Class 11: The Straight Line – Exercise 23.15 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Find the distance of the point $(4, 5)$ from the straight line $3x -5y + 7 = 0$ .

Answer:

Given equation: $3x -5y + 7 = 0$

Comparing with $ax + by + c = 0$ we get $a = 3, \ \ \ \ b = -5 , \ \ \ \ c = 7$

Therefore perpendicular distance from point $( 4, 5)$ from $3x -5y + 7 = 0$

$d = \Bigg|$ $\frac{ax_1+ by_1 + c}{\sqrt{a^2+b^2}}$ $\Bigg| = \Bigg|$ $\frac{3 \times 4 - 5 \times 5 + 7}{\sqrt{(3)^2+(-5)^2}}$ $\Bigg| =$ $\frac{6}{\sqrt{34}}$

Thus the required distance is $\frac{6}{\sqrt{34}}$

$\\$

Question 2: Find the perpendicular distance of the line joining the points $( \cos \theta , \sin \theta )$ and $( \cos \theta , \sin \theta )$ from the origin.

Answer:

The equation of the line joining $( \cos \theta, \sin \theta)$ and $( \cos \phi, \sin \phi)$ :

$y - \sin \theta =$ $\frac{\sin \phi - \sin \theta}{\cos \phi - \cos \theta}$ $( x - \cos \theta)$

$\Rightarrow (\cos \phi - \cos \theta ) y - \sin \theta (\cos \phi - \cos \theta ) = (\sin \phi - \sin \theta) x - \cos \theta (\sin \phi - \sin \theta )$

$\Rightarrow (\sin \phi - \sin \theta) x - (\cos \phi - \cos \theta ) y + \sin \theta \cos \phi - \cos \theta \sin \phi = 0$

Comparing with $ax + by + c = 0$ we get $a = (\sin \phi - \sin \theta), \ \ \ \ b = - (\cos \phi - \cos \theta ) , \ \ \ \ c = (\sin \theta \cos \phi - \cos \theta \sin \phi)$

Therefore,

$d = \Bigg|$ $\frac{(\sin \phi - \sin \theta)(0)- (\cos \phi - \cos \theta )(0) + (\sin \theta \cos \phi - \cos \theta \sin \phi)}{\sqrt{(\sin \phi - \sin \theta)^2+(- (\cos \phi - \cos \theta ))^2}}$ $\Bigg|$

$= \Bigg|$ $\frac{ \sin ( \theta - \phi) }{ \sqrt{ \sin^2 \phi + \sin^2 \theta - 2 \sin \phi \sin \theta+ \cos^2 \phi + \cos^2 \theta- 2 \cos \phi \cos \theta } }$ $\Bigg|$

$= \Bigg|$ $\frac{\sin ( \theta - \phi)}{ \sqrt{2 - 2 \cos ( \theta - \phi)} }$ $\Bigg|$

$=$ $\frac{1}{\sqrt{2}}$ $\Bigg|$ $\frac{\sin ( \theta - \phi)}{ \sqrt{1 - \cos ( \theta - \phi)} }$ $\Bigg|$

$=$ $\frac{1}{\sqrt{2}}$ $\Bigg|$ $\frac{\sin ( \theta - \phi)}{ \sqrt{2 \sin^2 ( \frac{\theta - \phi}{2} )} }$ $\Bigg|$

$=$ $\frac{1}{\sqrt{2}}$ $\Bigg|$ $\frac{2 \sin ( \frac{\theta - \phi}{2} ) \cos ( \frac{\theta - \phi}{2} )}{ \sqrt{2} \sin ( \frac{\theta - \phi}{2} ) }$ $\Bigg|$

$= \Big| \cos \Big($ $\frac{\theta - \phi}{2}$ $\Big) \Big|$

Thus the required distance is $\cos \Big($ $\frac{\theta - \phi}{2}$ $\Big)$

$\\$

Question 3: Find the length of the perpendicular from the origin to the straight line joining the two points whose coordinates are $(a \cos \alpha, a \sin \alpha)$ and $(a \cos \beta, a \sin \beta)$ .

Answer:

The equation of the line joining $( a \cos \alpha, a \sin \alpha)$ and $( a \cos \beta, a \sin \beta )$ :

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

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

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

$\Rightarrow x \cot \Big($ $\frac{ \alpha + \beta }{2}$ $\Big) + y - a \sin \alpha - a \cos \alpha \cot \Big($ $\frac{ \alpha + \beta }{2}$ $\Big) = 0$

Comparing with $ax + by + c = 0$ we get

$a = \cot \Big($ $\frac{ \alpha + \beta }{2}$ $\Big) , \ \ \ \ b = - 1 , \ \ \ \ c = - a \sin \alpha - a \cos \alpha \cot \Big($ $\frac{ \alpha + \beta }{2}$ $\Big)$

Therefore the perpendicular distance from (0,0) is:

$d = \Bigg|$ $\frac{ \cot (\frac{ \alpha + \beta }{2}) ( 0) -1 (0 ) - a \sin \alpha - a \cos \alpha \cot ( \frac{ \alpha + \beta }{2} ) }{ \sqrt{ \cot^2 ( \frac{\alpha + \beta}{2} ) + 1 } }$ $\Bigg|$

$= \Bigg|$ $\frac{ a \sin \alpha + a \cos \alpha \cot ( \frac{ \alpha + \beta }{2} ) }{ \sqrt{ \mathrm{cosec}^2 ( \frac{\alpha + \beta}{2} ) } }$ $\Bigg|$

$= \Big| \sin (\frac{ \alpha + \beta }{2}) \sin \alpha + \cos \alpha \cos (\frac{ \alpha + \beta }{2}) \Big|$

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

$= a \cos \Big($ $\frac{ \beta - \alpha }{2}$ $\Big)$

$= a \cos \Big($ $\frac{\alpha - \beta }{2}$ $\Big)$

$\\$

Question 4: Show that the perpendiculars let fall from any point on the straight line $2 x + 11 y -5 = 0$ upon the two straight lines $24 x + 7 y = 20$ and $4 x - 3 y - 2 = 0$ are equal to each other.

Answer:

Given equations:

$24 x + 7 y = 20$ … … … … … i)

$4 x - 3 y - 2 = 0$ … … … … … ii)

Let $P(a, b)$ be the point on $2 x + 11 y -5 = 0$

Therefore, the distance $d_1$ of $P( a, b)$ from line i)

$d_1 = \Bigg|$ $\frac{24a+7b-20}{\sqrt{24^2 + 7^2}}$ $\Bigg|= \Bigg|$ $\frac{24a+7b-20}{25}$ $\Bigg|$ … … … … … iii)

Similarly, the distance $d_2$ of $P( a, b)$ from line ii)

$d_2 = \Bigg|$ $\frac{4a-3b-2}{\sqrt{3^2 + (-4)^2}}$ $\Bigg|= \Bigg|$ $\frac{4a-3b-2}{5}$ $\Bigg|$ … … … … … iv)

Since $P(a, b)$ is on $2x + 11y - 5 = 0$

$\Rightarrow 2a + 11b - 5 = 0 \Rightarrow b =$ $\frac{5-2a}{11}$

Substituting the value of $b$ in iii) and iv) we get

$d_1 = \Big|$ $\frac{24a+7 \times (\frac{5-2a}{11})-20}{25}$ $\Big| = \Big|$ $\frac{50a-37}{55}$ $\Big|$

$d_2 = \Big|$ $\frac{4a-3 \times (\frac{5-2a}{11}) - 2}{5}$ $\Big| = \Big|$ $\frac{50a-37}{55}$ $\Big|$

$\therefore d_1 = d_2$

Hence the perpendicular drawn from any point on the straight line $2 x + 11 y -5 = 0$ upon the two straight lines $24 x + 7 y = 20$ and $4 x - 3 y - 2 = 0$ are equal to each other.

$\\$

Question 5: Find the distance of the point of intersection of the lines $2 x + 3 y =21$ and $3 x - 4y +11 =0$ from the line $8 x + 6 y + 5 = 0$ .

Answer:

Given lines:

$2 x + 3 y =21$ … … … … … i)

$3 x - 4y +11 =0$ … … … … … ii)

Solving i) and ii) we get the point of intersection as $( 3, 5)$

Comparing $8 x + 6 y + 5 = 0$ with $ax + by + c = 0$ we get $a = 8, \ \ \ \ b = 6 , \ \ \ \ c = 5$

Therefore perpendicular distance from point $( 3, 5)$ from $8 x + 6 y + 5 = 0$

$d = \Bigg|$ $\frac{ax_1+ by_1 + c}{\sqrt{a^2+b^2}}$ $\Bigg| = \Bigg|$ $\frac{8 \times 3 + 6 \times 4 + 5}{\sqrt{(8)^2+(6)^2}}$ $\Bigg| =$ $\frac{59}{\sqrt{10}}$

Thus the required distance is $\frac{59}{\sqrt{10}}$

$\\$

Question 6: Find the length of the perpendicular from the point $(4, -7)$ to the line joining the origin and the point of intersection of the line $2x -3y + 14=0$ and $5 x + 4y -7 =0$ .

Answer:

Given lines:

$2x -3y + 14=0$ … … … … … i)

$5 x + 4y -7 =0$ … … … … … ii)

Solving i) and ii) we get the point of intersection as $\Big($ $\frac{-35}{23}$ $,$ $\frac{84}{23}$ $\Big)$

Therefore the equation of line passing between $(0,0)$ and $\Big($ $\frac{-35}{23}$ $,$ $\frac{84}{23}$ $\Big)$ :

$y - 0 = \Big($ $\frac{\frac{84}{23} - 0}{\frac{-35}{23}-0}$ $\Big) ( x - 0)$

$\Rightarrow y =$ $\frac{-12}{5}$ $x$

Question 7: What are the points on x-axis whose perpendicular distance from the straight line $\frac{x}{a}$ $+$ $\frac{y}{b}$ $=1$ is $a$ ?

Answer:

Let the point $P(t, 0)$ be on the x-axis.

Given line: $\frac{x}{a}$ $+$ $\frac{y}{b}$ $=1$

Comparing with $ax + by + c = 0$ we get $a =$ $\frac{1}{a}$ $, \ \ \ \ b =$ $\frac{1}{b}$ $, \ \ \ \ c = -1$

Therefore perpendicular distance from point $( t,0)$

$d = \Bigg|$ $\frac{ax_1+ by_1 + c}{\sqrt{a^2+b^2}}$ $\Bigg| = \Bigg|$ $\frac{\frac{1}{a} \times t + \frac{1}{b} \times (0) -1}{\sqrt{(\frac{1}{a})^2+(\frac{1}{b})^2}}$ $\Bigg| = a$

Squaring both sides we get

$a^2 ($ $\frac{1}{a^2}$ $+$ $\frac{1}{b^2}$ $) =$ $\frac{t^2}{a^2}$ $+1 -$ $\frac{2t}{a}$

$1 +$ $\frac{a^2}{b^2}$ $=$ $\frac{t^2}{a^2}$ $+1 -$ $\frac{2t}{a}$

$a^2b^2 + a^2 = b^2t^2 + a^2b^2 - 2ab^2 t$

$b^2 t^2 - 2ab^2 t - a^2 = 0$

$t =$ $\frac{2ab^2 \pm \sqrt{a^2b^4 + b^2 a^4}}{2b^2}$

$=$ $\frac{a}{b}$ $( b \pm \sqrt{a^2 +b^2})$

Hence the required points on x-axis are

$\Big($ $\frac{a}{b}$ $( b + \sqrt{a^2 +b^2}), 0 \Big )$ and $\Big($ $\frac{a}{b}$ $( b - \sqrt{a^2 +b^2}), 0 \Big )$

$\\$

Question 8: Show that the product of perpendiculars on the line $\frac{x}{a}$ $\cos \theta +$ $\frac{y}{b}$ $\sin \theta =1$ from the points $(\pm \sqrt{a^2 - b^2}, 0)$ is $b^2$ .

Answer:

Let $d_1$ be the perpendicular distance from $(\sqrt{a^2 - b^2}, 0)$ on $\frac{x}{a}$ $\cos \theta +$ $\frac{y}{b}$ $\sin \theta =1$

$d_1 = \Bigg|$ $\frac{ \frac{\cos \theta}{a} (\sqrt{a^2 - b^2}) + \frac{\sin \theta}{b} (0) - 1 }{ \sqrt{ \frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta }{b^2} } }$ $\Bigg|$ $= b \Bigg|$ $\frac{(\sqrt{a^2 - b^2}) ( \cos \theta) - a}{ \sqrt{b^2\cos^2 \theta + a^2 \sin^2 \theta} }$ $\Bigg|$

Let $d_2$ be the perpendicular distance from $( - \sqrt{a^2 - b^2}, 0)$ on $\frac{x}{a} \cos \theta +\frac{y}{b} \sin \theta =1$

$d_1 = \Bigg|$ $\frac{ \frac{\cos \theta}{a} ( - \sqrt{a^2 - b^2}) + \frac{\sin \theta}{b} (0) - 1 }{ \sqrt{ \frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta }{b^2} } }$ $\Bigg|$ $= b \Bigg|$ $\frac{(\sqrt{a^2 - b^2}) ( \cos \theta) + a}{ \sqrt{b^2\cos^2 \theta + a^2 \sin^2 \theta} }$ $\Bigg|$

$\therefore d_1 \times d_2 = \Bigg|$ $\frac{(\sqrt{a^2 - b^2}) ( \cos \theta) - a}{ \sqrt{b^2\cos^2 \theta + a^2 \sin^2 \theta} }$ $\Bigg| \times \Bigg|$ $\frac{(\sqrt{a^2 - b^2}) ( \cos \theta) + a}{ \sqrt{b^2\cos^2 \theta + a^2 \sin^2 \theta} }$ $\Bigg|$

$= b^2 \Bigg|$ $\frac{(a^2 - b^2) \cos^2 \theta - a^2}{b^2 \cos^2 \theta + a^2 \sin^2 \theta}$ $\Bigg|$

$= b^2 \Bigg|$ $\frac{a^2 ( \cos^2 \theta - 1) - b^2 \cos^2 \theta}{b^2 \cos^2 \theta + a^2 \sin^2 \theta}$ $\Bigg|$

$= b^2 \Bigg|$ $\frac{- a^2 \sin^2 \theta - b^2 \cos^2 \theta}{b^2 \cos^2 \theta + a^2 \sin^2 \theta}$ $\Bigg|$

$= b^2$

$\\$

Question 9: Find the perpendicular distance from the origin of the perpendicular from the point $(1, 2)$ upon the straight line $x-\sqrt{3}y + 4 = 0$ .

Answer:

Given line: $x-\sqrt{3}y + 4 = 0$ … … … … … i)

Therefore line perpendicular to line i) is $\sqrt{3} x + y + \lambda = 0$

The perpendicular passes through $( 1,2 )$ . Therefore $\lambda = -\sqrt{3}-2$

Therefore the equation of the perpendicular line is

$\sqrt{3} x + y -\sqrt{3}-2 = 0$ … … … … … ii)

Now the perpendicular distance from $(0,0)$ to line ii)

$d = \Big|$ $\frac{\sqrt{3}(0) + 1 (0) - ( \sqrt{3}+2)}{\sqrt{3+1}}$ $\Big| = \Big|$ $\frac{\sqrt{3}+2}{2}$ $\Big|$

Hence the required distance is $\frac{\sqrt{3}+2}{2}$ .

$\\$

Question 10: Find the distance of the point $(1, 2)$ from the straight line with slope $5$ and passing through the point of intersection of $x + 2y =5$ and $x - 3 y =7$ .

Answer:

Given lines:

$x + 2y =5$ … … … … … i)

$x - 3 y =7$ … … … … … ii)

Solving i) and ii) we get the point of intersection as $\Big($ $\frac{29}{5}$ $,$ $\frac{-2}{5}$ $\Big )$

The equation of line passing through $\Big($ $\frac{29}{5}$ $,$ $\frac{-2}{5}$ $\Big )$ and slope of $5$ :

$y - \Big($ $\frac{-2}{5}$ $\Big) = 5 \Big( x -$ $\frac{29}{5}$ $\Big)$

$\Rightarrow 5y + 2 = 25 x - 145$

$\Rightarrow 25x - 5y - 147 = 0$

Therefore perpendicular distance from $( 1,2)$ on $25x - 5y - 147 = 0$

$\Rightarrow d = \Big|$ $\frac{25 \times ( 1) - 5 ( 2) - 147}{ \sqrt{25^2 + 5^2}}$ $\Big| =$ $\frac{132}{5\sqrt{26}}$

Hence the required distance is $\frac{132}{5\sqrt{26}}$

$\\$

Question 11: What are the points on y-axis whose distance from the line $\frac{x}{3}$ $+$ $\frac{y}{4}$ $= 1$ in $4$ units?

Answer:

Let $P(0, t)$ be the point on y-axis.

Given line: $\frac{x}{3}$ $+$ $\frac{y}{4}$ $= 1$

Comparing with $ax + by + c = 0$ we get $a =$ $\frac{1}{3}$ $, \ \ \ \ b =$ $\frac{1}{4}$ $, \ \ \ \ c = -1$

$\therefore \Bigg|$ $\frac{\frac{1}{3}( 0) + \frac{1}{4}(t) - 1}{\sqrt{\frac{1}{3^2}+\frac{1}{4^2}}}$ $\Bigg| = 4$

$\Rightarrow \Big|$ $\frac{t-4}{5}$ $\Big| =$ $\frac{4}{3}$

$\Rightarrow t-4 = \pm$ $\frac{20}{3}$

$\Rightarrow t = 4 \pm$ $\frac{20}{3}$

Hence the required points on y-axis are $\Big( 0, 4 +$ $\frac{20}{3}$ $\Big)$ and $\Big( 0, 4 -$ $\frac{20}{3}$ $\Big)$

$\\$

Question 12: In the $\triangle ABC$ with vertices $A (2, 3), B (4, -1)$ and $C (1, 2)$ find the equation and the length of the altitude from vertex $A$ .

Answer:

Given vertices: $A (2, 3), B (4, -1)$ and $C (1, 2)$

Equation of BC:

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

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

Therefore equation perpendicular to i) is

$x - y + \lambda = 0$ … … … … … ii)

Since this line passes thought $A ( 2, 3), \lambda = 1$

Therefore the equation of the perpendicular is $x - y + 1 = 0$

Perpendicular distance of $A(2,3)$ from $BC$ :

$d = \Big|$ $\frac{1(2) + 1( 3) - 3}{\sqrt{1^2+1^2}}$ $\Big| = \sqrt{2}$

Hence the required distance is $\sqrt{2}$

$\\$

Question 13: Show that the path of a moving point such that its distances from two lines $3x - 2y = 5$ and $3x+2y +7 = 0$ are equal is a straight line.

Answer:

Let $P( h, k)$ be the point such that its distances from two lines $3x - 2y = 5$ and $3x+2y +7 = 0$ are equal. Therefore

$\Big|$ $\frac{3h-2k-5}{\sqrt{3^2+(-2)^2}}$ $\Big| = \Big|$ $\frac{3h+2k-5}{\sqrt{3^2+2^2}}$ $\Big|$

$\Rightarrow | 3h - 2k - 5 | = | 3h+ 2k - 5|$

$\Rightarrow ( 3h - 2k - 5 ) = \pm ( 3h+ 2k - 5)$

Consider $+$ ve sign

$3h - 2k - 5 = 3h+ 2k - 5 \hspace{0.5cm} \Rightarrow 4k = 0 \hspace{0.5cm} \Rightarrow k = 0$

Consider $-$ ve sign

$3h - 2k - 5 = -3h- 2k + 5 \hspace{0.5cm} \Rightarrow 6h=10 \hspace{0.5cm} \Rightarrow 3h = 5$

Therefore the equations are $3x=5$ and $y = 0$

These are also straight lines.

$\\$

Question 14: If sum of perpendicular distances of a variable point $P (x, y)$ from the lines $x+y - 5 = 0$ and $3x-2y+7=0$ is always $10$ . Show that $P$ must move on a line.

Answer:

Let $P(h, k)$ be the point such that sum of the perpendicular distance from $P$ to the given lines is $10$ . Therefore

$\Big|$ $\frac{h+k-5}{\sqrt{1^2+1^2}}$ $\Big| + \Big|$ $\frac{3h-2k+7}{\sqrt{3^2+2^2}}$ $\Big| = 10$

$\Big|$ $\frac{h+k-5}{\sqrt{2}}$ $\Big| + \Big|$ $\frac{3h-2k+7}{\sqrt{13}}$ $\Big| = 10$

When both are $+$ ve

$\frac{h+k-5}{\sqrt{2}}$ $+$ $\frac{3h-2k+7}{\sqrt{13}}$ $= 10$

$\Rightarrow ( \sqrt{13} + 3\sqrt{2}) h + ( \sqrt{13}- 2\sqrt{2})k + 7\sqrt{2}- 5\sqrt{13} - 10\sqrt{26} = 0$

This is a straight line.

Similarly, when both are $-$ ve

$-$ $\frac{h+k-5}{\sqrt{2}}$ $-$ $\frac{3h-2k+7}{\sqrt{13}}$ $= 10$

$\Rightarrow ( \sqrt{13} + 3\sqrt{2}) h - ( \sqrt{13}+ 2\sqrt{2})k + 7\sqrt{2}- 5\sqrt{13} + 10\sqrt{26} = 0$

This is a straight line as well.

Similarly, the other two combinations are also straight lines.

$\\$

Question 15: If the length of the perpendicular from the point $(1, 1)$ to the line $ax -by + c = 0$ be unity, show that $\frac{1}{c}$ $+$ $\frac{1}{a}$ $-$ $\frac{1}{b}$ $=$ $\frac{c}{2ab}$