### Straight Lines Question 8

#### Question 8 - 30 January - Shift 1

A straight line cuts off the intercepts $OA=a$ and $OB=b$ on the positive directions of $x$-axis and $y-$ axis respectively. If the perpendicular from origin $O$ to this line makes an angle of $\frac{\pi}{6}$ with positive direction of $y$-axis and the area of $\triangle O A B$ is $\frac{98}{3} \sqrt{3}$, then $a^{2}-b^{2}$ is equal to:

(1) $\frac{392}{3}$

(2) 196

(3) $\frac{196}{3}$

(4) 98

## Show Answer

#### Answer: (1)

#### Solution:

#### Formula: Perpendicular / Normal form

Equation of straight line : $\frac{x}{a}+\frac{y}{b}=1$

Or $x \cos \frac{\pi}{3}+y \sin \frac{\pi}{3}=p$

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

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

Comparing both : $a=2 p, b=\frac{2 p}{\sqrt{3}}$

Now area of $\triangle OAB=\frac{1}{2} \cdot ab=\frac{98}{3} \cdot \sqrt{3}$

$\frac{1}{2} \cdot 2 p \cdot \frac{2 p}{\sqrt{3}}=\frac{98}{3} \cdot \sqrt{3}$

$p^{2}=49$

$$\begin{aligned} & a^2-b^2=4 p^2-\frac{4 p^2}{3}=\frac{2}{3} 4 p^2 \\ & =\frac{8}{3} \cdot 49=\frac{392}{3} \end{aligned}$$

$\therefore$ The answer is $\frac{392}{3}$