Question

If the normal form of the equation of a straight line $4 x+3 y+2=0$ is $x \cos \alpha+y \sin \alpha=p$ and its intercept form is $\frac{x}{a}+\frac{y}{b}=1$, then $\frac{p \sec \alpha}{a b}=$

• A. $-\frac{1}{2}$
• B. $\frac{3}{2}$
• C. $-\frac{3}{2}$
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

Given, $4 x+3 y+2=0$
$\Rightarrow -4 x-3 y=2$
$\frac{-4 x}{5}-\frac{3}{5} y=\frac{2}{5}$
$\because \cos \alpha=\frac{-4}{5}, \sin \alpha=\frac{-3}{5}$ and $P=\frac{2}{5}$
$\frac{x}{2 /-4}+\frac{y}{2 /-3}=1$
$\Rightarrow a=\frac{-2}{4}, b=\frac{-2}{3}$
$\therefore \frac{P \sec \alpha}{a b}=\frac{\frac{2}{5} \times\left(\frac{-5}{4}\right)}{\left(\frac{-2}{4}\right)\left(\frac{-2}{3}\right)}$
$=\frac{-3}{2}$