Question

Reduce the line $2x-3y+5=0$ in normal form and hence find perpendicular distance from the origin and angle made by the perpendicular with the positive x-axis.

• A. $\frac{2x}{\sqrt{13}}-\frac{3y}{\sqrt{13}}=\frac{5}{\sqrt{13}},p=\frac{5}{\sqrt{13}},\alpha =-{\tan }^{-1}\left(\frac{2}{3}\right)$
• B. $\frac{2x}{\sqrt{13}}-\frac{3y}{\sqrt{13}}=-\frac{5}{\sqrt{13}},p=\frac{5}{\sqrt{13}},\alpha =-{\tan }^{-1}\left(\frac{3}{2}\right)$
• C. $-\frac{2x}{\sqrt{13}}+\frac{3y}{\sqrt{13}}=\frac{5}{\sqrt{13}},p=\frac{5}{\sqrt{13}},\alpha ={\tan }^{-1}\left(\frac{2}{3}\right)$
• D. $-\frac{2x}{\sqrt{13}}+\frac{3y}{\sqrt{13}}=\frac{5}{\sqrt{13}},p=\frac{5}{\sqrt{13}},\alpha =-{\tan }^{-1}\left(\frac{3}{2}\right)$

Answer 1

The correct option is B $\frac{2x}{\sqrt{13}}-\frac{3y}{\sqrt{13}}=-\frac{5}{\sqrt{13}},p=\frac{5}{\sqrt{13}},\alpha =-{\tan }^{-1}\left(\frac{3}{2}\right)$

$2x-3y+5=0$
$\Rightarrow \frac{2x}{\sqrt{13}}-\frac{3y}{\sqrt{13}}+\frac{5}{\sqrt{13}}=0$
Hence, perpendicular distance is $p=\frac{5}{\sqrt{13}}$
And slope is $\frac{\frac{2}{\sqrt{13}}}{\frac{3}{\sqrt{13}}}=\frac{2}{3}.$
Hence, angle of perpendcular is $-{\tan }^{-1}\frac{3}{2}.$