Question

A line passing through $P(-2,3)$ meets the axes in $A$ and $B$. lf $P$ divides $AB$ in the ratio of $3:4$ then the perpendicular distance from $(1,1)$ to the line is

• A. $\frac{9}{\sqrt{5}}$
• B. $\frac{7}{\sqrt{5}}$
• C. $\frac{8}{\sqrt{5}}$
• D. $\frac{6}{\sqrt{5}}$

Answer 1

Answer: (C)

$\frac{8}{\sqrt{5}}$

Line passing through $P(-2,3)$ is,

$y-3=m(x+2)$

So, x-intercept $:x=-2-\frac{3}{m},y=0$

y-intercept $:y=2m+3,x=0$

So, P divides AB in ratio of $3:4$

$A=(-2-\frac{3}{m},0)$ $B=(0,2m+3)$

$P:\left(\frac{4x}{7},\frac{3y}{7}\right)$

$P:\left(\frac{4}{7}(-2-\frac{3}{m}),\frac{3}{7}(2m+3)\right)=(-2,3)$

By comparing, $\frac{-8}{7}-\frac{12}{7m}=-2$

$\frac{-12}{7m}=\frac{-6}{7}$

$m=2$

Equation of line is $y=2x+7$

So, distance from $(1,1)$ is

$d=\left|\frac{1-2-7}{\sqrt{1+4}}\right|=\left|\frac{+8}{\sqrt{5}}\right|$

$=\frac{8}{\sqrt{5}}$