Class 10: Equation of a Line – Exercise 14(e) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Point $P$ divides the line segment joining the points $A (8, 0) \ and \ B (16, -8)$ in the ratio $3 : 5$ . Find the co-ordinates of point $P$ . Also, find the equation of the line through $P$ and parallel to $3x + 5y = 7$ .

Answer:

Given $P$ divides $A (8, 0) \ and \ B (16, -8)$ in the ratio $3 : 5$

Ratio: $m_1:m_2 = 3 : 5$

Let the coordinates of the point $P \ be \ (x, y)$ . Therefore

$x =$ $\frac{3 \times 16+5 \times 8}{3+5}$ $=$ $\frac{88}{8}$ $= 11$

$y =$ $\frac{3 \times (-8)+5 \times 0}{3+5}$ $=$ $\frac{-24}{8}$ $= -3$

Therefore $P = (11, -3)$

Equation of the line given : $3x+5y=7 \Rightarrow slope = m =$ $\frac{-3}{5}$

Therefore the required equation is

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

$y -(-3) =$ $\frac{-3}{5}$ $(x-11)$ or $5y +3x=18$

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Question 2: The line segment joining the points $A (3, -4) \ and \ B (-2, 1)$ is divided in the ratio $1 : 3$ at point $P$ in it. Find the co-ordinates of $P$ . Also, find the equation of the line through $P$ and perpendicular to the line $5x - 3y = 4$ .

Answer:

Given $P$ divides $A (3,-4) \ and \ B (-2,1)$ in the ratio $1:3$

Ratio: $m_1:m_2 = 1:3$

Let the coordinates of the point $P \ be \ (x, y)$ . Therefore

$x =$ $\frac{1 \times (-2)+3 \times 3}{1+3}$ $=$ $\frac{7}{4}$

$y =$ $\frac{1 \times 1+3 \times (-4)}{1+3}$ $=$ $\frac{-11}{4}$

Therefore $P = ($ $\frac{7}{4}$ $,$ $\frac{-11}{4}$ $)$

Equation of the line given : $5x - 3y = 4 \Rightarrow slope = m_1 =$ $\frac{5}{3}$

Therefore the slope of the line perpendicular to the above line $=$ $\frac{-3}{5}$

Therefore the required equation is

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

$y -$ $\frac{-11}{4}$ $=$ $\frac{-3}{5}$ $(x-$ $\frac{7}{4}$ $)$ or $10y +6x+17 = 0$

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Question 3: A line $5x + 3y + 15 = 0$ meets $y-axis$ at point $P$ . Find the co-ordinates of point $P$ . Find the equation of a line through $P$ and perpendicular to $x - 3y + 4 = 0$ .

Answer:

At $y-axis, \ x = 0 \Rightarrow y = -5$

Therefore the coordinate of $P = (0, -5)$

$x - 3y + 4 = 0$ $\Rightarrow slope =$ $\frac{1}{3}$

Therefore the slope of a like perpendicular to this line $= -3$

Hence the line passing through $(0, -5)$ with a slope of $-3$ is

$y-(-5) = -3(x-0) \Rightarrow y+3x+5 = 0$

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Question 4: 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 5: 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 $M = ($ $\frac{3+0}{2}$ $,$ $\frac{0+3}{2}$ $) = ($ $\frac{3}{2}$ $,$ $\frac{3}{2}$ $)$

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Question 6: $(1, 5) \ and \ (-3, -1)$ are the co-ordinates of vertices $A \ and \ C$ respectively of rhombus $ABCD$ . Find the equations of the diagonals $AC \ and \ BD$ .

Answer:

Midpoint $M = ($ $\frac{-3+1}{2}$ $,$ $\frac{-1+5}{2}$ $) = (-1, 2)$

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

Equation of $AC$ :

$y - 5 =$ $\frac{3}{2}$ $(x-1) \Rightarrow 3x-2y+7=0$

Slope of $BD = -$ $\frac{2}{3}$

Therefore equation of $BD$ :

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

$\\$

Question 7: Show that $A (3, 2), B (6, -2) \ and \ C (2, -5)$ can be vertices of a square. Find the co-ordinates of its fourth vertex $D$ , if $ABCD$ is a square. Without using the co-ordinates of vertex $D$ , find the equation of side $AD$ of the square and the equation of diagonal $BD$ .

Answer:

Mid point of $AC =$ $($ $\frac{3+2}{2}$ $,$ $\frac{-5+2}{2}$ $) = ($ $\frac{5}{2}$ $,$ $\frac{-3}{2}$ $)$

Let the coordinate of $D$ be $(x, y)$

In a square, the diagonals bisect each other. Therefore

$\frac{x+6}{2}$ $=$ $\frac{5}{2}$ $\Rightarrow x = -1$

$\frac{y-2}{2}$ $=$ $\frac{-3}{2}$ $\Rightarrow y = -1$

Hence $D$ is $(-1, -1)$

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

Since $AB \perp AD$ , slope of $AD =$ $\frac{3}{4}$

Hence the equation of $AD$ :

$y - 2 =$ $\frac{3}{4}$ $(x-3) \Rightarrow 4y-3x+1 = 0$

Slope of $BD =$ $\frac{-2+1}{6+1}$ $= -$ $\frac{1}{7}$

Hence the equation of $BD$ :

$y +2 = -$ $\frac{1}{7}$ $(x-6) \Rightarrow 7y+x+8=0$

$\\$

Question 8: A line through origin meets the line $x = 3y + 2$ at right angles at point $X$ . find the co-ordinates of point $X$ .

Answer:

Given $x = 3y + 2$ … … … … (i)

Slope of line $x = 3y + 2$ is $\frac{1}{3}$

Slope of perpendicular $= -3$

The equation of a line passing through $(0,0)$ and having slope $3$ is

$y - 0 = -3 (x-0) \Rightarrow y = -3x$ … … … … (i)

Solving equations (i) and (ii)

$x= 3 (-3x) + 2 \Rightarrow x =$ $\frac{1}{5}$

Hence $y = -$ $\frac{3}{5}$

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Question 9: A straight line passes through the point $(3, 2)$ and the portion of this line, intercepted between the positive axes, is bisected at this point. Find the equation of the line.

Answer:

Let y-intercept be $(0, y)$ and x-intercept be $(x, 0)$

Given $(3,2)$ is the mid point of $(0,y)$ and $(x,0)$ . Therefore:

$y-intercept = (0, 4)$

$x-intercept = (6, 0)$

Slope of line $=$ $\frac{0-4}{6-0}$ $= -$ $\frac{2}{3}$

Equation of line:

$y-0 = -$ $\frac{2}{3}$ $(x-6) \Rightarrow 3y+2x=12$

$\\$

Question 10: Find the equation of the line passing through the point of intersection of $7x + 6y = 71 \ and \ 5x - 8y = -23$ ; and perpendicular to the line $4x -2y = 1$ .

Answer:

Solve equations

$7x + 6y = 71$ … … … … (i)

$5x - 8y = -23$ … … … … (ii)

Multiply (i) by 4 and (ii) by 3 and then add the equations, we get

$43x = 215 \Rightarrow x = 5$

Substituting $x = 5$ in (i) we get

$7(5)+6y=71 \Rightarrow y = 6$

Therefore the intercept is $(5, 6)$

Sloe of line $4x -2y = 1$ is $2$

Therefore the slope of perpendicular $= -$ $\frac{1}{2}$

Hence the equation of the perpendicular:

$y - 6 = -$ $\frac{1}{2}$ $(x-5)$

$2y -12 = -x+5$

$2y+x=17$

$\\$

Question 11: Find the equation of the line which is perpendicular to the line $\frac{x}{a}$ $-$ $\frac{y}{b}$ $=1$ at the point where this line meets $y-axis$ .

Answer:

Slope of line $\frac{x}{a}$ $-$ $\frac{y}{b}$ $=1$ is $\frac{b}{a}$

Therefore slope of line perpendicular to given line $= -$ $\frac{a}{b}$

$y-intercept = b$

Therefore the equation of line passing through (0,b) and having slope of $-$ $\frac{a}{b}$ is:

$y - b =$ $-$ $\frac{a}{b}$ $(x-0)$

$by-b^2 = -ax$

$ax+by=b^2$

$\\$

Question 12: $O (0, 0), A (3, 5) \ and \ B (-5, -3)$ are the vertices of a triangle $OAB$ . Find:

(i) The equation of the median of triangle $OAB$ through vertex $O$

(ii) The equation of altitude of triangle $OAB$ through vertex $B$

Answer:

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

Therefore the equation of median of $\triangle OAB$ through $O$ is

(i) $y - 0 =$ $\frac{1-0}{-1-0}$ $(x-0) \Rightarrow y = -x \ or \ y+x = 0$

(ii) Slope of $OA =$ $\frac{5-0}{3-0}$ $=$ $\frac{5}{3}$

Slope of line perpendicular to $OA = -$ $\frac{3}{5}$

Therefore the equation of altitude of $\triangle OAB$ through $B$

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

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

$5y+15 = -3x-15$

$5y+3x+30 = 0$

$\\$

Question 13: Determine whether the line through points $(-2, 3) \ and \ (4, 1)$ is perpendicular to the line $3x = y + 1$ . Does line $3x = y +1$ bisect the line segment joining the two given points?

Answer:

Slope of line passing through $(-2,3)$ and $(4,1) =$ $\frac{1-3}{4-(-2)}$ $= -$ $\frac{1}{3}$

Slope of $3x = y + 1$ is $3$

Slope of perpendicular $= -$ $\frac{1}{3}$

Therefore line passing through $(-2, 3)$ and $(4, 1)$ is perpendicular to $3x = y + 1$

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

Substituting $(1, 2)$ in $3x=y+1$ we get that it satisfies the equation. Therefore $3x=y+1$ bisects the line joining $(-2, 3)$ and $(4, 1)$

$\\$

Question 14: Given a straight line $x \cos 30^o+y \sin 30^o=2$ . Determine the equation of the other line which is parallel to its and passes through $(4, 3)$ .

Answer:

Given $x \cos 30^o + y \sin 30^o = 2$

Slope of this line $=$ $-$ $\frac{ \cos 30^o}{ \sin 30^o}$ $= -\sqrt{3}$

Equation of line with slope $-\sqrt{3}$ and passing through $(4, 3)$ is

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

$y+\sqrt{3} x= 4 \sqrt{3}+3$

$\\$

Question 15: Find the value of $k$ such that the line $(k-2)x+(k+3)y-5=0$ is:

(i) Perpendicular to the line $2x - y + 7 = 0$ (ii) Parallel to it.

Answer:

Given $(k-2)x+(k+3)y-5=0$

Slope of this line $= -$ $\frac{k-2}{k+3}$

Slope of line $2x-y+7 = 0$ is $2$

Slope of line perpendicular to this line $= -$ $\frac{1}{2}$

(i) If perpendicular

$-$ $\frac{1}{2}$ $= -$ $\frac{k-2}{k+3}$

$k+3 = 2k-4$

$k= 7$

(ii) If parallel

$2 = -$ $\frac{k-2}{k+3}$

$2k+6 = -k +2$

$3k = -4 \Rightarrow k = -$ $\frac{4}{3}$

$\\$

Question 16: The vertices of a triangle $ABC$ are $A (0, 5), B (-1, -2) \ and \ C (11, 7)$ . Write down the equation of $BC$ . Find:

(i) The equation of the line through $A$ and perpendicular to $BC$ .

(ii) The co-ordinates of the point $P$ , where the perpendicular through $A$ , as obtained in (i.), meets $BC$ .

Answer:

(i) Slope of $BC =$ $\frac{7-(-2)}{11-(-1)}$ $=$ $\frac{9}{4}$ $=$ $\frac{3}{4}$

Slope of line perpendicualr to $BC = -$ $\frac{4}{3}$

Therefore equation of line passing through $A (0,5)$ with slope $-$ $\frac{4}{3}$ is:

$y-5 = -$ $\frac{4}{3}$ $(x-0)$

$3y-15 = -4x$

$3y+4x=15$ … … … … (i)

(ii) Equation of $BC$

$y-7 =$ $\frac{3}{4}$ $(x-11)$

$4y - 28 = 3x - 33$

$4y-3x= -5$ … … … … (ii)

Solving (i) and (ii) we get $x = 3$ and $y = 1$ .

Therefore $P$ is $(3, 1)$

$\\$

Question 17: 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 18: $P (3, 4), Q (7, -2) \ and \ R (-2, -1)$ 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 19: $A ((8, -6), B (-4, 2) \ and \ C (0, -10)$ are vertices of a triangle $ABC$ . If $P$ is the mid-point of $AC$ , use co-ordinate geometry to show that $PQ$ is parallel to $BC$ . Give a special name to quadrilateral $PBCQ$ .

Answer:

Coordinates of $P =$ $($ $\frac{-4+8}{2}$ $,$ $\frac{2-6}{2}$ $)$ $= (2, -2)$

Coordinates of $Q =$ $($ $\frac{8+0}{2}$ $,$ $\frac{-6-10}{2}$ $)$ $= (4,-8)$

Slope of $PQ =$ $\frac{-8-(-2)}{4-2}$ $=$ $\frac{-6}{2}$ $= -3$

Slope of $BC =$ $\frac{-10-2}{0-(-4)}$ $=$ $\frac{-12}{4}$ $= -3$

Therefore $PB \parallel BC$ .

$PQBC$ is a trapezoid.

$\\$

Question 20: A line $AB$ meets the $x-axis$ at point $A$ and $y-axis$ at point $B$ . The point $P (-4, -2)$ divides the line segment $AB$ internally such that $AP : PB = 1 : 2$ . Find:

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

(ii) Equation of the line through $P$ and perpendicular to $AB$ .

Answer:

$AP:PB = 1:2$

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

Therefore $-4 =$ $\frac{1 \times x+2 \times 0}{3}$ $\Rightarrow x = -6$

Similarly, $-2 =$ $\frac{1 \times 0+2 \times y}{3}$ $\Rightarrow y= -3$

Therefore $B(-6, 0)$ and $A (0, -3)$

(ii) Slope of $AB =$ $\frac{-3-0}{0-(-6)}$ $=$ $\frac{-3}{6}$ $=$ $\frac{-1}{2}$

Slope of line perpendicular to $AB = 2$

Therefore the equation of line passing through $P(-4, -2)$ with slope $2$ :

$y - (-2) = 2(x-(-4))$

$y+2 = 2(x+4)$

$y = 2x+6$

$\\$

Question 21: 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 22: 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 23: 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 24: 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 25: The ordinate of a point lying on the line joining points $(6, 4) \ and \ (7, -5) \ is \ -23$ . Find the co-ordinates of that point.

Answer:

Let the ordinate of a point lying on the line joining points $(6, 4) \ and \ (7, -5) \ is \ -23$ be $(x, -23)$

Equation of line passing through $(6,4)$ and $(7, -5)$

$y - 4 =$ $\frac{-5-4}{7-6}$ $(x-6)$

$y-4 = -9(x-6)$

$y+9x=58$

Therefore if $y = -23$ , then

$-23+9x=58 \Rightarrow x = 9$

Therefore the point is $(9, -23)$

$\\$

Question 26: 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 27: $A \ and \ B$ 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 28: 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 29: $ABCD$ 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]