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(C) Given equation: $4x + 3y + 2 = 0$ or $4x + 3y = -2$. To convert to normal form $x \cos \alpha + y \sin \alpha = p$,divide by $\sqrt{4^2 + 3^2} = 5

Vedclass · Accepted Answer

$\frac{-3}{2}$

Vedclass · Answer

(C) Given equation: $4x + 3y + 2 = 0$ or $4x + 3y = -2$. To convert to normal form $x \cos \alpha + y \sin \alpha = p$,divide by $\sqrt{4^2 + 3^2} = 5$. Since $p$ must be positive,divide by $-5$: $\frac{-4}{5}x - \frac{3}{5}y = \frac{2}{5}$. Thus,$\cos \alpha = \frac{-4}{5}$,$\sin \alpha = \frac{-3}{5}$,and $p = \frac{2}{5}$. For intercept form $\frac{x}{a} + \frac{y}{b} = 1$,rewrite $4x + 3y = -2$ as $\frac{x}{-2/4} + \frac{y}{-2/3} = 1$. So,$a = \frac{-1}{2}$ and $b = \frac{-2}{3}$. Now,calculate $\frac{p \sec \alpha}{ab} = \frac{p}{ab \cos \alpha} = \frac{2/5}{(-1/2 	imes -2/3) 	imes (-4/5)} = \frac{2/5}{(1/3) 	imes (-4/5)} = \frac{2/5}{-4/15} = \frac{2}{5} 	imes \frac{-15}{4} = \frac{-3}{2}$.

If the normal form of the equation of a straight line $4x + 3y + 2 = 0$ 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}{ab} = $

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