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

Reference Material for Preparation

Question 1: Find the equation of a line whose: $y-intercept = 2 \ and \ slope = 3$ ,

Answer:

$y-intercept = 2 \ and \ slope = 3$

Since $y-intercept=2$ , the corresponding point on $y-axis = (0,2)$

Given Slope $m = 3$

Therefore $m = 3, (x_1, y_1)=(0,2)$

Required equation of the line: $(y-y_1)=m(x-x_1)$

$y-2=3(x-0)$

$\Rightarrow 3x-y+2=0$

$\Rightarrow y = 3x+2$

$\\$

Question 2: Find the equation of a line whose: $y-intercept = -1 \ and \ inclination = 45^o$

Answer:

$y-intercept = -1 \ and \ slope = tan \ 45^o = 1$

Since $y-intercept=-1$ , the corresponding point on $y-axis = (0,-1)$

Therefore $m = 1, (x_1, y_1)=(0,-1)$

Required equation of the line: $(y-y_1)=m(x-x_1)$

$y-(-1)=1(x-0)$

$\Rightarrow y+1=x$

$\Rightarrow x-y=1$

$\\$

Question 3: Find the equation of the line whose slope is $-\frac{4}{3}$ and which passes through $(-3, 4)$ .

Answer:

$slope = m =-\frac{4}{3}$

Therefore $m =-\frac{4}{3}, (x_1, y_1)=(-3,4)$

Required equation of the line: $(y-y_1)=m(x-x_1)$

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

$\Rightarrow 3y-12=-4x-12$

$\Rightarrow 4x+3y=0$

$\\$

Question 4: Find the equation of the line passing through $(5, 4)$ and makes an angle of $60^o$ with the positive direction of $x-axis$ .

Answer:

$slope = m =tan \ 60^o = \sqrt{3}$

Therefore $m =\sqrt{3}, (x_1, y_1)=(5,4)$

Required equation of the line: $(y-y_1)=m(x-x_1)$

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

$\Rightarrow \sqrt{3}x-y=5\sqrt{3}-4$

$\Rightarrow y=\sqrt{3}x+4-5\sqrt{3}$

$\\$

Question 5: Find the equation of the line passing through:

i) $(0, 1) \ and \ (1, 2)$

ii) $(-1, -4) \ and \ (3, 0)$

Answer:

i) $(0, 1) \ and \ (1, 2)$

Slope $= m = \frac{2-1}{1-0} = \frac{1}{1} = 1$

Required equation of the line: $(y-y_1)=m(x-x_1)$

$\Rightarrow y-1=1(x-0)$

$\Rightarrow y-1=x$

$\Rightarrow y=x+1$

ii) $(-1, -4) \ and \ (3, 0)$

Slope $= m = \frac{0-(4)}{3-(-1)} = \frac{4}{4} = 1$

Required equation of the line: $(y-y_1)=m(x-x_1)$

$\Rightarrow y-0=1(x-3)$

$\Rightarrow y=x-3$

$\\$

Question 6: The co-ordinates of two points $P \ and \ Q$ are $(2, 6) \ and \ (-3, 5)$ respectively. Find:

i) The gradient of $PQ$ ;

ii) The equation of $PQ$

iii) The co-ordinates of the point where $PQ$ intersects the $x-axis$ .

Answer:

$P \ and \ Q$ are $(2, 6) \ and \ (-3, 5)$ respectively

Slope or PQ $= m = \frac{5-6}{-3-2} = \frac{-1}{-5} = \frac{1}{5}$

Required equation of the line: $(y-y_1)=m(x-x_1)$

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

$\Rightarrow 5y-25=x+3$

$\Rightarrow 5y-3=28$

$\\$

Question 7: The co-ordinates of two points $A \ and \ B$ are $(-3, 4) \ and \ (2, -1)$ . Find:

i) The equation of $AB$ ;

ii) The co-ordinates of the point where the line $AB$ intersect the $y-axis$ .

Answer:

$A \ and \ B$ are $(-3, 4) \ and \ (2, -1)$

Slope or AB $= m = \frac{-1-4}{2-(-3)} = \frac{-5}{5} = -1$

Required equation of the line: $(y-y_1)=m(x-x_1)$

$y-(-1)= -1(x-2)$

$\Rightarrow y+1=-x+2$

$\Rightarrow y+x=1$

$x-intercept \Rightarrow y =0$

Therefore when $y =0, x = 1$

Hence $x-intercept = (1,0)$

$\\$

Question 8: The figure given alongside shows two straight lines $AB \ and \ CD$ intersecting each other at point $P (3, 4)$ . Find the equations of $AB \ and \ CD$ .

Answer:

$P (3, 4)$

Slope of $AB=m_1= tan \ 45^o = 1$

Slope of $DC=m_2= tan \ 60^o = \sqrt{3}$

Equation of $AB:$

$y-4=1(x-3)$

$\Rightarrow y = x+1$

Equation of $DC:$

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

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

$\\$

Question 9: 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 10: The following figure shows a parallelogram $ABCD$ whose side $AB$ is parallel to the $x-axis$ , and vertex $C = (7, 5)$ . Find the equations of $BC \ and \ CD$ .

Answer:

$AB$ is parallel to $x-axis$

Slope of $BC=m_1= tan \ 60^o = \sqrt{3}$

Equation of $BC:$

$y-5=\sqrt{3}(x-7)$

$\Rightarrow y =\sqrt{3}+5-7\sqrt{3}$

Slope of $DC=m_2= tan \ 0^o = 0$

Equation of $DC:$

$y-5=0(x-7)$

$\Rightarrow y=5$

$\\$

Question 11: Find the equation of the straight line passing through origin and the point of intersection of the lines $x + 2y = 7$ and $x-y=4$

Answer:

Solving $x + 2y = 7$ and $x-y=4$ we get $x = 5, y = 1$

Hence point of intersection $= (5, 1)$

Slope of $m= tan \ 0^o = 0$

Slope of $= m = \frac{1-0}{5-0} =\frac{1}{5}$

$y-0=\frac{1}{5}(x-0)$

$\Rightarrow x-5y=0$

$\\$

Question 12: In triangle $ABC$ , the co-ordinates of vertices $A, B \ and \ C$ are $(4, 7), (-2, 3) \ and \ (0, 1)$ respectively. Find the equation of median through vertex $A$ . Also, find the equation of the line through vertex $B$ and parallel to $AC$ .

Answer:

$A = (4, 7), B = (-2, 3) \ and \ C = (0, 1)$

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

$D =(\frac{-2+0}{2}, \frac{3+1}{2})=(-1,2)$

Slope of $AD = m = \frac{2-7}{-1-4} =\frac{-5}{-5}=1$

Equation of $AD:$

$y-7=1(x-4)$

$\Rightarrow y-x=3$

Slope of $AC = m = \frac{7-1}{4-0} =\frac{6}{4}=\frac{3}{2}$

Equation of the line through vertex $B$ and parallel to $AC$

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

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

$\Rightarrow 2y-3x=12$

$\\$

Question 13: $A, B \ and \ C$ have co-ordinates $(0, 3), (4, 4) \ and \ (8, 0)$ respectively. Find the equation of the line through $A$ and perpendicular to $BC$ .

Answer:

$A(0, 3), B(4, 4) \ and \ C(8, 0)$

Slope of $BC = m = \frac{0-4}{8-4} =\frac{4}{-4}=-1$

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

Therefore the equation of line perpendicular to BC and passing through A:

$y-3=1(x-0)$

$\Rightarrow y=x+3$

$\\$

Question 14: Find he equation of the perpendicular dropped from the point $(-1, 2)$ onto the line joining the points $(1, 4) \ and \ (2, 3)$ .

Answer:

$A(1, 4), B(2,3) \ \& C(-1,2)$

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

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

Therefore the equation of line perpendicular to AB and passing through C:

$y-2=1(x+1)$

$\Rightarrow y=x+3$

$\\$

Question 15: Find the equation of the line, whose:

i) $x-intercept = 5 \ and \ y-intercept = 3$

ii) $x-intercept = - 4 \ and \ y-intercept = 6$

iii) $x-intercept = - 8 \ and \ y-intercept = - 4$

Answer:

i) Points given are $(5,0) \ and \ (0,3)$

Slope $= \frac{3-0}{0-5}= -\frac{3}{5}$

Equation of line:

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

$\Rightarrow 5y-15=-3x$

$\Rightarrow 5y+3x=15$

ii) Points given are $(-4,0) \ and \ (0,6)$

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

Equation of line:

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

$\Rightarrow 2y-12=3x$

$\Rightarrow 2y=3x+12$

iii) Points given are $(-8,0) \ and \ (0,-4)$

Slope $m= \frac{-4-0}{0-(-8)}= \frac{-4}{8} = \frac{-1}{2}$

Equation of line:

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

$\Rightarrow 2y+8=-x$

$\Rightarrow x+2y+8=0$

$\\$

Question 16: Find the equation of line whose slope is $6$ and $x-intercept \ is \ 6$ .

Answer:

Slope $= -\frac{5}{6}$ , Intercept $= (6,0)$

Equation of the line:

$y-0=-\frac{5}{6}(x-6)$

$\Rightarrow 6y=-5x+30$

$\Rightarrow 6y+5x=30$

$\\$

Question 17: Find the equation of the line with $x - intercept \ 5$ and a point on it $(-3, 2)$

Answer:

Given point are $(5,0) \ and \ (-3,2)$

Slope $m= \frac{2-0}{-3-5}= \frac{-2}{8} = \frac{-1}{4}$

Equation of the line:

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

$\Rightarrow 4y-8=-x-3$

$\Rightarrow 4y+x=5$

$\\$

Question 18: Find the equation of the line through $(1, 3)$ and making an intercept of $5$ on the $y-axis$ .

Answer:

Given point are $(1,3) \ and \ (0,5)$

Slope $m= \frac{5-3}{0-1}= \frac{2}{-1} = -2$

Equation of the line:

$y-5=-2(x-0)$

$\Rightarrow y+2x=5$

$\\$

Question 19: Find the equations of the lines passing through point $(-2, 0)$ and equally inclined to the co-ordinate axes.

Answer:

Given $(-2, 0)$

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

Equation of line:

$y-0=1(x+2)$

$\Rightarrow y = x+2$

Also Slope $= m = tan \ 135^o = -1$

Equation of line:

$y-0=-1(x+2)$

$\Rightarrow y + x+2= 0$

$\\$

Question 20: 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 21: Write down the equation of the line whose gradient is -and which passes through point $P$ , where $P$ divides the line segment joining $A (4, -8) \ and \ B (12, 0)$ in the ratio $3:1$

Answer:

$m=-\frac{2}{5}$

Ratio: $m_1:m_2 = 3:1$

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

Therefore

$x = \frac{3 \times 12+1 \times 4}{1+2} =10$

$y = \frac{3\times 0+1 \times (-8)}{1+2} = -2$

Therefore $P = (10, -2)$

Equation of line:

$y-(-2)=-\frac{2}{5}(x-10)$

$\Rightarrow 5y+2x=10$

$\\$

Question 22: $A (1, 4), B (3, 2) \ and \ C (7, 5)$ 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:

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

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

$\Rightarrow 3y+3x=22$

$\\$

Question 23: $A (7, -1), B (4, 1) \ and \ C (-3, 4)$ are the vertices of a triangle $ABC$ . Find the equation of a line through the vertex $B$ and the point $P$ in $AC$ ; such that $AP : CP = 2 : 3$ .

Answer:

$A(7, -1) \ and \ (-3,4)$

Ratio: $m_1:m_2 = 2:3$

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

Therefore

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

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

Therefore $P = (3, 1)$

Slope $m= \frac{1-1}{3-4}= 0$

Equation of the line:

$y-1=0(x-3)$

$\Rightarrow y=1$

$\\$

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Class 10: Equation of a Line – Exercise 14(c)

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