OP Malhotra Class 11 Maths Solutions Chapter 16 The Straight Line Ex 16(d)

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.