 }\) …(4)
and p = 5
On dividing eqn. (4) by eqn. (3) ; we have
tan θ = –\(\frac { 12 }{ 5 }\)
Here cos θ < 0 and sin θ > 0
∴ θ lies in 2nd quadrant.
⇒ tan θ = -tan α = tan (π – α)
⇒ θ = π – α = π – tan^-1\(\left(\frac{12}{5}\right)\)
⇒ θ = 180° – 67°23′ = 112°37′
Thus eqn. (3) becomes;
x cos 112°37′ + y sin 112°37′ = 5

Question 9.
If Ax + By = C and x cos α + y sin α = p represent the same line, find p in terms of A, B, C.
Solution:
Given eqns. of lines are
Ax + By = C …(1)
and x cos α + y sin α = p …(2)
Since eqns. (1) and (2) represents same line.
∴ \(\frac{A}{\cos \alpha}\) = \(\frac{B}{\sin \alpha}\) = \(\frac{\mathrm{C}}{p}\)
⇒ AP = C cos α …(1)
and BP = C sin α …(2)
On squaring and adding eqn. (1) and (2) ; we have
(A² + B²)p² = C²(cos² α + sin² α) = C²

Question 10.
Show that (2, – 1) and (1, 1) are on opposite sides of 3x + 4y = 6.
Solution:
The eqn. of given line be
3x + 4y – 6 = 0 …(1)
putting the point (2, – 1) in L.H.S of eqn. (1) we have, 3 × 2 + 4( – 1) – 6 = – 4 < 0 Again putting the point (1, 1) in L.H.S of eqn. (1), we have, 3 × 1 + 4 × 1 – 6 = 1 > 0
Since the results are of opposite sign and hence the given points lies on opposite sides of given line (1).

Question 11.
The sides of a straight triangles are given by the equations 3x + 4y = 10, 4x – 3y = 5, and 7x + y + 10 = 0; show that the origin lies within the triangle.
Solution:
Sol. Given eqns. of lines are
3x + 4y = 10 …(1)
4x – 3y = 5 …(2)
and 7x + y + 10 = 0 …(3)
Line (1) meets x-axis at \(\mathrm{B}\left(0, \frac{5}{2}\right)\)
and y-axis at \(\mathrm{A}\left(\frac{10}{3}, 0\right)\)
line (2) meets coordinate axes at \(\mathrm{c}\left(\frac{5}{4}, 0\right)\) and \(\mathrm{D}\left(0, \frac{-5}{3}\right)\).
and line (3) meets coordinate axes at \(E\left(\frac{-10}{7}, 0\right)\) and F(0, – 10).
Clearly (0, 0) lies within the △PQR.

Question 12.
Find by calculation whether the points (13, 8),(26, – 4) lie in the same, adjacent, or opposite angles formed by the straight lines 5x + 6y – 12 = 0, and 10x + 11 y – 217 = 0.
Solution:
Given eqns. of straight lines are
5x + 6y – 112 = 0 …(1)
10x + 11y – 217 = 0 …(2)
putting the coordinates (13, 8) in L.H.S of eqn. (1); we have
5 × 13 + 6 × 8 – 112 = 113 – 112 = 1 > 0
and the coordinates (13, 8) in L.H.S of eqn. (2) ; we have
10 × 13 + 11 × 8 – 217 = 218 – 217 = 1 > 0
Putting the coordinates (26, – 4) in L.H.S of eqn. (1); we have
5 × 26 + 6( – 4) – 112 = 106 – 112 = – 6 < 0
Also, putting the coordinates (26, – 4) in L.H.S of eqn. (2) ; we have,
10 × 26 + 11 × (- 4) – 217 = 216 – 217 = – 1 < 0
Hence, both points (13, 8) and (26, – 4) lies on opposite sides of both lines.

Parents can use OP Malhotra Maths Class 11 Solutions Chapter 9 Complex Numbers Ex 9(c) to provide additional support to their children.

S Chand Class 11 ICSE Maths Solutions Chapter 9 Complex Numbers Ex 9(c)

Question 1.
If (- 2 + \(\sqrt{-3}\)) (- 3 + 2\(\sqrt{-3}\)) = a + bi, find the real numbers a and b with values of a and b, also find the modulus of a + bi.
Solution:
(- 2 + \(\sqrt{-3}\)) (- 3 + 2\(\sqrt{-3}\)) = a + bi
⇒ (- 2 + \(\sqrt{-3}\)i)(- 3 + 2\(\sqrt{-3}\)i) = a + ib
⇒ 6 – 4\(\sqrt{-3}\)i – 3\(\sqrt{-3}\)i – 6 = a + ib
⇒ – 7\(\sqrt{-3}\)i = a + ib
On comparing real and imaginary parts on both sides, we get
a = 0 ; b = – 7\(\sqrt{-3}\)
∴ |a + ib | = |0 + (- 7\(\sqrt{-3}\))i| = 7\(\sqrt{-3}\)

Question 2.
Find the modulus of (1 – i)^-2 + (1 + i)^-2.
Solution:
(1 – i)^-2 + (1 + i)^-2 = \(\frac{1}{(1-i)^2}+\frac{1}{(1+i)^2}\)
= \(\frac{1}{1-1-2 i}+\frac{1}{1-1+2 i}\)
= \(\frac{-1}{2 i}+\frac{1}{2 i}\) = 0
∴ |1(1 – i)^-2 + (1 + i)^-2| = |0| = o

Question 3.
If z = 6 + 8i, verify that
(i) | z | = | \(\bar { z }\) |
(ii) – | z | < Re (z) ≤ | z |
(iii) – | z | < Im (z) ≤ | z |
(iv) z^-1 = \(\frac{\bar{z}}{|z|^2}\)
Solution:
Given z = 6 + 8i ∴ \(\bar { z }\) = \(\overline{6+8 i}\) = 6 – 8i
(i) | z | = 6 + 8i = \(\sqrt{6^2+8^2}\)
= \(\sqrt{36+64}=\sqrt{100}\) = 10
|z| = |6 – 8i| = \(\sqrt{6^2+(-8)^2}\)
= \(\sqrt{36+64}=\sqrt{100}\) = 10
∴ |z| = |\(\bar { z }\)|

(ii) since – 10 ≤ 8 ≤ 10
⇒ – | z | ≤ Re (z) ≤ | z |

(iii) since – 10 ≤ 8 ≤ 10
⇒ – | z | ≤ Im (z) ≤ | z |

(iv) z^-1 = (6 + 8i)^-1 = \(\frac{1}{6+8 i} \times \frac{6-8 i}{6-8 i}\)
= \(\frac{6-8 i}{36+64}=\frac{6-8 i}{100}\)
and \(\frac{\bar{z}}{|z|^2}=\frac{6-8 i}{\left[\sqrt{6^2+8^2}\right]^2}=\frac{6-8 i}{100}\)
∴ z^-1 = \(\frac{\bar{z}}{|z|^2}\)

Question 4.
If z₁ = 3 + 4i, z₂ = 8 – 15i, verify that
(i) | – z₁| = | z₁|
(ii) | z²₁| = |z₁|²
(iii) |z₁z₂| = |z₁| |z₂|
(iv) \(\left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|}\)
(v) | z₁| + z₂| < | z₁| + | z₂ |
(vi) | z₂ – z₁ | > \(|| z_2|-| z_1||\)
(vii) |z₁ + z₂|² + |z₁ – z₂|² = 2(|z₁|² + | z₂|²).
Solution:
z₁ = 3 + 4i; z₂ = 8 – 15i
(i) – z₁ = – (3 + 4i) = – 3 – 4i
∴ |- z₁ | = \(\sqrt{(-3)^2+(-4)^2}=\sqrt{9+16}\) = 5
| z, | = \(\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}\) = 5
∴ |- z₁ | = | z₁|

(ii) z₁² = (3 + 4i)² = 9 – 16 + 24i
= – 7 + 24i
∴ \(\left|z_1^2\right|=\sqrt{(-7)^2+24^2}\)
= \(\sqrt{49+576}=\sqrt{625}\) = 25
|z₁|² = \(\left(\sqrt{3^2+4^2}\right)^2\)
= 9 + 16 = 25
∴ \(\left|z_1^2\right|=\left|z_1\right|^2\)

(iii) z₁z₂ = (3 + 4i) (8 – 15i)
= 24 – 45i + 32i + 60
⇒ z₁z₂ = 84 – 13i
∴ |z₁z₂| = \(\sqrt{84^2+(-13)^2}\)
= \(\sqrt{7056+169}=\sqrt{7225}\) = 85
Here, | z₁ | | z₂ | = | 3 + 4i | | 8 – 15i |
= \(\sqrt{3^2+4^2} \sqrt{8^2+(-15)^2}\)
= \(\sqrt{9+16} \sqrt{64+225}\)
= \(\sqrt{25} \sqrt{289}\) = 5 x 17 = 85
∴ \(\left|z_1 z_2\right|=\left|z_1\right|\left|z_2\right|\)

(vi) |z₂ – z₁ | = | 8 – 15i – 3 – 4i | = | 5 – 19i |
= \(\sqrt{5^2+(-19)^2}=\sqrt{25+361}\)
= \(\sqrt{386}\) … (1)
\(|| z_2|-| z_1||=\left|\sqrt{8^2+(-15)^2}-\sqrt{3^2+4^2}\right|\)
= | 17 – 5 | = 12
From (1) and (2); we have
\(\left|z_2-z_1\right|>|| z_2|-| z_1||\) [∵ \(\sqrt{386}\) > 12]

(vii) z₁ + z₂ = 3 + 4i + 8 – 15i = 11 – 11i
and z₁ – z₂ = 3 + 4i – 8 + 15i = – 5 + 19i
L.H.S = | z₁ + z₂|² + | z₁ – z₂|
= | 11 – 11i |² + | – 5 + 19i |²
= \(\left(\sqrt{11^2+(-11)^2}\right)^2+\left(\sqrt{(-5)^2+19^2}\right)^2\)
= \((\sqrt{121+121})^2+(\sqrt{25+361})^2\)
= \((11 \sqrt{2})^2+(\sqrt{386})^2\)
= 242 + 386 = 628

R.H.S = 2\(\left(\left|z_1\right|^2+\left|z_2\right|^2\right)\)
= 2 \(\left(\left(\sqrt{3^2+4^2}\right)^2+\left(\sqrt{8^2+(-15)^2}\right)^2\right)\)
= 2 (25 + 289) = 2 x314 = 628
∴ L.H.S = R.H.S.

Question 5.
Find the modulus of the following using the property of modulus.
(i) (3 + 4i) (8 – 6i)
(ii) \(\frac{8+15 i}{8-6 i}\)
(iii) \(\frac{3+2 i}{2-5 i}+\frac{3-2 i}{2+5 i}\)
(iv) \(\frac{(2-3 i)(4+5 i)}{(1-4 i)(2-i)}\)
Solution:
(i) Let z = (3 + 4i) (8 – 6i)
∴ | z | = |(3 + 4i)(8 – 6i)|
= | 3 + 4i | | 8 – 6i | [∵ |z₁z₂ | = | z₁ | | z₂ |]
= \(\sqrt{3^2+4^2} \sqrt{8^2+(-6)^2}\)
= \(\sqrt{9+16} \sqrt{64+36}\) = 5 x 10 = 50

(ii) Let z = \(\frac{8+15 i}{8-6 i}\)

Question 6.
Let z be a complex number such that \(\left|\frac{z-5 i}{z+5 i}\right|\) = 1, then show that z is purely real.
Solution:
Let z = x + iy

On squaring both sides, we have
x² + (y – 5)² = x² + (y + 5)²
⇒ x² + y² + 25 – 10y = x² + y² + 25 + 10y
⇒ 20y = 0
⇒ y = 0
∴ z = x, which is purely real.

Question 7.
Find the complex number z satisfying the equation \(\left|\frac{z-12}{z-8 i}\right|=\frac{5}{3},\left|\frac{z-4}{z-8}\right|\) = 1.
Solution:
Let z = x + iy
Given \(\left|\frac{z-12}{z-8 i}\right|=\frac{5}{3}\) ⇒ \(\left|\frac{x+i y-12}{x+i y-8 i}\right|=\frac{5}{3}\)
⇒ \(\frac{|(x-12)+i y|}{|x+i(y-8)|}=\frac{5}{3}\) [∵ \(\left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|}\) ]
\(\frac{\sqrt{(x-12)^2+y^2}}{\sqrt{x^2+(y-8)^2}}=\frac{5}{3}\)
On squaring both sides ; we have
⇒ \(\frac{(x-12)^2+y^2}{x^2+(y-8)^2}=\frac{25}{9}\)
⇒ 9 [(x- 12)² + y²] = 25 [x² + (y – 8)²]
⇒ 9 [x² – 24x + 144 + y²] = 25 [x² + y² – 16y + 64]
⇒ 16 (x² + y²) – 400y + 216x + 304 = 0
⇒ 2x² + 2y² – 50y + 27x + 38 = 0 …(1)
Also, \(\left|\frac{z-4}{z-8}\right|\) = 1 ⇒ \(\left|\frac{x-4+i y}{x-8+i y}\right|\) = 1
⇒ \(\sqrt{(x-4)^2+y^2}=\sqrt{(x-8)^2+y^2}\)
On squaring both sides ; we have
(x – 4)² + y² = (x – 8)² + y²
⇒ x² – 8x + 16 = x² – 16x + 64
⇒ 8x – 48 = 0
⇒ x = 6
∴ from (1); we have
2 x 6² + 2y² – 50y + 27 x 6 + 38 = 0
⇒ 72 + 2y² – 50y + 162 + 38 = 0
⇒ 2y² – 50y + 272 = 0
⇒ y² – 25y + 136 = 0
∴ y = \(\frac{25 \pm \sqrt{625-544}}{2}=\frac{25 \pm 9}{2}\)
⇒ y = 17, 8
Thus required complex numbers are 6 + 17i and 6 + 8i.

Question 8.
If z is a complex number such that
| z – 1 | = | z + 1|, show that Re (z) = 0.
Solution:
Given |z – 1 | = |z + 1|; where z = x + iy
⇒ |x + iy – 1 | = |x + iy + 1|
⇒ |(x – 1) + zy | = |(x + 1) + iy |
⇒ \(\sqrt{(x-1)^2+y^2}=\sqrt{(x+1)^2+y^2}\)
On squaring both sides ; we have
(x – 1)² + y² = (x + 1)² + y²
⇒ x² – 2x + 1 = x² + 2x + 1
⇒ 4x = 0
⇒ x = 0
⇒ Re (z) = 0

Question 9.
Solve: | z | + z = 2 + i, where z is a complex number.
Solution:
Given | z | + z = 2 + i; where z = x + iy
⇒ \(\sqrt{x^2+y^2}\) + x + iy = 2 + i
Comparing real and imaginary parts on both sides ; we have
\(\sqrt{x^2+y^2}\) + x = 2 …(1)
and y = 1 … (2)
∴ from (1) and (2); we have
\(\sqrt{x^2+1}\) = 2 – x ;
on squaring both sides, we have
⇒ x² + 1 = (2 – x)²
⇒ x² + 1 = 4 + x² – 4x
⇒ 3 – 4x = 0 ⇒ x = \(\frac { 3 }{ 4 }\)
∴ required complex number z = \(\frac { 3 }{ 4 }\) + 1

The availability of OP Malhotra Maths Class 11 Solutions Chapter 9 Complex Numbers Ex 9(b) encourages students to tackle difficult exercises.

S Chand Class 11 ICSE Maths Solutions Chapter 9 Complex Numbers Ex 9(b)

Question 1.
In each of the following find r + s, r – s, rs, \(\frac { r }{ s }\) if r denotes the first complex number and s denotes the second complex number :
(i) 3 + 7i, i
(ii) – i, 5 + 2 i
(iii) 3i, 1 – i
(iv) – 7, – 1 – 3i
(v) 7 + 3i, 3i – 7
Solution:
(i) Given r = 3 + 7i and s = i
∴ r + s = 3 + 7i + i = 3 + 8i;
rs = (3 + 7i) i = 3i + 7i² = 3i – 7
r – s = 3 + 7i – i = 3 + 6i;
\(\frac{r}{s}=\frac{3+7 i}{i} \times \frac{i}{i}=\frac{3 i-7}{-1}\) = 7 – 3 i

(ii) Given r = – i; s = 5 + 2i
∴ r + s = – i + 5 + 2i = 5 + i;
r – s = – i – 5 – 2i = – 5 – 3i
r . s = – i (5 + 2i) = – 5i – 2i² = – 5i + 2
\(\frac{r}{s}=\frac{-i}{5+2 i} \times \frac{5-2 i}{5-2 i}=\frac{-i(5-2 i)}{5^2-(2 i)^2}\)
= \(\frac{-5 i-2}{25+4}=-\frac{5}{29} i-\frac{2}{29}\)

(iii) Given r = 3i and s = 1 – i
∴ r + s = 3i + 1 – i = 1 + 2i;
r – s = 3i – 1 + i = 4i – 1
rs = 3i (1 – i) = 3i – (- 3) = 3i + 3
\(\frac{r}{s}=\frac{3 i}{1-i} \times \frac{1+i}{1+i}=\frac{3 i(1+i)}{1^2-i^2}\)
= \(\frac{3 i+3 i^2}{1+1}=\frac{3}{2} i-\frac{3}{2}\)

(iv) Given r = – 7 and s = – 1 – 3i
∴ r + s = – 7 – 1 – 3i = – 8 – 3i;
r – s = – 7 + 1 + 3i = – 6 + 3i
rs = – 7 (- 1 – 3i) = 7 + 21i
and \(\frac{r}{s}=\frac{-7}{-1-3 i}=\frac{7}{3 i+1} \times \frac{3 i-1}{3 i-1}\)
= \(\frac{7(3 i-1)}{-9-1}=\frac{21 i-7}{-10}=\frac{-21}{10} i+\frac{7}{10}\)

(v) Given r = 7 + 3i; s = 3i – 7
∴ r + s = 7 + 3i + 3i – 7 = 6i;
r – s = 7 + 3i – 3i + 7 = 14
rs = (7 + 3i) (3i – 7) = (3i)² – 7²
= – 9 – 49 = – 58
\(\frac{r}{s}=\frac{7+3 i}{3 i-7} \times \frac{3 i+7}{3 i+7}=\frac{(3 i+7)^2}{-9-49}\)
= \(\frac{-9+49+42 i}{-58}=\frac{40+42 i}{-58}\)
= \(\frac{-20}{29}-\frac{21}{29} i\)

Question 2.
Solve each of the following equations for real x and y :
(i) (x + y) + (3 – 2i) = 1 + 4i
(ii) (x + yi) – (7 + 4i) = 3 – 5i
(iii) 2x + yi = 1 + (2 + 3i)
(iv) x + 2yi = i – (- 3 + 5i)
Solution:
(i) Given, (x + yi) + (3 – 2i) = 1 + 4i
⇒ (x + 3) + i (y – 2) = 1 + 4i
On comparing real and imaginary parts on both sides ; we have
x + 3 = 1 ⇒ x = – 2
y – 2 = 4 ⇒ y = 6

(ii) Given (x + yi) – (7 + 4i) = 3 – 5i
⇒ (x – 7) + i (y – 4) = 3 – 5i
On comparing real and imaginary parts on both sides ; we have
x – 7 = 3 ⇒ x = 10
and y – 4 = – 5 ⇒ y = – 1

(iii) Given 2x + yi = 1 + (2 + 3i)
⇒ 2x + yi = 3 + 3i
On comparing real and imaginary parts on both sides, we have
2x = 3 ⇒ x = \(\frac { 3 }{ 2 }\) and y = 3

(iv) Given x + 2yi = i – (- 3 + 5i) = i + 3 – 5i
⇒ x + 2yi = 3 – 4i
On comparing real and imaginary parts on both sides,
x = 3 and 2y = – 4 ⇒ y = – 2

Question 3.
Determine the conjugate and the reciprocal of each complex number given below :
(i) i
(ii) i³
(iii) 3 – i
(iv) \(\sqrt{-1}-3\)
(v) \(\sqrt{-9}-3\)
Solution:
(i) Conjugate of i = \(\bar{i}\) = – i
and Reciprocal of i = \(\frac{1}{i}=\frac{-i^2}{i}\)= – i

(ii) Conjugate of i³ = \(\bar{i}\)³ = (- i)³ = – i³
= – i² x i = i
and Reciprocal of i³ = \(\frac{1}{i^3}\)
= \(\frac{1}{-i} \times \frac{i}{i}=\frac{i}{-i^2}\) = i

(iii) Conjugate of (3 – i) = \(\overline{3-i}\) = 3 + i

Question 4.
Simplify :
(i) (3 – 7i)²
(ii) \(\left(\frac{-1}{2}-\frac{\sqrt{3}}{2} i\right)^2\)
(iii) (9 + 4i) (\(\frac { 3 }{ 2 }\) – i ) (9 – 4i)
Solution:
(i) (3 – 7i)² = 3² x (7i)² – 42i
= 9 – 49 – 42i
= – 40 – 42i

(ii) \(\left(\frac{-1}{2}-\frac{\sqrt{3}}{2} i\right)^2\)
= \(\left(-\frac{1}{2}\right)^2+\left(-\frac{\sqrt{3}}{2} i\right)^2+\frac{\sqrt{3}}{2} i\)
= \(\frac{1}{4}+\frac{3}{4} i^2+\frac{\sqrt{3}}{2} i=\frac{1}{4}-\frac{3}{4}+\frac{\sqrt{3}}{2} i\)
= \(\frac{-1}{2}+\frac{\sqrt{3}}{2} i\)

(iii) (9 + 4i)(\(\frac { 3 }{ 2 }\) – i)(9 – 4i)
= (9 + 4i)(9 – 4i)(\(\frac { 3 }{ 2 }\) – i)
= [81 – (4i)²](\(\frac { 3 }{ 2 }\) – i)
= (81 + 16) (\(\frac { 3 }{ 2 }\) – i)
= 97(\(\frac { 3 }{ 2 }\) – i) = \(\frac { 291 }{ 2 }\) – 97i

Question 5.
Determine real values of x and y for which each statement is true.
(i) \(\frac { x+y }{ i }\) + x – y + 4 = 0
(ii) – (x + 3y)i + (2x – y + 1) = \(\frac { 8 }{ i }\)
(iii) (x – yi) = \(\frac{2+i}{1+i}\)
(iv) (3 – 4i) + (x + yi) = 1 + 0i
(v) (x – yi)(2 + 3i) = \(\frac{x-2 i}{1-i}\)
(vi) (x⁴ + 2xi) – (3x² + yi) = (3 – 5i) + (1 + 2yi)
Solution:
(i) Given \(\frac { x+y }{ i }\) + x – y + 4 = 0
⇒ (x + y) + i (x – y + 4) = 0 = 0 + i0
On comparing real and imaginary parts on both sides ; we have
x + y = 0 …(1)
and x – y = – 4 …(2)
On adding eqn. (1) and (2); we have 2x = – 4
⇒ x = – 2 from (1) ; y = 2

(ii) Given – (x + 3y) i + (2x – y + 1) = \(\frac { 8 }{ i }\)
⇒ (x + 3y) + (2x – y + 1) i = 8 [∵ i² = – 1]
On comparing real and imaginary parts on both sides ; we have
x + 3y = 8 …(1)
2x – y = – 1 …(2)
eqn. (1) + 3 x eqn. (2); we have
7x = 5 ⇒ x = \(\frac { 5 }{ 7 }\)
∴ from (1); \(\frac { 5 }{ 7 }\) + 3y =8
⇒ 3y = 8 – \(\frac { 5 }{ 7 }\) = \(\frac { 51 }{ 7 }\) ⇒ y = \(\frac { 17 }{ 7 }\)

(iii) Given (x – yi) = \(\frac{2+i}{1+i} \times \frac{1-i}{1-i}\)
= \(\frac{(2+i)(1-i)}{1^2-i^2}\)
⇒ (x – yi) = \(\frac{2-2 i+i+1}{1+1}=\frac{3-i}{2}\)
= \(\frac { 3 }{ 2 }\) – \(\frac { i }{ 2 }\)
On comparing real and imaginary parts on both sides; we have
x = \(\frac { 3 }{ 2 }\) and – y = – \(\frac { 1 }{ 2 }\) ⇒ y = \(\frac { 1 }{ 2 }\)

(iv) Given (3 – 4i) (x + iy) = 1 + 0i
⇒ x + iy = \(\frac{1}{3-4 i} \times \frac{3+4 i}{3+4 i}=\frac{3+4 i}{3^2-(4 i)^2}\)
⇒ x + iy = \(\frac{3+4 i}{9+16}=\frac{3}{25}+\frac{4}{25}\)i
On comparing real and imaginary parts on both sides ; we have
x = \(\frac { 3 }{ 25 }\) and y = \(\frac { 4 }{ 25 }\)

(v) Given (x – yi) (2 + 3i) = \(\frac{x-2 i}{1-i}\)
⇒ (x – yi) (2 + 30 (1 – i) = x – 2i
⇒ (x – yi) (2 – 2i + 3i + 3) = x – 2i
⇒ (x – yi) (5 + i) = x – 2i
⇒ 5x + xi – 5yi + y = x – 2i
On comparing real and imaginary parts on both sides ; we have
5x + y = x ⇒ 4x + y = 0 …(1)
x – 5y = – 2 …(2)
∴ from (1) ; y = -4x
∴ from (2); x + 20x = – 2 ⇒ x = \(\frac { -2 }{ 21 }\)
∴ from (1) ; y = – 4x = \(\frac { 8 }{ 21 }\)

(vi) Given
(x⁴ + 2xi) – (3x² + yi) = (3 – 5i) + 1 + 2yi
⇒ (x⁴ – 3x²) + i (2x – y) = 4 + (2y – 5) i
On comparing real and imaginary parts on both sides, we have
x⁴ – 3x² = 4
⇒ x⁴ – 3x² – 4 = 0
⇒ (x² – 4) (x² + 1) = 0
⇒ x = ± 2, ± i
Since x, y ∈ R ⇒ x = ± 2
Also 2x – y = 2y – 5 ⇒ 2x – 3y = – 5 …(1)
When x = 2
∴ from (1); 4 – 3y = – 5 ⇒ y = 3
When x = – 2
∴ from(1) ; – 4 – 3y = – 5 ⇒ y = \(\frac { 1 }{ 3 }\)

Question 6.
Write the conjugate of (6 + 5i)².
Solution:
Let z = (6 + 5i)² = 36 – 25 + 60i = 11 + 60i
∴ \(\bar { z }\) = \(\overline{11+60 i}\) = 11 – 60i

Question 7.
Write the additive inverse of the following:
(i) – 2 + 3 i
(ii) 3 – 4i
Solution:
(i) Let z = – 2 + 3i
∴ additive inverse of z = – z
= -(-2 + 3i)
= 2 – 3i

(ii) Let z = 3 – 4i
∴ additive inverse of z = – z = – (3 – 4i)
= – 3 + 4i

Question 8.
Find the multiplicative inverse of each of the following complex numbers when it exists.
(i) 2 + 2i
(ii) 0 + 0i
(iii) – 7 + 0i
(iv) – 16
(v) \(\frac{i}{1+i}\)
(vi) (1 +1)²
(vii) \(\frac{3+4 i}{4-5 i}\)
(viii) (6 + 5i)²
(ix) \(\frac{(2+3 i)(3+2 i) i}{5+i}\)
Solution:
(i) Let z = 2 + 2i
∴ Multiplicative inverse of z = \(\frac { 1 }{ z }\)
= \(\frac{1}{2+2 i} \times \frac{2-2 i}{2-2 i}=\frac{2-2 i}{2^2-(2 i)^2}\)
= \(\frac{2-2 i}{4+4}=\frac{2-2 i}{8}=\frac{1}{4}-\frac{i}{4}\)

(ii) Let z = – 7 + 0i = – 7
∴Multiplicative inverse of z = \(\frac { 1 }{ z }\)
= – \(\frac { 1 }{ 7 }\) = – \(\frac { 1 }{ 7 }\) + 0i

(iii) Let z = 0 + 0i = 0
∴ Multiplicative inverse of z does not exists, since z = 0

(iv) Let z = – 16
∴ Multiplicative inverse of z = \(\frac { 1 }{ z }\)
= \(\frac{1}{-16}=\frac{-1}{16}\) + 0i

(v) Let z = \(\frac{i}{1+i}\)
∴ Multiplicative inverse of z = \(\frac { 1 }{ z }\)
= \(\frac { 1 }{ z }\)
= – i + 1 = 1 – i

(vi) Let z = (1 + i)² = 1 + i2 + 2i
= 1 – 1 + 2i = 2i
∴ Multiplicative inverse of z = \(\frac { 1 }{ z }\)
= \(\frac{1}{2 i} \times \frac{i}{i}=\frac{i}{2 i^2}=\frac{i}{-2}\) = 0 – \(\frac { 1 }{ 2 }\)i

(vii) Let z = \(\frac{3+4 i}{4-5 i}\)

(viii) Let z = (6 + 5i)² = 36 – 25 + 60i = 11 + 60i

(ix) Let z = \(\frac{(2+3 i)(3+2 i) i}{5+i}\)
= \(\frac{(2 i-3)(2 i+3)}{5+i}=\frac{(2 i)^2-3^2}{5+i}\)
⇒ z = \(\frac{-4-9}{5+i}=\frac{-13}{5+i}\)
∴ Multiplicative inverse of z = \(\frac { 1 }{ z }\)
= \(\frac{5+i}{-13}=\frac{-5}{13}-\frac{1}{13}\)i

Question 9.
Simplify :
(i) (1 + i)^-1
(ii) \(\sqrt{-\frac{49}{25}} \sqrt{-\frac{1}{9}}\)
(iii) \(\sqrt{-64} \cdot(3+\sqrt{-361})\)
(iv) (3 – 7i)²
(v) \(\left(-\frac{1}{2}-\frac{\sqrt{3}}{2} i\right)^2\)
(vi) \(\frac{(1-i)^3}{\left(1-i^3\right)}\)
(vii) \(\left(\frac{1+i}{1-i}\right)^{4 n+1}\) (n, is a + ve integer)
(viii) \(\frac{\sqrt{(5+12 i)}+\sqrt{(5-12 i)}}{\sqrt{(5+12 i)}-\sqrt{(5-12 i)}}\)
Solution:
(i) (1 + i)^-1 = \(\frac{1}{1+i} \times \frac{1-i}{1-i}\)
= \(\frac{1-i}{1^2-i^2}=\frac{1-i}{2}=\frac{1}{2}-\frac{i}{2}\)

(ii) \(\sqrt{-\frac{49}{25}} \sqrt{-\frac{1}{9}}=\left(\frac{7}{5} i\right)\left(\frac{1}{3} i\right)\)
= \(\frac { 7 }{ 15 }\)i² = – \(\frac { 7 }{ 15 }\)

(iii) \(\sqrt{-64} \cdot(3+\sqrt{-361})\) = 8i (3 + 19i)
= 24i + 152i² = 24i – 152

(iv) (3 – 7i)² = 3² + (7i)² – 42i
= 9 – 49 – 42i = – 40 – 42i

(v) \(\left(-\frac{1}{2}-\frac{\sqrt{3}}{2} i\right)^2=\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^2\)
= \(\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2} i\right)^2+2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} i\)
= \(\frac{1}{4}-\frac{3}{4}+\frac{\sqrt{3}}{2} i=-\frac{1}{2}+\frac{\sqrt{3}}{2} i\)

(vi) \(\frac{(1-i)^3}{\left(1-i^3\right)}=\frac{1^3-i^3-3 i(1-i)}{1-i^2 \times i}\)
= \(\frac{1+i-3 i-3}{1+i}=\frac{-2-2 i}{1+i}\)
= \(\frac{-2(1+i)}{1+i}\) = – 2

(vii) \(\left(\frac{1+i}{1-i}\right)^{4 n+1}=\left[\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right]^{4 n+1}\)
= \(\left[\frac{(1+i)^2}{1^2-i^2}\right]^{4 n+1}=\left[\frac{1+i^2+2 i}{1+1}\right]^{4 n+1}\)
= \(\left[\frac{1-1+2 i}{2}\right]^{4 n+1}=i^{4 n+1}\)
= i⁴ⁿ, i = (i⁴)ⁿ x i = 1ⁿ x i = i

(viii) 5 + 12i = 9 – 4+ 12i
= 3² + (2i)² + 2 x 3 x 2i
= (3 + 2i)²
and 5 – 12i = 9 – 4 – 12i
= 3² + (2i)² – 2 x 3 x 2i
= (3 – 2i)²
∴ \(\frac{\sqrt{5+12 i}+\sqrt{5-12 i}}{\sqrt{5+12 i}-\sqrt{5-12 i}}=\frac{\sqrt{(3+2 i)^2}+\sqrt{(3-2 i)^2}}{\sqrt{(3+2 i)^2}-\sqrt{(3-2 i)^2}}\)
= \(\frac{3+2 i+3-2 i}{3+2 i-(3-2 i)}\)
= \(\frac{6}{4 i}=\frac{3}{2 i} \times \frac{i}{i}=\frac{3 i}{-2}\) = – \(\frac { 3 }{ 2 }\)i

Question 10.
Prove that \(\left[\left(\frac{3+2 i}{2-5 i}\right)+\left(\frac{3-2 i}{2+5 i}\right)\right]\) is rational.
Solution:

Question 11.
Show that \(\frac{1+2 i}{3+4 i} \times \frac{1-2 i}{3-4 i}\) is real.
Solution:
\(\frac{1+2 i}{3+4 i} \times \frac{1-2 i}{3-4 i}=\frac{1^2-(2 i)^2}{3^2-(4 i)^2}\) = \(\frac{1+4}{9+16}=\frac{5}{25}=\frac{1}{5}\)
which is clearly a real number.

Question 12.
Perform the indicated operation and give your answer in the form x + yi, where x and y are real numbers and i = \(\sqrt{-1}\).
Solution:
(i) (3 + 4i)^-1
(ii) \(\frac{2-\sqrt{-25}}{1-\sqrt{-16}}\)
(iii) \(\frac{5+2 i}{-1+\sqrt{3} i}\)
(iv) \(\frac{5-3 i}{6+i}\)
(v) \((\sqrt{5}-7 i)(\sqrt{5}-7 i)^2+(-2+7 i)^2\)
Solution:

Question 13.
If x + yi = \(\frac{u+v i}{u-v i}\) prove that x² + y² = 1.
Solution:
Given x + iy = \(\frac{u+v i}{u-v i}\)
Taking modulus on both sides; we have
|x + iy| = \(\left|\frac{u+v i}{u-v i}\right|=\frac{|u+v i|}{|u-v i|}\) [∵ \(\left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|}\) ]
⇒ \(\sqrt{x^2+y^2}=\frac{\sqrt{u^2+v^2}}{\sqrt{u^2+\left(-v^2\right)}}\)
⇒ \(\sqrt{x^2+y^2}\) = 1
On squaring both sides ; we have
x² + y² = 1

Question 14.
Prove that:
\([4+3 \sqrt{-20}]^{\frac{1}{2}}+[4-3 \sqrt{-20}]^{\frac{1}{2}}\) = 6.
Solution:

Question 15.
Express the following in the form a + bi :
(i) \(\sqrt{\frac{5(2+i)}{2-i}}\)
(ii) \(\frac{(3-i)^2}{2+i}\)
(iii) (1 + i)^-3
(iv) \(\left(\frac{4 i^3-i}{2 i+1}\right)^2\)
(v) \(\frac{i-1}{i+1}\)
(vi) \(\frac{2+i}{(3-i)(1+2 i)}\)
(vii) \(\frac{5}{2 i-7 i^2}\)
Solution:

Question 16.
Prove that \(\left(\frac{-1+i \sqrt{3}}{2}\right)^3\) is a positive integer.
Solution:
\(\left(\frac{-1+i \sqrt{3}}{2}\right)^3\)
= \(\frac{1}{8}\left[(-1)^3+(i \sqrt{3})^3-3 \sqrt{3} i(-1+i \sqrt{3})\right]\)
= \(\frac{1}{8}[-1-3 \sqrt{3} i+3 \sqrt{3} i+9]=\frac{8}{8}\) = 1
which is a positive integer.

Question 17.
If one of the values of x of the equation 2x² – 6x + k = 0 be \(\frac { 1 }{ 2 }\) (a + 5i), find the values of a and k.
Solution:
Given eqn. be 2x² – 6x + k = 0
⇒ x = \(\frac{6 \pm \sqrt{36-8 k}}{4}=\frac{6 \pm 2 \sqrt{9-2 k}}{4}\)
⇒ x = \(\frac{3 \pm \sqrt{9-2 k}}{2}\)
Since one of the values of x is given to be \(\frac { 1 }{ 2 }\)(a + 5i) which is only possible
if 9 – 2k = – 25 ⇒ 2k = 34 ⇒ k = 17
∴ x = \(\frac{3 \pm 5 i}{2}=\frac{1}{2}(3 \pm 5 i)\)
Hence required value of a be 3.

Question 18.
Define conjugate complex numbers and show that their sum and product are real numbers.
Solution:
Let z = x + iy be any complex number where x, y ∈ R
Then conjugate of z = \(\bar{z}=\overline{x+i y}\)
= x – iy
∴ z + \(\bar { z }\) = x + iy + x – iy = 2x, which is a real number
and z\(\bar { z }\) = (x + iy) (x – iy) = x² – (i y)²
= x² + y²
which is again a real number, since x and y are reals.

Question 19.
If \(\bar { z }\) = – z ≠ 0, show that z is necessarily a purely imaginary number.
Solution:
Let z = x + iy where x, y ∈ R, x, y are not both zero.
∴ \(\bar { z }\) = x – iy
Also, given \(\bar { z }\) = – z
⇒ x – iy = – (x + iy)
⇒ x – iy = – x – iy
⇒ 2x = 0 ⇒ x = 0
∴ z = iy which is purely an imaginary number.

Question 20.
z and z’ are complex numbers such that their product zz’ = 3 – 4i.
Given that z’ is 5 + 3i, express z in the form a + bi where a and b are rational numbers.
Solution:
Given z, z’ ∈ C
s.t zz’ = 3 – 4z …(1)
Also, z’ = 5 + 3z
∴ from (1); we have
z = \(\frac{3-4 i}{5+3 i} \times \frac{5-3 i}{5-3 i}=\frac{(3-4 i)(5-3 i)}{5^2-(3 i)^2}\)
⇒ z = \(\frac{15-9 i-20 i-12}{25+9}=\frac{3-29 i}{34}\)
⇒ z = \(\frac{3}{34}-\frac{29}{34}\) i

Question 21.
If a + bi = \(\frac{(x+i)^2}{2 x^2+1}\), prove that a² + b² = \(\frac{\left(x^2+1\right)^2}{\left(2 x^2+1\right)^2}\).
Solution:

Question 22.
Let z₁ = 2 – i, z₂ = – 2 + i, find
(i) Re \(\left(\frac{z_1 z_2}{\bar{z}_1}\right)\),
(ii) Im \(\left(\frac{1}{z_1 \bar{z}_2}\right)\)
Solution:
Given z₁ = 2 – i, z₂ = – 2 + i
∴ \(\bar{z}_1=\overline{2-i}\) = 2 + i ;
\(\bar{z}_2=\overline{-2+i}\) = 2 + i ;

Question 23.
If z₁ = 3 + 5i and z₂ = 2 – 3i, then verify that \(\overline{\left(\frac{z_1}{z_2}\right)}=\frac{\bar{z}_1}{\bar{z}_2}\).
Solution:

Question 24.
If x = – 2 – \(\sqrt{3}\)i, where i = \(\sqrt{-1m}\), find that value of 2x⁴ + 5x³ + 7x² – x + 41.
Solution:
Given x = – 2 – \(\sqrt{3}\)i
⇒ x + 2 = – \(\sqrt{3}\)i
On squaring both sides ; we have
(x + 2)² = (- \(\sqrt{3}\)i)² ⇒ x² + 4x + 4 = – 3
⇒ x² + 4x + 7 = 0 … (1)
Now 2x⁴ + 5x³ + 7s2 – x + 41 = 2x² (x² + 4x + 7) – 3x³ – 7x² – x + 41
= 2x² (x² + 4x + 7) – 3x (x² + 4x + 7) + 5x² – 20x + 41
= 2x² (x² + 4x + 7) – 3x (x² + 4x + 7) + 5 (x² – 4x + 7) + 6
= 2x² x 0 – 3x × 0 + 5 x 0 + 6
= 6 [using (1)]

Question 25.
If z = – 3 + \(\sqrt{-2}\), then prove that z⁴ + 5z³ + 8z² + 7z + 4 is equal to – 29.
Solution:
Given z = – 3 + \(\sqrt{-2}\) = – 3 + \(\sqrt{2}\) i
⇒ z + 3 = \(\sqrt{2}\) i; On squaring both sides ; we have
(z + 3)² = (\(\sqrt{2}\) i)² ⇒ z² + 6z + 9 = – 2
⇒ z² + 6z+ 11 =0 …(1)
∴ z⁴ + 5z³ + 8z² + 7z + 4 = z² (z² + 6z + 11) – z³ – 3z² + 7z + 4
= z²(z² + 6Z+ 11) – z (z² + 6z + 11) + 3z² + 18z + 4
= z² (z² + 6z + 11) – z (z² + 6z + 11) + 3 (z² + 6z + 11) – 29
= z² x 0 – z x 0 + 3 x 0 – 29 [using eqn. (1)]
= – 29

Students often turn to Class 11 ISC Maths S Chand Solutions Chapter 16 The Straight Line Ex 16(c) to clarify doubts and improve problem-solving skills.

S Chand Class 11 ICSE Maths Solutions Chapter 16 The Straight Line Ex 16(c)

Question 1.
write down the equation of the straight line cutting off intercepts a and b from the axes where
(i) a = – 2, b = 3
(ii) a = 5, b = – 6
(iii) a = –\(\frac { k }{ m }\), b = k
Solution:
(i) Let the eqn. of straight line having intercepts a and b on axes be
\(\frac { x }{ a }\) + \(\frac { y }{ b }\) = 1 …(1)
given a = – 2, b = 3
∴ eqn. (1) becomes ;
\(\frac{x}{-2}\) + \(\frac{y}{3}\) = 1 ⇒ – 3x + 2y = + 6
⇒ 3x – 2y + 6 = 0

(ii) The eqn. of straight line having intercepts a and b on axes be given by
\(\frac { x }{ a }\) + \(\frac { y }{ b }\) = 1 …(1)
Here a = 5 and b = – 6
∴ eqn. (1) becomes ;
\(\frac { x }{ 5 }\) + \(\frac { y }{ -6 }\) = 1
⇒ 6x – 5y – 30 = 0

(iii) The eqn. of straight line cut off intercepts a and b on axes be given by
\(\frac { x }{ a }\) + \(\frac { y }{ b }\) = 1 …(1)
Here a = –\(\frac { k }{ m }\) and b = k
∴ eqn. (1) becomes; \(\frac{x}{\frac{-k}{m}}+\frac{y}{k}=1\)
⇒ -mx + y = k ⇒ mx – y + k = 0

Question 2.
Determine the x-intercept ‘a’ with the y intercept ‘b’ of the following lines. Sketch each.
(i) 3x + 5y – 15 = 9,
(ii) x – y – 7 = 0.
Solution:
(i) Given eqn. of line
3x + 5y – 15 = 0
it can be written as 3x + 5y = 15
⇒ \(\frac { x }{ 5 }\) + \(\frac { y }{ 3 }\) = 1
On comparing with \(\frac { x }{ a }\) + \(\frac { y }{ b }\) = 1
Here a = 5 and b = 3

(ii) Given eqn. of line be x – y – 7 = 0
⇒ x – y = 7 ⇒ ⇒ \(\frac { x }{ 7 }\) + \(\frac { y }{ -7 }\) = 1
On comparing with \(\frac { x }{ a }\) + \(\frac { y }{ b }\) = 1
∴ a = 7 and b = – 7

Question 3.
Find the equation of the line which makes equal intercepts on the axes and passes through the point (2, 3).
Solution:
Since the line makes equal intercepts on axes and let the length of intercept on axes be a
Let its eqn. be
\(\frac { x }{ a }\) + \(\frac { y }{ a }\) = 1
⇒ x + y = a …(1)
Since eqn. (1) passes through the point (2, 3)
∴ 2 + 3 = a ⇒ a = 5
Thus eqn. (1) becomes ; x + y = 5 be the required eqn. of line.

Question 4.
Write down the equation of the line which makes an intercept of 2a on the x-axis and 3a on the y-axis. Given that the line passes through the point (14, – 9), find the numerical value of a.
Solution:
The required eqn. of line which makes an intercept of 2a on x-axis and 3a on y-axis be given by
\(\frac { x }{ 2a }\) + \(\frac { y }{ 3a }\) = 1
⇒ 3x + 2y = 6a …(1)
It is given that line (1) passes through the point (14, – 9).
∴ 3 × 14 + 2 × (-9) = 6a ….(1)
⇒ 24 = 6a ⇒ a = 4

Question 5.
Find the equation of the straight line which passes through the point (5, 6) and has intercept on the axes equal in magnitude but opposite in sign.
Solution:
Let the eqn. of straight line in intercept form be given by
\(\frac { x }{ a }\) + \(\frac { y }{ b }\) = 1 …(1)
since intercept on the axes equal in magnitude but opposite in sign
∴ b = -a Thus eqn. (1) becomes;
\(\frac { x }{ a }\) + \(\frac { y }{ b }\) = 1 ⇒ x – y = a ….(2)
line (2) passes through the point (5, 6).
∴ 5 – 6 = a ⇒ a = – 1
Thus eqn. (2) becomes; x – y = – 1 be the required line.

Question 6.
A straight line passes through (2, 3) and the portion of the line intercepted between the axes is bisected at this point. Find its equation.
Solution:
Let the eqn. of line using intercept form be
\(\frac { x }{ a }\) + \(\frac { y }{ b }\) = 1 …(1)
line (1) meets coordinate axes at A(a, 0) and B(0, b).
Let P be the mid-point of AB.
∴ Coordinates of P are \(\left(\frac{a}{2}, \frac{b}{2}\right)\).

Also given coordinates of P are (2, 3).
∴ \(\frac { a }{ 2 }\) = 2 ⇒ a = 4
and \(\frac { b }{ 2 }\) = 3 ⇒ b = 6
Thus eqn. (1) becomes ; \(\frac { x }{ 4 }\) + \(\frac { y }{ 6 }\) = 1
⇒ 3x + 2y – 12 = 0
be the required eqn. of line.

Question 7.
Show that the three points (5, 1),(1, – 1) and (11, 4) lie on a straight line. Further find
(i) its intercepts on the axes;
(ii) the length of the portion of the line intercepted between the axes;
(iii) the slope of the line.
Solution:
Using two point form, eqn. of line through the points (5, 1) and (1, – 1) be given by
y – 1 = \(\frac{-1-1}{1-5}\) (x – 5)
⇒ y – 1 = \(\frac{1}{2}\) (x – 5)
⇒ x – 2y – 3 = 0 …(1)
Now the point (11, 4) lies on eqn. (1).
if 11 – 2 × 4 – 3 = 0 if 0 = 0, which is true.
Hence the given points (5, 1),(1, – 1) and (11, 4) lies on same straight line.

(i) eqn. (1) can be written as ;
x – 2y = 3 ⇒ \(\frac{x}{3}\) + \(\frac{y}{\frac{-3}{2}}\) = 1 …(2)
So line (2) cut off intercept 3 and \(\frac{-3}{2}\) on axes.

(ii) Clearly line (2) intersects x-axis at A(3, 0) and B\(\left(0, \frac{-3}{2}\right)\)
∴ required length of the portion of the line intercept between the axes

(iii) eqn. (1) can be written as ;
y = \(\frac{x}{2}\) – \(\frac{3}{2}\)
On comparing with y = mx + b
we have m = \(\frac{1}{2}\) and b = \(\frac{-3}{2}\)
∴ slope of line = \(\frac{1}{2}\)

Question 8.
Find the equation of the straight line which passes through the point (3, – 2) and cuts off positive intercepts on the x and y-axes which are in the ratio 4 : 3.
Solution:
Let the eqn. of straight line be
\(\frac { x }{ a }\) + \(\frac { y }{ b }\) = 1 …(1)
where a > 0, b > 0
Since the intercepts on coordinates axes are in the ratio 4 : 3. Let the intercepts on x axis and y-axis are 4a and 3a.

∴ eqn. (1) becomes ; \(\frac { x }{ 4a }\) + \(\frac { y }{ 3a }\) = 1
⇒ 3x + 4y = 12a …(2)
Since line (2) passes through the point (3, – 2).
∴ 9 – 8 = 12a ⇒ 12a = 1
∴ eqn. (1) becomes ; 3x + 4y = 1 be the required line.

Question 9.
Find the equation of the straight line at a distance of 3 units from the origin such that the perpendicular from the origin to the line makes an angle α, given by the equation tan α = \(\frac { 5 }{ 12 }\), with the positive direction of the axis of x.
Solution:
Here p = ⊥ distance from (0, 0) on required line = 3
given tan α = \(\frac { 5 }{ 12 }\)
∴ sin α = \(\frac { 5 }{ 13 }\)
and cos α = \(\frac { 12 }{ 13 }\)

Thus using normal form, eqn. of line be given by x cos α + y sin α = p
⇒ x × \(\frac { 12 }{ 13 }\) + y + \(\frac { 5 }{ 13 }\) = 3
⇒ 12x + 5y = 39

Question 10.
Show on a diagram the position of the straight line x cos 30° + y sin 30° = 2 in relation to the co-ordinate axes, indicating clearly which angle is 30° and which length is 2 units. Find
(i) the equation of the straight line parallel to that given and passing through the point (4, 3);
(ii) the length of the perpendicular from the origin on to this line ;
(iii) the distance between the two parallel straight lines.
Solution:
Given eqn. of straight line be
x cos 30° + y sin 30° = 2
x \(\frac{\sqrt{3}}{2}\) + \(\frac{y}{2}\) = 2 ⇒ √3x + y = 4 …(1)
line (1) intersect x-axis at A \(\left(\frac{4}{\sqrt{3}}, 0\right)\) and B (0, 4).

(i) from (1); y = -√3x + 4
∴ slope of line (1) = -√3
∴ slope of line parallel to line (1) = -√3
Thus eqn. of straight line parallel to line (1) and passing through the point (4, 3) be given by
y – 3 = -√3(x – 4)
⇒ √3x + y = 3 + 4√3
⇒ \(\frac{\sqrt{3}}{2} x\) + \(\frac{y}{2}\) = \(\frac{3}{2}\) + 2√3
⇒ x cos 30° + y sin 30° = \(\frac{3}{2}\) + 2√3 …(2)

(ii) On comparing with
x cos α + y sin α = p
∴ length of ⊥ from origin on line (2)
= \(\frac{3}{2}\) + 2√3

(iii) Distance between parallel lines = ⊥ distance of any point (x₁, y₁) on line (1) to line (2)

Question 11.
A straight line \(\frac{x}{a}\) – \(\frac{y}{b}\) = 1 passes through the point (8, 6) and cuts off a triangle of area 12 units from the axes of coordinates. Find the equations of the straight line.
Solution:
Given eqn. of straight line be
\(\frac{x}{a}\) – \(\frac{y}{b}\) = 1 …(1)
eqn. (1) passes through the point (8, 6).
∴ \(\frac{8}{a}\) – \(\frac{6}{b}\) = 1
⇒ 8b – 6a = ab …(2)
Further line (1) passes through A(a, 0) and B(0, – b) and also line (1) passes through the point P(8, 6).

From (2) and (3); we have
8b – 6 × \(\frac { 24 }{ b }\) = 24 ⇒ 8b² – 24b = 144
⇒ b² – 3b – 18 = 0 ⇒ (b – 6)(b + 3) = 0
⇒ b = 6, – 3
When b = 6 ∴ from (2); a = \(\frac { 24 }{ 6 }\) = 4
Thus eqn. (1) becomes; \(\frac { x }{ 4 }\) – \(\frac { y }{ 6 }\) = 1 be the required line.
When b = – 3 ∴ from (2); a = \(\frac { 24 }{ -3 }\) = – 8
Thus eqn. (1) becomes ;
\(\frac { x }{ -8 }\) – \(\frac { y }{ -3 }\) = 1
⇒ \(\frac { x }{ 8 }\) – \(\frac { y }{ 3 }\) = 1
which is the required line.

Question 12.
A straight line passes through the points (a, 0) and (0, b). The length of the line segment contained between the axes is 13 and the product of the intercepts on the axes is 60 . Calculate the values of a and b and find the equation of the straight line.
Solution:
Given |AB| = 13 units
⇒ \(\sqrt{(a-0)^2+(0-b)^2}\) = 13
⇒ a² + b² = 169 ….(1)
Also product of the intercepts on the axes be 60 .
∴ ab = 60 …(2)
(a+b)² = a² + b² + 2ab
=169 + 2 × 60 = 289
⇒ a + b = ± 17

Case-I : When a + b = 17 ⇒ b = 17 – a
∴ from (2); a(17 – a) = 60
⇒ a² – 17a + 60 = 0
⇒ (a – 5)(a – 12) = 0
⇒ a = 5, 12
When a = 5 ⇒ b = 12
When a = 12 ⇒ b = 5
Thus eqn. of lines (using intercept form) be given by
\(\frac { x }{ 5 }\) – \(\frac { y }{ 12 }\) = 1
0r \(\frac { x }{ 12 }\) – \(\frac { y }{ 5 }\) = 1

Case-II : When a + b = – 17
⇒ b = – 17 – a
∴ from (2); a( – 17 – a) = 60
⇒ a² + 17a + 60 = 0
⇒ (a + 5) (a + 12) = 0
⇒ a = – 5, – 12
When a = – 5 ⇒ b = – 12
When a = – 5 ⇒ b = – 12
When a = – 12 ⇒ b = – 5
Thus the corresponding eqns. of straight lines are
\(\frac { x }{ -5 }\) + \(\frac { y }{ -12 }\) = 1
0r \(\frac { x }{ -12 }\) + \(\frac { y }{ -5 }\) = 1

Regular engagement with OP Malhotra Maths Class 11 Solutions Chapter 9 Complex Numbers Ex 9(a) can boost students confidence in the subject.

S Chand Class 11 ICSE Maths Solutions Chapter 9 Complex Numbers Ex 9(a)

Express each of the following in the form b or bi, where b is a real number:

Question 1.
3i . 2
Solution:
3i . 2 = 6i

Question 2.
i (- i)
Solution:
i (- i) = – i² = – (- 1) = 1 + 0i

Question 3.
– i (- i)
Solution:
– i (- i) = i² = – 1

Question 4.
5i (- 8i)
Solution:
5i (- 8i) = – 40i² = – 40 (- 1) = 40

Question 5.
\(\frac { 20i }{ 4 }\)
Solution:
\(\frac { 20i }{ 4 }\) = 5i

Question 6.
\(\sqrt{-25}\)
Solution:
\(\sqrt{-25}=\sqrt{5 \times 5} \sqrt{-1}\) = 5i

Question 7.
\(\sqrt{-8}\)
Solution:
\(\sqrt{-8}=\sqrt{2 \times 2} \sqrt{-2}=2 \sqrt{2} i\)

Question 8.
\(\sqrt{\frac{-1}{3}}\)
Solution:
\(\sqrt{\frac{-1}{3}}=\sqrt{\frac{1}{3}} i=\sqrt{\frac{1}{3} \times \frac{3}{3}} i=\frac{\sqrt{3}}{3} i\)

Question 9.
\(\frac{1}{2} \sqrt{\frac{-3}{4}}\)
Solution:
\(\frac{1}{2} \sqrt{\frac{-3}{4}}=\frac{1}{2} \times \frac{1}{2} \sqrt{3} i=\frac{\sqrt{3}}{4} i\)

Question 10.
\(\frac { 6 }{ -i }\)
Solution:
\(\frac{6}{-i}=\frac{6}{-i} \times \frac{i}{i}=\frac{6 i}{-i^2}=\frac{6 i}{-(-1)}\) = 6i

Question 11.
\(\sqrt{-144}\)
Solution:
\(\sqrt{-144}=\sqrt{12 \times 12} \sqrt{-1}\) = 12i

Question 12.
\(\frac { x }{ i }\)
Solution:
\(\frac{x}{i}=\frac{x}{i} \times \frac{i}{i}=\frac{i x}{i^2}\) = – i x

Question 13.
i¹³
Solution:
i¹³ = i¹² i = (i⁴)³ i = 1³ i = i [∵ i⁴ = 1]

Question 14.
i²⁸
Solution:
i²⁸ = (i⁴)⁷ = 1 [∵ i⁴ = (i²)² = (- 1)² = 1]

Question 15.
i¹⁸
Solution:
i¹⁸ = (i⁴)⁴ i² = 1⁴ x (- 1) = – 1

Question 16.
i²³
Solution:
i²³ = (i⁴)⁵ i³ = 1⁵ x i² x i = 1 x (- 1) x i = – i

Question 17.
\(\sqrt{-4}+\sqrt{-16}-\sqrt{-25}\)
Solution:
\(\sqrt{-4}+\sqrt{-16}-\sqrt{-25}\) = 2i + 4i – 5i
= 6i – 5i = i

Question 18.
\(\sqrt{-20}+\sqrt{-12}\)
Solution:
\(\sqrt{-20}+\sqrt{-12}=2 \sqrt{5} i+2 \sqrt{3} i\)
= 2(\(\sqrt{5}+\sqrt{2}\))i

Question 19.
– \(\sqrt{\frac{-7}{4}}-\sqrt{\frac{-1}{7}}\)
Solution:

Question 20.
\(\frac{\sqrt{-2}}{\sqrt{-8}}\)
Solution:
\(\frac{\sqrt{-2}}{\sqrt{-8}}=\frac{\sqrt{2} i}{2 \sqrt{2} i}=\frac{1}{2}\)

Question 21.
\(\frac{1}{i}+\frac{1}{i^2}+\frac{1}{i^3}+\frac{1}{i^4}\)
Solution:
\(\frac{1}{i}+\frac{1}{i^2}+\frac{1}{i^3}+\frac{1}{i^4}=\frac{1}{i}+\frac{1}{-1}+\frac{1}{-i}+1\)
= \(\frac{1}{i}-1-\frac{1}{i}+1\) = 0

Question 22.
\(\frac{1}{i}-\frac{1}{i^2}+\frac{1}{i^3}-\frac{1}{i^4}\)
Solution:
\(\frac{1}{i}-\frac{1}{i^2}+\frac{1}{i^3}-\frac{1}{i^4}=\frac{1}{i}+1-\frac{1}{i}-1\) = 0 [∵ i² = – 1 and i⁴ = 1]

Question 23.
i + 2i² + 3i³ + i⁴
Solution:
i + 2i² + 3i³ + i⁴ = i + 2 (- 1) + 3i (- 1) + 1 = i – 2 – 3i + 1 = – 2i – 1

Question 24.
\(\left[i^{18}+\left(\frac{1}{i}\right)^{25}\right]^3\)
Solution:

Question 25.
\(\sqrt{\frac{-x}{4}}+\sqrt{\frac{-x}{16}}-\sqrt{\frac{-x}{64}}\),
where x is a positive real number.
Solution:

Question 26.
\(\sqrt{-5 x^8}-\sqrt{-20 x^8}+\sqrt{-45 x^8}\), where x a positive real number.
Solution:
\(\sqrt{-5 x^8}-\sqrt{-20 x^8}+\sqrt{-45 x^8}\)
= \(\sqrt{5} x^4 i-2 \sqrt{5} x^4 i+3 \sqrt{5} x^4 i\)
= \(2 \sqrt{5} i x^4\)

Question 27.
If i = \(\sqrt{-1}\) prove the following :
(x + 1 + i) (x + 1 – i) (x – 1 – i) = x⁴ + 4.
Solution:
L.H.S = (x + 1 + i) (x + 1 – i) (x – 1 + i) (x – 1 – i)
= [(x + 1)² – i²] [(x – 1)² – i²]
= [x² + 2x + 1 + 1] [x² – 2x + 1 + 1]
= (x² + 2x + 2) (x² – 2x + 2)
= (x² + 2 + 2x) (x² + 2 – 2x)
= (x² + 2)² – (2x)²
= x⁴ + 4x² + 4 – 4x²
= x⁴ + 4 = R.H.S