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 $\triangle OAB=\frac{1}{2} \cdot ab=\frac{98}{3} \cdot \sqrt{3}$
$\frac{1}{2} \cdot 2 p \cdot \frac{2 p}{\sqrt{3}}=\frac{98}{3} \cdot \sqrt{3}$
$p^{2}=49$
$$\begin{aligned} & \mathrm{a}^2-\mathrm{b}^2=4 \mathrm{p}^2-\frac{4 \mathrm{p}^2}{3}=\frac{2}{3} 4 \mathrm{p}^2 \ & =\frac{8}{3} \cdot 49=\frac{392}{3} \end{aligned}$$
$\therefore$ The answer is $\frac{392}{3}$
