Class 10: Equation of a Line – ICSE Board Problems – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: A line intersects $x-axis$ at point $(-2, 0)$ and cuts off an intercept of $3$ units from the positive side of $y-axis$ . Find the equation of the line. [1992]

Answer:

$x-intercept = (-2, 0)$

$y-intercept = (0, 3)$

Equation of line

$y-3 =$ $\frac{3-0}{0-(-2)}$ $(x-0)$

$y-3 =$ $\frac{3}{2}$ $x$

$2y -6 = 3x$

$2y = 3x+6$

$\\$

Question 2: Find the equation of a line passing through the point $(2, 3)$ and having the $x-intercept$ of $4$ units. [2002]

Answer:

$x-intercept = (4, 0)$

Equation of line passing through $(2,3)$ and $(4, 0)$

$y - 0 =$ $\frac{0-3}{4-2}$ $(x-2)$

$y = -$ $\frac{3}{2}$ $(x-4)$

$2y = -3x+12$

$2y + 3x = 12$

$\\$

Question 3: The given figure (not drawn to scale) shows two straight lines $AB \ and \ CD$ . If equation of the line $AB$ is $y =x+1$ and equation of $CD$ is $y = \sqrt{3}x-1$ . Write down the inclination of lines $AB \ and \ CD$ ; also, find the angle between $AB \ and \ CD$ [1989]

Answer:

$AB: y = x+1$

$CD: y = \sqrt{3}x - 1$

Slope of $AB = 1$

$\tan \alpha_1 = 1 \Rightarrow \alpha_1 = 45^o$

Slope of $CD = \sqrt{3}$

$\tan \alpha_2 = 1 \Rightarrow \alpha_2 = 60^o$

Therefore $45^o+(180^o-60^o) + \theta = 180^o$

$\Rightarrow \theta = 15^o$

$\\$

Question 4: Write down the equation of the line whose gradian is $\frac{2}{3}$ and which passes through $P$ , where $P$ divides the line segment joining $A (-2, 6) \ and \ B (3, -4)$ in the ratio $2 : 3$ [2001]

Answer:

Given $P$ divides the line segment joining $A (-2, 6) \ and \ B (3, -4)$ in the ratio $2 : 3$

Let the coordinates of $P = (x, y)$

Therefore

$x =$ $\frac{2 \times 3 + 3 \times (-2)}{2+3}$ $=$ $\frac{6-6}{5}$ $= 0$

$y =$ $\frac{2 \times (-4) + 3 \times 6}{2+3}$ $=$ $\frac{-8+18}{5}$ $= 2$

$\therefore P(0,2)$

Equation of a line passing through $P(0,2)$ with slope $\frac{3}{2}$

$y-2 =$ $\frac{3}{2}$ $(x-0)$

$2y - 4 = 3x \Rightarrow 2y = 3x+4$

$\\$

Question 5: Point $A \ and \ B$ have co-ordinates $(7, 3) \ and \ (1, 9)$ respectively. Find:

(i) The slope of $AB$

(ii) The equation of perpendicular bisector of the line segment $AB$

(iii) The value of $p \ of \ (-2, p)$ lies on it [2008]

Answer:

(i) Slope of $AB =$ $\frac{9-(-3)}{1-7}$ $=$ $\frac{12}{-6}$ $= -2$

(ii) Mid point of $AB = ($ $\frac{7+1}{2}$ $,$ $\frac{-3+9}{2}$ $) = (4,3)$

Therefore equation of line passing through $(4,3)$ and slope $\frac{1}{2}$ is

$y - 3 =$ $\frac{1}{2}$ $(x-4)$

$2y -6 = x-4$

$2y= x+2$

(iii) $p \ of \ (-2, p)$ lies on it

Therefore $2(p) = (-2)+2 \Rightarrow p = 0$

$\\$

Question 6: $A \ and \ B$ are two points on the $x-axis \ and \ y-axis$ respectively. $P (2, -3)$ is the mid-point of $AB$ . Find the

(i) Co-ordinates of $A \ and \ B$

(ii) Slope of line $AB$

(iii) Equation of line $AB$ [2010]

Answer:

Let $A(x,0)$ and $B(0, y)$

$P(2, -3)$ is the mid point

(i) Therefore $2 =$ $\frac{0+x}{2}$ $\Rightarrow x = 4$

$-3 =$ $\frac{y+0}{2}$ $\Rightarrow y = -6$

Hence $A(4,0)$ and $B(0, -6)$

(ii) Slope of $AB =$ $\frac{-6-0}{0-4}$ $=$ $\frac{-6}{-4}$ $=$ $\frac{3}{2}$

(iii) Equation of $AB$

$y = 0 =$ $\frac{3}{2}$ $(x-4)$

$2y = 3x-12$

$\\$

Question 7: The equation of a line is $3x + 4y - 7 = 0$ . Find:

(i) Slope of the line.

(ii) The equation of a line perpendicular to the given line and passing through the intersection of the lines $x -y + 2 = 0$ and $3x + y- 10 = 0$ [2010]

Answer:

(i) Slope $= -$ $\frac{3}{4}$

(ii) Slope of perpendicular $=$ $\frac{4}{3}$

For point of intersection solve $x -y + 2 = 0$ and $3x + y- 10 = 0$

$y = 4$ and $x = 2$

Therefore intersection $= (2, 4)$

Therefore equation of line

$y-4 =$ $\frac{4}{3}$ $(x-2)$

$3y-12 = 4x-8$

$3y = 4x+4$

$\\$

Question 8: $ABCD$ is a parallelogram where $A (x, y), B (5, 8), C (4, 7) \ and \ D (2, -4)$ . Find:

(i) Co-ordinates of $A$

(ii) Equation of diagonal $BD$ [2011]

Answer:

(i) Mid point of $BD = ($ $\frac{5+2}{2}$ $,$ $\frac{-4+8}{2}$ $)= ($ $\frac{7}{2}$ $, 2)$

Therefore we have $A(x, 0), O($ $\frac{7}{2}$ $, 2)$ and $C(4, 7)$

$O$ is the mid point of $AC$ as well (diagonals of a parallelogram bisect each other)

Hence $\frac{x+4}{2}$ $=$ $\frac{7}{2}$ $\Rightarrow x = 3$

and $\frac{y+7}{2}$ $= 2 \Rightarrow y = -3$

Hence $A ( 3, -3)$

(ii) Equation of $BD$

$y - 8 =$ $\frac{-4-8}{2-5}$ $(x-5)$

$y-8 = 4(x-5)$

$y - 8 = 4x - 20$

$y + 12 = 4x$

$\\$

Question 9: From the given figure, find:

(i) The co-ordinates of $A, B, \ and \ C$ .

(ii) The equation of the line through $A$ and parallel to $BC$ . [2004]

Answer:

Slope of $BC =$ $\frac{0-2}{3-(-1)}$ $=$ $\frac{-2}{4}$ $= -$ $\frac{1}{2}$

The equation of line parallel to $BC$ and passing through $A(2,3)$

$y-3 = -$ $\frac{1}{2}$ $(x-2)$

$2y-6 = -x+2$

$2y+x=8$

$\\$

Question 10: $P (3, 4), Q (7, -2) \ and \ R (-2, -1)$ are the vertices of triangle $PQR$ . Write down the equation of the median of the triangle through $R$ . [2005]

Answer:

Mid point of $PQ =$ $($ $\frac{3+7}{2}$ $,$ $\frac{4-2}{2}$ $)$ $= (5,1)$

Therefore equation passing through $(5,1)$ and $P(-2,-1)$ is

$y-1 =$ $\frac{-1-1}{-2-5}$ $(x-5)$

$y-1 =$ $\frac{2}{7}$ $(x-5)$

$7y-7 = 2x-10$

$7y-2x+3=0$

$\\$

Question 11: Find the value of $k$ for which the lines $kx - 5y + 4 = 0$ and $5x - 2y + 5 = 0$ are perpendicular to each other. [2003]

Answer:

Slope of $kx - 5y + 4 = 0$ $\Rightarrow m_1 =$ $\frac{k}{5}$

Slope of $5x - 2y + 5 = 0$ $\Rightarrow m_2 =$ $\frac{5}{2}$

Since the two lines are perpendicular, $m_1. m_2 = -1$

$\Rightarrow$ $\frac{k}{5}$ $.$ $\frac{5}{2}$ $= -1$

$\Rightarrow k = -2$

$\\$

Question 12: A straight line passes through the points $P (-1, 4) \ and \ Q (5, -2)$ . It intersects the co-ordinate axes at points $A \ and \ B$ . $M$ is the mid-point of the line segment $AB$ . Find:

The equation of line

The co-ordinates of $A \ and \ B$

The co-ordinates of $M$ [2003]

Answer:

$P (-1, 4) \ and \ Q (5, -2)$

The equation of the line:

$y - 4 =$ $\frac{-2-4}{5-(-1)}$ $(x+1)$

$\Rightarrow y-4 = -1 (x+1)$

$\Rightarrow y+x=3$

The $x-intercept A = (3,0)$ and the $y-intercept B = (0,3)$

The coordinate of $M$ $= ($ $\frac{3+0}{2}$ $,$ $\frac{0+3}{2}$ $) = ($ $\frac{3}{2}$ $,$ $\frac{3}{2}$ $)$

$\\$

Question 13: If the lines $y = 3x + 7 \ and \ 2y + px = 3$ are perpendicular to each other, find the value of $p$ . [2006]

Answer:

Given equation is $y = 3x + 7$

$\Rightarrow Slope (m_1) = 3$

Given equation is $2y + px = 3$

$\Rightarrow 2y=-px+3$

$\Rightarrow y =$ $\frac{-p}{2}$ $x+$ $\frac{3}{2}$

$\Rightarrow Slope (m_2) =$ $\frac{-p}{2}$

Since they are perpendicular, $m_1 . m_2 =-1$

$\Rightarrow 3.$ $\frac{-p}{2}$ $=-1$

$\Rightarrow p=$ $\frac{2}{3}$

$\\$

Question 14: The line through $A (-2, 3) \ and \ B (4, b)$ is perpendicular to the line $2x - 4y = 5$ . Find the value of $b$ [2012]

Answer:

Slope of $AB =$ $\frac{b-3}{4-(-2)}$ $=$ $\frac{b-3}{6}$

Given equation is $2x - 4y = 5$

$\Rightarrow 4y=2x-5$

$\Rightarrow y =$ $\frac{1}{2}$ $x-$ $\frac{5}{4}$

$\Rightarrow Slope (m_2) =$ $\frac{1}{2}$

Since they are perpendicular, $m_1 . m_2 =-1$

$\Rightarrow$ $\frac{b-3}{6}$ $.$ $\frac{1}{2}$ $=-1$

$\Rightarrow p=-9$

$\\$

Question 15: i) Find the equation of the line passing through $(5, -3)$ and parallel to $x - 3y = 4$ .

ii) Find the equation of the line parallel to the line $3x + 2y = 8$ and passing through the point $(0, 1)$ [2007]

Answer:

i) Given Point $(x_1, y_1)=(5,-3)$

Given equation is $x - 3y = 4$

$\Rightarrow 3y=x-4$

$\Rightarrow y =$ $\frac{1}{3}$ $x-4$

$\Rightarrow Slope (m) =$ $\frac{1}{3}$

Equation of a line with slope $m$ and passing through $(x_1, y_1)$ is

$y-y_1=m(x-x_1)$

$\Rightarrow y-(-3)=$ $\frac{1}{3}$ $(x-5)$

$\Rightarrow 3y+9=x-5$

$\Rightarrow x-3y-14=0$

ii) Given Point $(x_1, y_1)=(0,1)$

Given equation is $3x+2y=8$

$\Rightarrow 2y=-3x+8$

$\Rightarrow y =$ $\frac{-3}{2}$ $x+4$

$\Rightarrow Slope (m) =$ $\frac{-3}{2}$

Equation of a line with slope $m$ and passing through $(x_1, y_1)$ is

$y-y_1=m(x-x_1)$

$\Rightarrow y-1=$ $\frac{-3}{2}$ $(x-0)$

$\Rightarrow 2y-2=-3x$

$\Rightarrow 2y+3x=2$

$\\$

Question 16: i) Write down the equation of the line $AB$ , through $(3, 2)$ and perpendicular to the line $2y = 3x + 5$ .

ii) $AB$ meets the $x-axis \ at \ A$ and the $y-axis$ at $B$ . write down the co-ordinates of $A \ and \ B$ . Calculate the area of triangle $OAB$ , where $O$ is origin. [1995]

Answer:

i) Given Point $(x_1, y_1)=(3,2)$

Given equation is $2y=3x+5$

$\Rightarrow y =$ $\frac{3}{2}$ $x+5$

$\Rightarrow Slope (m) =$ $\frac{3}{2}$

Therefore slope of the new line $= m =$ $\frac{-1}{\frac{3}{2}}$ $= -\frac{2}{3}$

Equation of a line with slope $m$ and passing through $(x_1, y_1)$ is $y-y_1=m(x-x_1)$

$\Rightarrow y-2= -$ $\frac{2}{3}$ $(x-(-2))$

$\Rightarrow 3y-6=-2x+6$

$\Rightarrow 3y+2x=12$

ii) Equation of $AB$ is $\Rightarrow 3y+2x=12$

When $y = 0, x = 6$ . Therefore $A (6,0)$

When $c = 0, y =4$ . Therefore $B(0,4)$

Area of the triangle $=$ $\frac{1}{2}$ $\times 6 \times 4 = 12$ sq. units.

$\\$

Question 17: Find the value of a for the points $A (a, 3), B (2, 1) \ and \ C (5, a)$ are collinear. Hence, find the equation of the line. [2014]

Answer:

Given points $A (a, 3), B (2, 1) \ and \ C (5, a)$

Slope of $AB$ $=$ $\frac{1-3}{2-a}$ $=$ $\frac{-2}{2-a}$

Slope of $BC$ $=$ $\frac{a-1}{5-2}$ $=$ $\frac{a-1}{3}$

Because $A, B, \ and \ C$ are collinear:

$\frac{-2}{2-a}$ $=$ $\frac{a-1}{3}$

$-6=(2-a)(a-1)$

$-6=2a-2-a^2+a$

$-6=3a-a^2-2$

$a^2-3a-4=0$

$(a-4)(a+1)=0 \Rightarrow a=4, \ or \ -1$

$\\$

Question 18: In, $A = (3, 5), B = (7, 8) \ and \ C = (1, -10)$ . Find the equation of the median through $A$ . [2013]

Answer:

$A = (3, 5), B = (7, 8) \ and \ C = (1, -10)$

Let $D$ be the mid point of $BC$ . Therefore the coordinates of $D$ are

$D =($ $\frac{1+7}{2}$ $,$ $\frac{-10+8}{2}$ $)=(4, -1)$

Slope of $AD = m =$ $\frac{-1-5}{4-3}$ $=$ $\frac{-6}{1}$ $=-6$

Equation of $AD:$

$y-(-1)=-6(x-4)$

$\Rightarrow y+1=-6x+24$

$\Rightarrow y+6x=23$

$\\$

Question 19: The line through $P (5, 3)$ intersects $y-axis \ at \ Q$ .

i) Write the slope of the line.

ii) Write the equation of the line.

iii) Find the co-ordinates of $Q$ [2012]

Answer:

Given points $P(-2, 0) \ and \ Q(0, y)$

Slope $= m = tan \ 45^o = 1$

Equation of line:

$y-3=1(x-5)$

$\Rightarrow y = x-2$

When $x=0, y = -2$

Hence the co-ordinates of $Q = (0, -2)$

$\\$

Question 20: $A (1, 4), B (3, 2) \ and \ C (7, 5)$ are vertices of a triangle $ABC$ . Find:

i) The co-ordinates of the centroid of a triangle $ABC$ .

ii) The equation of a line through the centroid and parallel to $AB$ . [2002]

Answer:

Let $O$ be the centroid. Therefore the coordinates of $O$ are:

$O=($ $\frac{1+3+7}{3}$ $,$ $\frac{4+2+5}{3}$ $)=($ $\frac{11}{3}$ $,$ $\frac{11}{3}$ $)$

Slope $m=$ $\frac{2-4}{3-1}$ $=$ $\frac{-2}{2}$ $= -1$

Therefore the equation of a line parallel to $AB$ will pass through $($ $\frac{11}{3}$ $,$ $\frac{11}{3}$ $)$

Equation of the line:

$y-$ $\frac{11}{3}$ $=-1(x-$ $\frac{11}{3}$ $)$

$\Rightarrow 3y-11=-(3x-11)$

$\Rightarrow 3y+3x=22$

ICSE / ISC / CBSE Mathematics Portal for K12 Students

Mathematics Made Easy for School Students

Class 10: Equation of a Line – ICSE Board Problems

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