Class 11: The Straight Line – Exercise 23.5 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Note: Equation of a like is also defined as $y - y_1 = m ( x-x_1)$

Question 1: Find the equation of the straight through the following pair of points:

i) $(0, 0)$ and $(2,-2)$ ii) $(a, b)$ and $(a + c \sin \alpha, b + c \cos \alpha)$

iii) $(0, -a)$ and $( b, 0)$ iv) $(a, b)$ and $(a + b , a -b)$

v) $( a t_1, a/t_1)$ and $( a t_2, a/t_2)$ vi) $(a \cos \alpha, a\sin \alpha)$ and $(a \cos \beta, a\sin \beta)$

Answer:

i) Given points $A(0, 0)$ and $B(2,-2)$

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

Substituting in $y - y_1 = m ( x-x_1)$ we get the equation of $AB$ as:

$y - 0 = -1(x-0) \hspace{0.5cm} \Rightarrow y + x = 0$

ii) Given points $A(a, b)$ and $B(a + c \sin \alpha, b + c \cos \alpha)$

Slope of $AB = m =$ $\frac{b+c \cos \alpha - b}{a+c \sin \alpha - a}$ $=$ $\frac{c \cos \alpha}{c \sin \alpha}$ $= \cot \alpha$

Substituting in $y - y_1 = m ( x-x_1)$ we get the equation of $AB$ as:

$y - b = \cot \alpha (x-a) \hspace{0.5cm} \Rightarrow y - x \cot \alpha + a - b = 0$

iii) Given points $A(0, -a)$ and $B( b, 0)$

Slope of $AB = m =$ $\frac{0-(-a)}{b-0}$ $=$ $\frac{a}{b}$

Substituting in $y - y_1 = m ( x-x_1)$ we get the equation of $AB$ as:

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

$\Rightarrow by + ab = ax$

$\Rightarrow ax-by-ab=0$

iv) Given points $A(a, b)$ and $B(a + b , a -b)$

Slope of $AB = m =$ $\frac{a-b-b}{a+b-a}$ $=$ $\frac{a-2b}{b}$

Substituting in $y - y_1 = m ( x-x_1)$ we get the equation of $AB$ as:

$y - b =$ $\frac{a-2b}{b}$ $(x-a)$

$\Rightarrow yb-b^2=(a-2b)x-a(a-2b)$

$\Rightarrow yb-b^2 = (a-2b)x - a^2+2ab$

$\Rightarrow (a-2b)x - yb + b^2 + 2ab - a^2=0$

v) Given points $A( a t_1, a/t_1)$ and $B( a t_2, a/t_2)$

Slope of $AB = m =$ $\frac{\frac{a}{t_2} - \frac{a}{t_1}}{at_2 - at_1}$ $=$ $\frac{t_1-t_2}{t_1t_2(t_2-t_1)}$ $= -$ $\frac{1}{t_1t_2}$

Substituting in $y - y_1 = m ( x-x_1)$ we get the equation of $AB$ as:

$y -$ $\frac{a}{t_1}$ $=$ $\frac{-1}{t_1t_2}$ $(x-at_1)$

$\Rightarrow t_1t_2y- at_2 = -x + at_1$

$\Rightarrow x + t_1t_2y-a(t_1+t_2)=0$

vi) Given points $A(a \cos \alpha, a\sin \alpha)$ and $B(a \cos \beta, a\sin \beta)$

Slope of $AB = m =$ $\frac{a \sin \beta - a \sin \alpha}{a \cos \beta - a \cos \alpha}$ $=$ $\frac{\sin \beta - \sin \alpha}{ \cos \beta - \cos \alpha}$

Substituting in $y - y_1 = m ( x-x_1)$ we get the equation of $AB$ as:

$y - a \sin \alpha =$ $\frac{\sin \beta - \sin \alpha}{ \cos \beta - \cos \alpha}$ $(x-a \cos \alpha)$

$\Rightarrow y ( \cos \beta - \cos \alpha) - a \sin \alpha ( \cos \beta - \cos \alpha) = x ( \sin \beta - \sin \alpha) - a \cos \alpha ( \sin \beta - \sin \alpha)$

$\Rightarrow y ( \cos \beta - \cos \alpha) - x ( \sin \beta - \sin \alpha) = a \sin \alpha ( \cos \beta - \cos \alpha) - a \cos \alpha ( \sin \beta - \sin \alpha)$

$\Rightarrow y . 2 \sin \Big(\frac{\beta + \alpha}{2} \Big) \sin \Big(\frac{ \alpha - \beta}{2} \Big) - x . 2 \sin \Big(\frac{\beta - \alpha}{2} \Big) \cos \Big(\frac{ \beta + \alpha }{2} \Big) = a ( \sin \alpha \cos \beta - \cos \alpha \sin \beta)$

$\Rightarrow y . 2 \sin \Big(\frac{\beta + \alpha}{2} \Big) \sin \Big(\frac{ \alpha - \beta}{2} \Big) - x . 2 \sin \Big(\frac{\beta - \alpha}{2} \Big) \cos \Big(\frac{ \beta + \alpha }{2} \Big) = 2a \sin \Big(\frac{\alpha-\beta}{2} \Big) \cos \Big(\frac{ \alpha - \beta}{2} \Big)$

$\Rightarrow y \sin \Big(\frac{\beta + \alpha}{2} \Big)+ x \cos \Big(\frac{ \beta + \alpha }{2} \Big) = a \cos \Big(\frac{ \alpha - \beta}{2} \Big)$

$\\$

Question 2: Find the equation to the sides of the triangles the coordinates of whose angular points are respectively: (i) $(1, 4),(2, -3)$ and $(-1, -2)$ ii) $(0, 1), (2, 0)$ and $( -1, -2)$

Answer:

i) Given points $A(1, 4), B(2, -3)$ and $C(-1, -2)$

Slope of $AB = m_1 =$ $\frac{-3-4}{2-1}$ $=$ $\frac{-7}{1}$ $= -7$

Therefore equation of $AB$ :

$y - 4 = - 7 ( x-1) \hspace{0.5cm} \Rightarrow 7x + y - 11 = 0$

Slope of $BC = m_2 =$ $\frac{-2-(-3)}{-1-2}$ $=$ $\frac{1}{-3}$ $=$ $\frac{-1}{3}$

Therefore equation of $BC$ :

$y - (-3) =$ $\frac{-1}{3}$ $( x-2) \hspace{0.5cm} \Rightarrow 3y + 9 = - x + 2 \hspace{0.5cm} \Rightarrow x + 3y + 7 = 0$

Slope of $CA = m_3 =$ $\frac{4-(-2)}{1-(-1)}$ $=$ $\frac{6}{2}$ $= 3$

Therefore equation of $CA$ :

$y - 4 = 3 ( x-1) \hspace{0.5cm} \Rightarrow 3x-y+1 = 0$

ii) Given points $A(0,1), B(2,0)$ and $C(-1, -2)$

Slope of $AB = m_1 =$ $\frac{0-1}{2-0}$ $=$ $\frac{-1}{2}$

Therefore equation of $AB$ :

$y - 1 = \frac{-1}{2} ( x-0) \hspace{0.5cm} \Rightarrow x+2y-2 = 0$

Slope of $BC = m_2 =$ $\frac{-2-0}{-1-2}$ $=$ $\frac{-2}{-3}$ $=$ $\frac{2}{3}$

Therefore equation of $BC$ :

$y - 0 =$ $\frac{2}{3}$ $( x-2) \hspace{0.5cm} \Rightarrow 2x-3y-4=0$

Slope of $CA = m_3 =$ $\frac{1-(-2)}{0-(-1)}$ $=$ $\frac{3}{1}$ $= 3$

Therefore equation of $CA$ :

$y - 1 = 3 ( x-0) \hspace{0.5cm} \Rightarrow 3x-y+1 = 0$

$\\$

Question 3: Find the equations of the medians of a triangle, the the coordinates of the vertices are $(-1, 6), (- 3, - 9)$ and $(5, -8)$ .

Answer:

Given points $A(-1,6), B(-3,-9)$ and $C(5,-8)$ . Please refer to the figure shown.

Let $D, E \ \& \ F$ be the mid point of $AB, BC$ and $CA$ respectively. Therefore,

$D = \Big($ $\frac{-3-1}{2}$ $,$ $\frac{-9+6}{2}$ $\Big ) = \Big( -2,$ $\frac{-3}{2}$ $\Big )$

$E = \Big($ $\frac{5-3}{2}$ $,$ $\frac{-8-9}{2}$ $\Big ) = \Big( 1,$ $\frac{-17}{2}$ $\Big )$

$F = \Big($ $\frac{-1+5}{2}$ $,$ $\frac{6-8}{2}$ $\Big ) = ( 2, -1)$

Slope of $AE = m_1 =$ $\frac{\frac{-17}{2}-6}{1-(-1)}$ $=$ $\frac{-29}{4}$

Therefore equation of $AE$ :

$y - 6 =$ $\frac{-29}{4}$ $( x-(-1)) \hspace{0.5cm} \Rightarrow 4y-24=-29x-29 \hspace{0.5cm} \Rightarrow 29x+4y+5 = 0$

Slope of $BF = m_2 =$ $\frac{-1-(-9)}{2-(-3)}$ $=$ $\frac{8}{5}$

Therefore equation of $BF$ :

$y - (-9) =$ $\frac{8}{5}$ $( x-(-3)) \hspace{0.5cm} \Rightarrow 5y+45 = 8x+ 24 \hspace{0.5cm} \Rightarrow 8x-5y-21=0$

Slope of $CD = m_3 =$ $\frac{\frac{-3}{2} - (-8)}{-2-5}$ $=$ $\frac{-13}{14}$

Therefore equation of $CD$ :

$y - (-8) =$ $\frac{-13}{14}$ $( x-5) \hspace{0.5cm} \Rightarrow 14y + 112 = - 13x +65 \hspace{0.5cm} \Rightarrow 13x + 14y + 47 = 0$

$\\$

Question 4: Find the equations to the diagonals of the rectangle the equations of whose sides are $x = a, x = a', y = b$ and $y = b'$ .

Answer:

Please refer to the figure shown. Therefore the four points are $A(a, b), B ( a', b), C( a', b'), D( a , b')$

Slope of $AC = m_3 =$ $\frac{b'-b}{a'-a}$

Therefore equation of $AC$ :

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

$\Rightarrow (a'-a)y-a'b+ab=(b'-b)x-ab'+ab$

$\Rightarrow (b'-b)x-(a'-a)y - ab'+a'b = 0$

$\Rightarrow (a'-a)y - ( b'-b)x = ba'-ab'$

Slope of $BD = m_3 =$ $\frac{b'-b}{a-a'}$

Therefore equation of $BD$ :

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

$\Rightarrow (a-a')y - ab + ba' = ( b'-b) x - a'b' + a'b$

$\Rightarrow (b-b')x - ( a - a') y + ab - a'b'=0$

$\Rightarrow ( a - a') y + (b-b')x = a'b' - ab$

$\\$

Question 5: Find the equation of the side $BC$ of the $\triangle ABC$ whose vertices are $A(- 1, - 2), B (0, 1)$ and $C (2,0)$ respectively. Also, find the equation of the median through $A ( -1, -2)$ .

Answer:

Given points $A(-1,-2), B(0,1)$ and $C(2,0)$

Slope of $BC = m_1 =$ $\frac{0-1}{2-0}$ $=$ $\frac{-1}{2}$

Therefore equation of $BC$ :

$y - 1 =$ $\frac{-1}{2}$ $( x-0) \hspace{0.5cm} \Rightarrow 2y-2=-x \hspace{0.5cm} \Rightarrow x+ 2y - 2 = 0$

Mid point $D$ of $BC = \Big($ $\frac{0+2}{2}$ $,$ $\frac{1+0}{2}$ $\Big)= \Big( 1,$ $\frac{1}{2}$ $\Big)$

Slope of $AD = m_2 =$ $\frac{\frac{1}{2} -(-2)}{1-(-1)}$ $=$ $\frac{5}{4}$

Therefore equation of $AD$ :

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

$\Rightarrow y+2 =$ $\frac{5}{4}$ $( x+1) \Rightarrow 4y+ 8 = 5x + 5 \Rightarrow 5x - 4y - 3$

$\\$

Question 6: Using the concept of the equation of a line, prove that the three points $(-2, -2) (8, 2)$ and $( 3, 0)$ are collinear.

Answer:

Given points $A(-2,-2), B(8,2)$ and $C(3,0)$

Slope of $AC = m_1 =$ $\frac{0-(-2)}{3-(-2)}$ $=$ $\frac{2}{5}$

Therefore equation of $AC$ :

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

$\Rightarrow y+2 = \frac{2}{5} (x+2) \hspace{0.5cm} \Rightarrow 5y+10 = 2x + 4 \hspace{0.5cm} \Rightarrow 2x - 5y - 6 = 0$

Now check if $B( 8, 2)$ satisfy the equation

$\therefore 2 ( 8) - 5(2) - 6 = 0$

$\Rightarrow 16-10-6 = 0$

Therefore $A, B, C$ are collinear.

$\\$

Question 7: Prove that the line $y -x+2=0$ , divides the join of points $(3,-1)$ and $(8,9)$ in the ratio $2:3$ .

Answer:

Let $y -x+2=0$ , divides the join of points $(3,-1)$ and $(8,9)$ at a point P in the ratio of $k:1$

$\therefore P = \Big($ $\frac{3+8k}{k+1}$ $,$ $\frac{-1+9k}{k+1}$ $\Big)$

Since $P$ lies on $y - x+2 = 0$ , it should satisfy the equation.

$\therefore$ $\frac{-1+9k}{k+1}$ $-$ $\frac{3+8k}{k+1}$ $+ 2 = 0$

$\Rightarrow -1 + 9k - 3 - 8 k + 2 k + 2 = 0$

$\Rightarrow 3k - 2 = 0$

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

Hence the line $y -x+2=0$ , divides the join of points $(3,-1)$ and $(8,9)$ in the ratio $2:3$ .

$\\$

Question 8: Find the equation to the straight line which bisects the distance between the points $(a, b), (a', b')$ and also bisects the distance between the points $( -a, b)$ and $(a', -b')$ .

Answer:

Given points $A(a, b), B(a', b')$ and $C( -a, b)$ and $D(a', -b')$

Mid point $P$ of $AB = \Big($ $\frac{a+a'}{2}$ $,$ $\frac{b+b'}{2}$ $\Big)$

Mid point $Q$ of $CD = \Big($ $\frac{-a+a'}{2}$ $,$ $\frac{b-b'}{2}$ $\Big)$

Slope of $PQ = m_1 =$ $\frac{\frac{b-b'}{2} - \frac{b+b'}{2} }{\frac{-a+a'}{2} - \frac{a+a'}{2}}$ $=$ $\frac{b-b'-b-b'}{-a+a'-a-a'}$ $=$ $\frac{-2b'}{-2a}$ $=$ $\frac{b'}{a}$

Therefore equation of $PQ$ :

$y -$ $\frac{b+b'}{2}$ $=$ $\frac{b'}{a}$ $\Big( x-$ $\frac{a+a'}{2}$ $\Big)$

$\Rightarrow 2y - b - b' =$ $\frac{b'}{a}$ $( 2x - a - a')$

$\Rightarrow 2ay - ab - ab' = 2b'x-ab'-a'b'$

$\Rightarrow 2b'x-2ay+ab-a'b'=0$

$\Rightarrow 2ay - 2b'x = ab - a'b'$

$\\$

Question 9: In what ratio is the line joining the points $(2, 3)$ and $(4, -5)$ divided by the line passing through the points $(6,8)$ and $(-3, -2)$ .

Answer:

Given points: $(6,8)$ and $(-3, -2)$

Slope of $AB =$ $\frac{-2-8}{-3-6}$ $=$ $\frac{-10}{-9}$ $=$ $\frac{10}{9}$

Therefore equation of $AB$ :

$y - 8 =$ $\frac{10}{9}$ $( x - 6) \hspace{0.5cm} \Rightarrow 9y - 72 = 10 x - 60 \hspace{0.5cm} \Rightarrow 10x - 9 y + 12 = 0$

Let $10x - 9 y + 12 = 0$ divides the line joining $(2, 3)$ and $(4, -5)$ in the ratio of $k:1$

$\therefore$ point of intersection $= \Big($ $\frac{4k+2}{k+1}$ $,$ $\frac{-5k+3}{k+1}$ $\Big)$

This point is on line $10x - 9 y + 12 = 0$

$\therefore 10 \Big($ $\frac{4k+2}{k+1}$ $\Big ) - 9 \Big($ $\frac{-5k+3}{k+1}$ $\Big) + 12 = 0$

$\Rightarrow 40k+20+45k-27+12k+12=0$

$\Rightarrow 97k + 5 = 0$

$\Rightarrow k = -$ $\frac{5}{97}$

Hence the required ratio is $5:97$ externally.

$\\$

Question 10: The vertices of a quadrilateral are $A(-2,6),8(1,2),C(10, 4)$ and $D(7, 8)$ . Find the equations of its diagonals.

Answer:

Given points: $A(-2,6), B(1,2) , C(10,4), D(7,8)$

Slope of $AC =$ $\frac{4-6}{10-(-2)}$ $=$ $\frac{-2}{12}$ $=$ $\frac{-1}{6}$

Therefore equation of $AC$ :

$y - 6 =$ $\frac{-1}{6}$ $( x - (-2)) \hspace{0.5cm} \Rightarrow 6y-36=-x-2 \hspace{0.5cm} \Rightarrow x+6y-34=0$

Slope of $BD =$ $\frac{8-2}{7-1}$ $=$ $\frac{6}{6}$ $= 1$

Therefore equation of $BD$ :

$y - 2 =$ $1 ( x - 1) \hspace{0.5cm} \Rightarrow x-y+1=0$

$\\$

Question 11: The length $L$ (in centimeters) of a copper rod is a linear function of its Celsius temperature $C$ . In an experiment, if $L = 124.942$ when $C = 20$ , and $L = 125.134$ when $C = 110$ , express $L$ in terms of $C$ .

Answer:

Given two points $( 124.942, 20)$ and $(125.134, 110)$

Slope of line $=$ $\frac{110-20}{125.134-124.942}$ $=$ $\frac{90}{0.192}$

Therefore the equation of line:

$y - 20 =$ $\frac{90}{0.192}$ $( x - 124.942)$

$\Rightarrow 0.192 y - 3.84 = 90x - 11244.78$

$\Rightarrow 90x = 0.192y + 11240.78$

$\Rightarrow x =$ $\frac{0.192}{90}$ $y +$ $\frac{11240.78}{90}$

$\Rightarrow x =$ $\frac{4}{1875}$ $y + 124.898$

$\\$

Question 12: The owner of a milk store finds that he can sell $980$ liters milk each week at ( $14$ per liter and $1220$ liters of milk each week at Rs. $16$ per liter. Assuming a linear relationship between selling price and demand, how many liters can he sell weekly at Rs. $17$ per liter.

Answer:

Given two points $( 14, 980)$ and $(16, 1220)$

Slope of line $=$ $\frac{1220-980}{16-14}$ $=$ $\frac{240}{2}$ $= 120$

Therefore the equation of line:

$y - 980 = 120(x-14)$ $\hspace{0.5cm} \Rightarrow y = 120x - 700$

When $x = 17, y = 120 \times 17 - 700 = 1340$

Hence the owner of the milk store would be able to sell $1340$ liters of milk at Rs. $17$ / liter.

$\\$

Question 13: Find the equation of the bisector of $\angle A$ of the triangle whose vertices are $A (4,3), B(0,0)$ and $C( 2, 3)$ .

Answer:

Let $AD$ be the bisector of $\angle BAC$ .

We know, $\frac{AB}{AC}$ $=$ $\frac{BD}{DC}$

Now $AB = \sqrt{(4-0)^2 + ( 3-0)^2} = \sqrt{25} = 5$

$AC = \sqrt{(4-2)^2 + ( 3-3)^2} = \sqrt{4} = 2$

$\therefore$ $\frac{BD}{DC}$ $=$ $\frac{5}{2}$

Therefore coordinates of $D = \Big($ $\frac{2 \times 5 + 0 \times 2}{5+2}$ $,$ $\frac{3 \times 5 + 0 \times 2}{5+2}$ $\Big) = \Big($ $\frac{10}{7}$ $,$ $\frac{15}{7}$ $\Big)$

Slope of $AD =$ $\frac{\frac{15}{7}-3}{\frac{10}{7}-4}$ $=$ $\frac{15-21}{10-28}$ $=$ $\frac{-6}{-18}$ $=$ $\frac{1}{3}$

Therefore equation of $AD$ :

$y - 3 =$ $\frac{1}{3}$ $( x - 4) \hspace{0.5cm} \Rightarrow 3y-9=x-4 \hspace{0.5cm} \Rightarrow x-3y+5=0$

$\\$

Question 14: Find the equations to the straight lines which go through the origin and trisect the portion of the straight line $3x+ y = 12$ which is intercepted between the axes of coordinates