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

The availability of step-by-step Class 11 ISC Maths S Chand Solutions Chapter 16 The Straight Line Ex 16(b) can make challenging problems more manageable.

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

Question 1.
State the equation of the line which has the y-intercept
(i) 2 and slope 7 ;
(ii) -3 and slope -4 ;
(iii) -1 and is parallel to y = 5x – 7;
(iv) 2 and is inclined at 45° to the x-axis ;
(v) -5 and is equally inclined to the axes.
Solution:
(i) Given y-intercept = b = 2 and m = slope of line = 7
By using slope intercept form, eqn. of line be given by
y = mx + b i.e. y = 7x + 2

(ii) Given y-intercept b = – 3 and
slope m = – 4
∴ using slope intercept form, eqn. of line be given by
y = mx + b i.e. y = – 4x – 3

(iii) Given y-intercept = – 1 = b eqn. of given line be y = 5x – 7
On comparing with y = mx + b where m = 5 and b = – 7
∴ slope of given line be 5 = m
Thus slope of line || to given line be 5 .
∴ eqn. of line be given
y = mx + b i.e. y = 5x -1

(iv) Given y-intercept of line be 2 i.e. b = 2 and given required line makes an angle 45° with the +ve direction of x-axis.
∴ θ = 45° ⇒ m = tan θ = tan 45° = 1
Thus by using slope intercept form, eqn. of line be given by
y = mx + b i.e. y = x + 2

(v) Given y-intercept of required line be -5 since required line is equally inclined to axes i.e. m = ± 1
Hence the required eqn. of line using slope intercept form be given by
y = mx + b i.e. y = ± x – 5

Question 2.
What will be the value of m and c if the straight line y = mx + c passes through the points (3, – 4) and (- 1, 2)?
Solution:
Given eqn. of straight line be
y = mx + c …(1)
eqn. (1) passes through the points (3, – 4) and (- 1, 2).
– 4 = 3m + c …(2)
2 = – m + c …(3)
eqn. (2) – eqn. (3) gives ;
-6 = 4m ⇒ m = – \(\frac { 3 }{ 2 }\)
∴ from (3); c = 2 + m = 2 – \(\frac { 3 }{ 2 }\) = \(\frac { 1 }{ 2 }\)

Question 3.
Find the equation of the straight line through the given point P and having the given slope m if
(i) P(- 4, 7), m = -√3;
(ii) P(- 1, – 5), m = \(\frac { -6 }{ 11 }\).
Solution:
(i) Using one point form, eqn. of line passing through the point (-4, 7) and having slope m = -√3 is given by
y – y₁ = m(x – x₁)
i.e. y – 7 = -√3(x + 4)
⇒ √3x + y = 7 – 4√3

(ii) Using one point form, eqn. of line passing through the point (-1, – 5) and having slope m = –\(\frac { 6 }{ 11 }\) is given by
y – y₁ = m(x – x₁)
⇒ y + 5 = –\(\frac { 6 }{ 11 }\) (x + 1)
⇒ 6x + 11y + 61 = 0

Question 4.
Find the equation of the line through the point (1, – 2) making an angle of 135° with the x-axis.
Solution:
Since the required line making an angle of 135° with x-axis
∴ θ = 135°
Thus slope of line m = tan θ = tan 135°
= tan (90° + 45°) = – cot 45° = – 1
Hence using one point form, eqn. of line through the point (1, – 2) and having slope -1 be given by
y – (- 2) = – 1(x – 1) ⇒ x + y + 1 = 0

Question 5.
Find the equation to the straight line passes through
(i) the origin and perpendicular to x + 2y = 4;
(ii) the point (4, 3) and parallel to 3x + 4y = 12;
(iii) the point (4, 5) and (a) parallel to, (b) perpendicular to 3x – 2y + 5 = 0.
Solution:
(i) Eqn. of given line be

Hence eqn. of line through the point (0, 0) and having slope 2 be given by
y – 0 = 2(x – 0) | ∵ y – y₁ = m(x – x₁)
⇒ y = 2x

(ii) equation of given line be
3x + 4y = 12 …(1)
∴ 4y = 12 – 3x ⇒ y = \(\frac{-3}{4}\) x + 3
Thus slope of line (1) = \(\frac{-3}{4}\)
Since required line is parallel to given line.
Thus slope of required line = slope of given line = \(\frac{-3}{4}\) [∵ m₁ = m₂]
Hence eqn. of required line which pass through the point (4, 3) and having slope \(\frac{-3}{4}\) is given by
y – 3 = \(\frac{-3}{4}\) (x – 4)
⇒ 3x + 4y – 24 = 0

(iii) (a) eqn. of given line be
3x – 2y + 5 = 0 …(1)
∴ slope of given line (1) = \(\frac{-3}{-2}\) = \(\frac{3}{2}\)
[∵ 2y = 3x + 5 ⇒ y = \(\frac{3x}{2}\) + 5]
Since required line is parallel to line (1)
Thus both lines have same slopes
∴ slope of required line = \(\frac{3}{2}\)
Thus, eqn. of line passing through the point (4, 5) and having slope \(\frac{3}{2}\) is given by
by y – 5 = \(\frac{3}{2}\) (x – 4) | y – y₁ = m (x – x₁)
⇒ 3x – 2y – 2 = 0

(b) Since required line is ⊥ to line (1).
∴ slope of required line = \(\frac{-1}{\text { slope of line (1) }}\) = \(\frac{-1}{\frac{3}{2}}\) = \(\frac{-2}{3}\)
Thus, eqn. of line through the point (4, 5) and having slope \(\frac{-2}{3}\) is given by
y – 5 = \(\frac{-2}{3}\) (x – 4)
⇒ 2x – 3y – 23 = 0

Question 6.
Find the equation to the line which is perpendicular to the line \(\frac{x}{a}\) – \(\frac{y}{b}\) = 1 at the point where it meets the x-axis.
Solution:
eqn. of given line be
\(\frac{x}{a}\) – \(\frac{y}{b}\) = 1 …(1)
eqn. (1) meets x-axis at y = 0
∴ from (1) ; x = a
Thus the required line passes through the point of intersection of line (1) and x-axis i.e. at point (a, 0)
On comparing eqn. (1) with y = mx + c
m = slope of line (1) = \(\frac{b}{a}\)
∴ slope of required line = \(-\frac{1}{\text { slope of line (1) }}\) = \(\frac{-1}{\frac{b}{a}}\) = \(\frac{-a}{b}\)
By using one point form, eqn. of line through the point (a, 0) and having slope \(\frac{-a}{b}\) is given by
y – 0 = \(\frac{-a}{b}\) (x – a) ⇒ ax + by = a²
which is the required eqn. of line.

Question 7.
Find the equation to the two lines through the point (4, 5) which make an acute angle of 45° with the line 2x – y + 7 = 0.
Solution:
Equation of given line be
2x – y + 7 = 0 ⇒ y = 2x + 7
∴ slope of given line = 2 = m₁
Let the slope of required line be m.
since required line makes an angle 45° with the given line.
∴ tan 45° = ±\(\frac{m_1-m}{1+m_1 m}\)
⇒ 1 = ±\(\left(\frac{2-m}{1+2 m}\right)\)
⇒ (1 + 2m) = ±(2 – m)
Taking +ve sign;
1 + 2m = 2 – m ⇒ 3m = 1 ⇒ m = \(\frac{1}{3}\)
Taking -ve sign ;
1 + 2m = – 2 + m
⇒ m = – 3
Thus the required eqn: of line passes through the point (4, 5) and having slope \(\frac{1}{3}\) is given by y – 5 =\(\frac{1}{3}\) (x – 4) |y – y₁ = m (x – x₁)
Also the required eqn. of line through the point (4, 5) and having slope -3 is given by y – 5 = – 3(x – 4) ⇒ 3x + y – 17 = 0

Question 8.
The line through A(4, 7) with gradient m meets the x-axis at P and the y-axis at R. The line through B(8, 3) with gradient \(\frac{-1}{m}\) meets the x-axis at Q and the y-axis at S. Find in terms of m, the coordinates of P, Q, R and S. Obtain expressions for OP.OQ and OR.OS, where O is the point (0, 0).
Solution:
Using one point form, eqn. of line through the point A(4, 7) with gradient m is given by
y – 7 = m(x – 4) …(1)
eqn. (1) meets x-axis at y = 0
∴ from (1); x = 4 – \(\frac{7}{m}\)
∴ Coordinates of P are \(\left(\frac{4 m-7}{m}, 0\right)\)
Since line (1) meets x-axis at P and R eqn. (1) meets y-axis at x = 0
∴ from (1); we have y = 7 – 4m
∴ Coordinates of R are (0, 7 – 4m)
Also eqn. of line through the point B(8, 3) and having slope –\(\frac { 1 }{ m }\) be given by
y – 3 = –\(\frac { 1 }{ m }\)(x – 8) …(2)
| y – y₁ = m (x – x₁)
Since eqn. (2) meets x-axis at Q and y-axis at S.
For intersection of line (2) with x-axis, we have y = 0
∴ from (2); x = 8 + 3m
∴ Coordinates of Q are (8 + 3m, 0)
For intersection of line (2) with y-axis we have x = 0
∴ from (2); y = 3 + \(\frac { 8 }{ m }\) = \(\frac{3 m+8}{m}\)

Question 9.
Write down the slopes of the lines
(i) joining P(1, 1) and Q(2, 3);
(ii) joining L(-p, q) and M(r, s);
(iii) parallel to the line joining A(-1, 5) and B(-6, – 7)
(iv) perpendicular to the line joining B(2, – 3) and S(- 4, 1).
Solution:

Question 10.
Find the equations of the lines joining the points
(i) A(1, 1) and B(2, 3);
(ii) L(a, b) and M(b, a);
(iii) P(3, 3) and Q(7, 6).
Solution:
(i) Using two point form, eqn. of line passing through the points A(1, 1) and B(2, 3) is given by

(ii) eqn. of line joining the points L(a, b) and M(b, a) is given by
y – b = \(\frac{a-b}{b-a}\)(x – a)
⇒ y – b = – (x – a)
⇒ x + y – b – a = 0
⇒ x + y = a + b

(iii) eqn. of line joining P(3, 3) and Q(7, 6) is given by
y – 3 = \(\frac{6-3}{7-3}\) (x – 3)

⇒ y – 3 = \(\frac{3}{4}\) (x – 3)
⇒ 4y – 12 = 3x – 9
⇒ 3x – 4y + 3 = 0 …(1)
eqn. (1) meets x-axis at y = 0
∴ from (1); x = – 1
and line (1) meets y-axis at x = 0
∴ from (1); y = \(\frac{3}{4}\)
Thus line (1) meets coordinate axes at (-1, 0) and \(\left(0, \frac{3}{4}\right)\)
Therefore length of the portion of line intercepted between coordinate axes

Question 11.
Given the vertices A(10, 4), B(- 4, 9) and C(-2, – 1) of △ABC, find
(i) the equation of the side AB;
(ii) the equation of the median through A;
(iii) the equation of the altitude through B;
(iv) the equation of the perpendicular bisector of the side AB.
Solution:
(i) Using two point form, eqn. of side AB is given by
y – 4 = \(\frac{9-4}{-4-10}\)(x – 10)
y – 4 = \(\frac{5}{-14}\)(x – 10)
⇒ – 14y + 56 = 5x – 50
⇒ 5x + 14y – 106 = 0

(ii) Let AD be the median through A on BC Then D be the mid-point of BC.
∴ Coordinates of D are \(\left(\frac{-4-2}{2}, \frac{9-1}{2}\right)\) i.e. D (- 3, 4)
Hence the required eqn. of the median through A(0, 4) is given by
y – 4 = \(\frac{4-4}{(-3-10)}\) (x – 10) ⇒ y = 4

(iii) Let BE be the altitude through B on side AC of △ABC.
slope of line AC = \(\frac{-1-4}{-2-10}\) = \(\frac{-5}{-12}\) = \(\frac{5}{12}\)
∴ slope of line BE = \(\frac{-1}{\text { slope of } A C}\) = \(\frac{-1}{\frac{5}{12}}\) = \(\frac{-12}{5}\)
Hence the required eqn. of BE be given by
y – 9 = –\(\frac{12}{5}\) (x + 4)
⇒ 12x + 5y + 3 = 0

(iv) Let CF be the ⊥ bisector of side AB
∴ Coordinates of F are \(\left(\frac{10-4}{2}, \frac{4+9}{2}\right)\)
i.e. \(F\left(3, \frac{13}{2}\right)\)
and slope of line AB = \(\frac{9-4}{-4-10}\) = \(\frac{5}{-14}\)
Thus slope of ⊥ bisector = \(\frac{-1}{\text { slope of } A B}\) = \(\frac{-1}{\frac{5}{-14}}\) = \(\frac{14}{5}\)
Hence using one point form, eqn. of line through point F\(\left(3, \frac{13}{2}\right)\) and having slope \(\frac{14}{5}\) is given by
y – \(\frac{13}{2}\) = \(\frac{14}{5}\) (x – 3)
⇒ 10y – 65 = 28x – 84
⇒ 28x – 10y – 19 = 0

Question 12.
The points A, B and C are (4, 0),(2, 2) and (0, 6) respectively. AB produced cuts the y-axis at P and CB produced cuts the x-axis at Q. Find the co-ordinates of the points P and Q. Find the equation of the straight line joining the mid-points of AC and OB (where O is the origin), and verify that this line passes through the midpoint of PQ.
Solution:
Given points are A(4, 0), B(2, 2) and C (0, 6)
The eqn. of line AB using two point form is given by

line (1) meets y-axis i.e. x = 0
∴ from (1); y = 4
Thus line AB cuts y-axis at P(0, 4).
eqn. of line CB is given by
y – 2 =\(\frac{6-2}{0-2}\) (x – 2)
⇒ y – 2 = – 2 (x – 2)
⇒ 2x + y = 6 …(2)
line CB meets x-axis i.e. y = 0
∴ from (2); x = 3
Hence line CB meets x-axis at Q(3, 0).
Mid-point of line AC and OB are
R \(\left(\frac{4+0}{2}, \frac{0+6}{2}\right)\) and S \(\left(\frac{0+2}{2}, \frac{0+2}{2}\right)\)
i.e. R(2, 3) and S(1, 1)
Thus required eqn. of straight line RS is given by
y – 3 = \(\frac{(1-3)}{1-2}\) (x – 2)
⇒ y – 3 = 2 (x – 2)
⇒ 2x – y – 1 = 0 …(3)
Mid-point of PQ is given by
T \(\left(\frac{0+3}{2}, \frac{4+0}{2}\right)\) i.e. T \(\left(\frac{3}{2}, 2\right)\)
Clearly T \(\left(\frac{3}{2}, 2\right)\) lies on eqn. (3)
∵ 2 × \(\frac { 3 }{ 2 }\) – 2 – 1 = 0
⇒ 0 = 0, which is true.
Thus the line joining the mid-points of AC and OB passes through the mid-point of PQ.

Question 13.
A line through the point (3, 0) meets the variable line y = tx at right angles at the point P. Find, in terms of t, the coordinates of P. Find the value of k for which P lies on the curve x² + y² = kx.
Solution:
The given eqn. of variable line be
y = tx …(1)
∴ slope of line be t.
Since the line which pass through point (3, 0) meet (1) at right angle at point P
∴ slope of required line = \(\frac{-1}{\text { slope of variable line }}\) = \(\frac{-1}{t}\)
Thus eqn. of line through the point (3, 0) and having slope –\(\frac{1}{t}\) is given by
y – 0 = –\(\frac{1}{t}\)(x – 3)
| ∵ y – y₁ = m(x – x₁)
x + ty = 3 …(2)
Clearly point P be the point of intersection of line (1) and (2).
From (1) and (2); we have
x + t² x = 3 ⇒ x = \(\frac{3}{1+t^2}\)
∴ from (1); y = \(\frac{3 t}{1+t^2}\)
Thus the coordinate of point P are

Question 14.
The point P is the foot of the perpendicular from A(0, t) to the line whose equation is y = tx. Determine
(i) the equation of the line AP,
(ii) the co-ordinates of P,
(iii) the area of △OAP, where O is the origin.
Solution:
(i) Eqn. of given line y = tx …(1)
and its slope be t

Thus eqn. of line AP be given by
y – t = \(\frac{-1}{t}\)(x – 0)
[∵ y – y₁ = m(x – x₁)]
⇒ ty – t² = – x
⇒ x + ty = t²

(ii) Clearly P be the point of intersection of line (1) and (2)
∴ x + t(tx) = t² ⇒ x(1 + t²) = t²
⇒ x = \(\frac{t^2}{1+t^2}\)
∴ from (1); y = \(\frac{t^3}{1+t^2}\)
Hence coordinates of point P are
\(\left(\frac{t^2}{1+t^2}, \frac{t^3}{1+t^2}\right)\)

(iii) ∴ required area of △OAP

Question 15.
Find the equation of the line joining the origin to the point of intersection of 4x + 3y = 8 and x + y = 1.
Solution:
Given eqn. of lines are
4x + 3y = 8 …(1)
x + y = 1 …(2)
For finding point of intersection (1) and (2). We solve eqn. (1) and (2) simultaneously eqn. (1) – 3 × eqn. (2) gives ;
x = 5
∴ from (2); y = 1 – 5 = – 4
∴ point of intersection of lines (1) and (2) be (5, – 4).
Hence using two point form, eqn. of line joining (0, 0) to point (5, – 4) is given by
y – 0 = \(\left(\frac{-4-0}{5-0}\right)\) (x – 0)
⇒ y = –\(\frac { 4 }{ 5 }\) x
4x + 5y = 0

Question 16.
Find the equation to the straight line which passes through the point of intersection of the lines 3x + 4y – 1 = 0 and 5x + 8y – 3 = 0 and is perpendicular to the line 4x – 2y + 3 = 0.
Solution:
Given equations of lines are
3x + 4y – 1 = 0 …(1)
5x + 8y – 3 = 0 …(2)
To find the point of intersection of lines (1) and (2)
we solve eqn. (1) and (2) simultaneously.
eqn. (2) – eqn. (1) × 2
– x – 1 = 0 ⇒ x = – 1
∴ from (1); we have
– 3 + 4y – 1 = 0 ⇒ y = 1
Thus coordinates of point of intersection of lines (1) and (2) are P(- 1, 1).
Also slope of given line 4x – 2y + 3
= \(\frac{-4}{-2}\) = 2
∴ slope of required line = \(\frac{-1}{\text { slope of given line }}\)
= \(\frac{-1}{2}\)
Thus using one point form, eqn. of line through the point P(- 1, 1) and having slope
–\(\frac{1}{2}\) be given by y – 1 =-\(\frac{1}{2}\)(x + 1)
⇒ x + 2y – 1 = 0