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

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

Answer:

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

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

Therefore perpendicular distance from point $\displaystyle ( 4, 5) \text{ from } 3x -5y + 7 = 0$

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

Therefore,

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

$\displaystyle \text{Comparing with } ax + by + c = 0$ we get

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

Given equations:

$\displaystyle 24 x + 7 y = 20$ … … … … … i)

$\displaystyle 4 x - 3 y - 2 = 0$ … … … … … ii)

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

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

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

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

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

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

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

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

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

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

$\displaystyle \therefore d_1 = d_2$

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

$\displaystyle \\$

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

Answer:

Given lines:

$\displaystyle 2 x + 3 y =21$ … … … … … i)

$\displaystyle 3 x - 4y +11 =0$ … … … … … ii)

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

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

Therefore perpendicular distance from point $\displaystyle ( 3, 5) \text{ from } 8 x + 6 y + 5 = 0$

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

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

$\displaystyle \\$

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

Answer:

Given lines:

$\displaystyle 2x -3y + 14=0$ … … … … … i)

$\displaystyle 5 x + 4y -7 =0$ … … … … … ii)

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

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

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

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

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

Answer:

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

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

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

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

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

Squaring both sides we get

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

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

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

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

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

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

Hence the required points on x-axis are

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

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

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

$\displaystyle = b^2$

$\displaystyle \\$

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

Answer:

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

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

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

Therefore the equation of the perpendicular line is

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

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

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

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

$\displaystyle \\$

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

Answer:

Given lines:

$\displaystyle x + 2y =5$ … … … … … i)

$\displaystyle x - 3 y =7$ … … … … … ii)

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

Given vertices: $\displaystyle A (2, 3), B (4, -1) \text{ and } C (1, 2)$

Equation of BC:

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

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

Therefore equation perpendicular to i) is

$\displaystyle x - y + \lambda = 0$ … … … … … ii)

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

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

Perpendicular distance of $\displaystyle A(2,3) \text{ from } BC$ :

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

Consider $\displaystyle +$ ve sign

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

Consider $\displaystyle -$ ve sign

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

Therefore the equations are $\displaystyle 3x=5 \text{ and } y = 0$

These are also straight lines.

$\displaystyle \\$

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

Answer:

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

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

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

When both are $\displaystyle +$ ve

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

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

This is a straight line.

Similarly, when both are $\displaystyle -$ ve

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

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

This is a straight line as well.

Similarly, the other two combinations are also straight lines.

$\displaystyle \\$

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

Answer:

Given the perpendicular distance from point $\displaystyle ( 1, 1)$ to the straight line $\displaystyle ax -by + c = 0$ is $\displaystyle 1$ . Therefore,