Class 11: The Straight Line – Exercise 23.7 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Note: The equation of the line whose length of the perpendicular from the origin is $\displaystyle p$ and the angle made by the perpendicular with the positive x-axis is given by $\displaystyle \alpha$ is given by $\displaystyle x \cos \alpha + y \sin \alpha = p$

Question 1: Find the equation of a line for which

i) $\displaystyle p = 5, \alpha = 60^{\circ}$ ii) $\displaystyle p = 4, \alpha = 150^{\circ}$ iii) $\displaystyle p = 8, \alpha = 225^{\circ}$ vi) $\displaystyle p = 8, \alpha = 300^{\circ}$

Answer:

i) $\displaystyle \text{Given } p = 5, \alpha = 60^{\circ}$

Therefore the equation of line in normal form:

$\displaystyle \therefore x \cos 60^{\circ} + y \sin 60^{\circ} = 5$

$\displaystyle \Rightarrow \frac{x}{2} + \frac{\sqrt{3}y}{2} = 5$

$\displaystyle \Rightarrow x + \sqrt{3} y = 10$

ii) $\displaystyle \text{Given } p = 4, \alpha = 150^{\circ}$

Therefore the equation of line in normal form:

$\displaystyle \therefore x \cos 150^{\circ} + y \sin 150^{\circ} = 4$

$\displaystyle \Rightarrow x \cos (180^{\circ}-30^{\circ}) + y \sin (180^{\circ}-30^{\circ}) = 4$

$\displaystyle \Rightarrow -x \cos 30^{\circ} + y \sin 30^{\circ} = 4$

$\displaystyle \Rightarrow \frac{-\sqrt{3} x}{2} + \frac{y}{2} = 4$

$\displaystyle \Rightarrow \sqrt{3}x - y +8=0$

iii) $\displaystyle \text{Given } p = 8, \alpha = 225^{\circ}$

Therefore the equation of line in normal form:

$\displaystyle \therefore x \cos 225^{\circ} + y \sin 225^{\circ} = 8$

$\displaystyle \Rightarrow x \cos (180^{\circ}+45^{\circ}) + y \sin (180^{\circ}+45^{\circ}) = 8$

$\displaystyle \Rightarrow -x \cos 45^{\circ} - y \sin 45^{\circ} = 8$

$\displaystyle \Rightarrow \frac{- x}{\sqrt{2}} - \frac{y}{\sqrt{2}} = 8$

$\displaystyle \Rightarrow x+y + 8\sqrt{2} = 0$

vi) $\displaystyle \text{Given } p = 8, \alpha = 300^{\circ}$

Therefore the equation of line in normal form:

$\displaystyle \therefore x \cos 300^{\circ} + y \sin 200^{\circ} = 8$

$\displaystyle \Rightarrow x \cos (360^{\circ}-60^{\circ}) + y \sin (360^{\circ}-60^{\circ}) = 8$

$\displaystyle \Rightarrow x \cos 60^{\circ} - y \sin 60^{\circ} = 8$

$\displaystyle \Rightarrow \frac{x}{2} - \frac{\sqrt{3}y}{2} = 8$

$\displaystyle \Rightarrow x-\sqrt{3} y = 16$

$\displaystyle \\$

Question 2: Find the equation of the line on which the length of the perpendicular segment from the origin to the line is $\displaystyle 4$ and the inclination of the perpendicular segment with the positive direction of x-axis is $\displaystyle 30^{\circ}$ .

Answer:

$\displaystyle \text{Given } p = 4, \alpha = 30^{\circ}$

Therefore the equation of line in normal form:

$\displaystyle \therefore x \cos 30^{\circ} + y \sin 30^{\circ} = 4$

$\displaystyle \Rightarrow \frac{\sqrt{3} x}{2} + \frac{y}{2} = 4$

$\displaystyle \Rightarrow \sqrt{3}x + y = 8$

$\displaystyle \\$

Question 3: Find the equation of the line whose perpendicular distance from the origin is $\displaystyle 4$ units and the angle which the normal makes with the positive direction of x-axis is $\displaystyle 15^{\circ}$ .

Answer:

$\displaystyle \text{Given } p = 4, \alpha = 15^{\circ}$

Therefore the equation of line in normal form:

$\displaystyle \therefore x \cos 15^{\circ} + y \sin 15^{\circ} = 4$

$\displaystyle \therefore x \cos ( 45^{\circ} - 30^{\circ}) + y \sin ( 45^{\circ} - 30^{\circ}) = 4$

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

$\displaystyle \Rightarrow (\sqrt{3}+1) x + (\sqrt{3}-1)y = 8\sqrt{2}$

$\displaystyle \\$

Question 4: Find the equation of the straight line at a distance of $\displaystyle 3$ units from the origin such that the perpendicular from the origin to the line makes an angle $\displaystyle \alpha$ given by $\displaystyle \tan \alpha = \frac{5}{12}$ with the positive direction of x-axis.

Answer:

$\displaystyle \text{Given } p = 3, \alpha = \tan^{-1} \Big( \frac{5}{12} \Big)$

$\displaystyle \therefore \tan \alpha = \frac{5}{12} \Rightarrow \frac{\sin \alpha}{\cos \alpha} = \frac{5}{12} \Rightarrow \frac{\sin \alpha}{\cos \alpha} = \frac{5/13}{12/13}$

$\displaystyle \therefore \sin \alpha = \frac{5}{13}$ and $\displaystyle \cos \alpha = \frac{12}{13}$

Therefore the equation of line in normal form:

$\displaystyle x \cos \alpha + y \sin \alpha = p$

$\displaystyle \Rightarrow x \Big( \frac{12}{13} \Big) + y \Big( \frac{5}{13} \Big) = 3$

$\displaystyle \Rightarrow 12x + 5y = 39$

$\displaystyle \\$

Question 5: Find the equation of the straight line on which the length of the perpendicular from the origin is $\displaystyle 2$ and the perpendicular makes an angle $\displaystyle \alpha$ with x-axis such that $\displaystyle \sin \alpha = \frac{1}{3}$ .

Answer:

$\displaystyle \text{Given } p = 2, \sin \alpha = \frac{1}{3}$

$\displaystyle \therefore \cos \alpha = \sqrt{ 1 - \sin^2 \alpha } = \sqrt{1 - \frac{1}{9}} = \frac{2\sqrt{2}}{3}$

Therefore the equation of line in normal form:

$\displaystyle x \cos \alpha + y \sin \alpha = p$

$\displaystyle \Rightarrow x \Big( \frac{2\sqrt{2}}{3} \Big) + y \Big( \frac{1}{3} \Big) = 2$

$\displaystyle \Rightarrow 2\sqrt{2}x + y = 6$

$\displaystyle \\$

Question 6: Find the equation of the straight line upon which the length of the perpendicular from the origin is $\displaystyle 2$ and the slope of this perpendicular is $\displaystyle \frac{5}{12}$ .

Answer:

$\displaystyle \text{Given } p = 2, \tan \alpha = \frac{5}{12}$

$\displaystyle \therefore \sin \alpha = \frac{5}{13}$ and $\displaystyle \cos \alpha = \frac{12}{13}$

Therefore the equations of line in normal form:

Case 1: $\displaystyle \alpha = \alpha$

$\displaystyle x \cos \alpha + y \sin \alpha = p$

$\displaystyle \Rightarrow x \Big( \frac{12}{13} \Big) + y \Big( \frac{5}{13} \Big) = 2$

$\displaystyle \Rightarrow 12x + 5y = 26$

Case 2: $\displaystyle \alpha = (180^{\circ}+ \alpha)$

$\displaystyle x \cos (180^{\circ}+ \alpha) + y \sin (180^{\circ}+ \alpha) = p$

$\displaystyle - x \cos \alpha - y \sin \alpha = 2$

$\displaystyle \Rightarrow -x \Big( \frac{12}{13} \Big) - y \Big( \frac{5}{13} \Big) = 2$

$\displaystyle \Rightarrow -12x - 5y = 26$

$\displaystyle \Rightarrow 12x + 5y - 26=0$

$\displaystyle \\$

Question 7: The length of the perpendicular from the origin to a line is $\displaystyle 7$ and the line makes an angle of $\displaystyle 15^{\circ}$ with the positive direction of y-axis. Find the equation of the line.

Answer:

$\displaystyle \text{Given } p = 7, \alpha = 30^{\circ}$

Therefore the equation of line in normal form:

$\displaystyle \therefore x \cos 30^{\circ} + y \sin 30^{\circ} = 7$

$\displaystyle \Rightarrow \frac{\sqrt{3} x}{2} + \frac{y}{2} = 7$

$\displaystyle \Rightarrow \sqrt{3} x + y = 14$

$\displaystyle \\$

Question 8: Find the value of $\displaystyle \theta$ and $\displaystyle p$ , if the equation $\displaystyle x \cos \theta + y \sin \theta = p$ is the normal form of the line $\displaystyle \sqrt{3}x +y + 2 = 0$ .

Answer:

Given equation is

$\displaystyle \sqrt{3}x +y + 2 = 0$

$\displaystyle \Rightarrow \sqrt{3}x +y = - 2$

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

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

Comparing this with the equation $\displaystyle x \cos \theta + y \sin \theta = p$ we get

$\displaystyle p = 1,$

$\displaystyle \cos \theta = \frac{-\sqrt{3}}{2} \Rightarrow \theta = 210^{\circ}$

$\displaystyle \\$

Question 9: Find the equation of the straight line which makes a triangle of area $\displaystyle 96\sqrt{3}$ with the axes and perpendicular from the origin to it makes an angle of $\displaystyle 30^{\circ}$ with y-axis.

Answer:

$\displaystyle \text{Given } p = 7, \alpha = 60^{\circ}$

Therefore the equation of line in normal form:

$\displaystyle \therefore x \cos 60^{\circ} + y \sin 60^{\circ} = p$

$\displaystyle \Rightarrow \frac{x}{2} + \frac{\sqrt{3}y}{2} = p$

$\displaystyle \Rightarrow \frac{x}{2p} + \frac{\sqrt{3}y}{2p} = 1$

$\displaystyle \text{Therefore x-intercept } = 2p \text{ and y-intercept } = \frac{2p}{\sqrt{3}}$

It is given that the area of $\displaystyle \triangle AOB = 96\sqrt{3}$

$\displaystyle \therefore \frac{1}{2} \times 2p \times \frac{2p}{\sqrt{3}} = 96\sqrt{3}$

$\displaystyle \Rightarrow p^2 = 144$

$\displaystyle \Rightarrow p = 12$

Therefore equation of $\displaystyle AB = x + \sqrt{3}y = 24$

$\displaystyle \\$

Question 10: Find the equation of a straight line on which the perpendicular from the origin makes an angle of $\displaystyle 30^{\circ}$ with x-axis and which forms a triangle of area $\displaystyle 50\sqrt{3}$ with the axes