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

Question 1: Find the equation of a line for which

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

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

Therefore the equation of line in normal form:

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

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

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

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

Therefore the equation of line in normal form:

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

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

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

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

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

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

Therefore the equation of line in normal form:

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

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

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

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

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

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

Therefore the equation of line in normal form:

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

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

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

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

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

$\\$

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

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

Therefore the equation of line in normal form:

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

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

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

$\\$

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

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

Therefore the equation of line in normal form:

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

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

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

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

$\\$

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

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

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

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

Therefore the equation of line in normal form:

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

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

$\Rightarrow 12x + 5y = 39$

$\\$

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

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

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

Therefore the equation of line in normal form:

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

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

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

$\\$

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

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

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

Therefore the equations of line in normal form:

Case 1: $\alpha = \alpha$

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

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

$\Rightarrow 12x + 5y = 26$

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

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

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

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

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

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

$\\$

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

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

Therefore the equation of line in normal form:

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

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

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

$\\$

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

Given equation is

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

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

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

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

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

$p = 1,$

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

$\\$

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

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

Therefore the equation of line in normal form:

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

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

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

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

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

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

$\Rightarrow p^2 = 144$

$\Rightarrow p = 12$

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

$\\$

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

Given $OL = p, \alpha = 30^{\circ}$

Therefore the equation of line in normal form:

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

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

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

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

It is given that the area of $\triangle AOB =$ $\frac{50}{\sqrt{3}}$

$\therefore$ $\frac{1}{2}$ $\times$ $\frac{2p}{\sqrt{3}}$ $\times 2p =$ $\frac{50}{\sqrt{3}}$

$\Rightarrow p^2 = 25$

$\Rightarrow p = 5$

Therefore equation of $AB = \sqrt{3} x + y = 10$