Question

A point equidistant from the lines $4x+3y+10=0,5x-12y+26=0$ and $7x+24y-50=0$ is

• A. $(1,–1)$
• B. $(0,0)$
• C. $(1,1)$
• D. $(0,1)$

Answer 1

The correct option is B $(0,0)$

Given equation are

$4x+3y+10=0$ …(i)

$5x-12y+26=0$ …(ii)

$7x+24y-50=0$ …(iii)

Let $(p,q)$ be the point which is equidistant from the given lines

$\because$ Distance of point $(x_1,y_1)$ from the line $ax+by+c=0$ is

$D=\frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}$

Distance of $(p,q)$ from equation (i) is

$d=\frac{|4p+3q+10|}{\sqrt{(4^2+3^2)}}=\frac{|4p+3q+10|}{5}$ …(iv)

Distance of $(p,q)$ from equation (ii) is

$d=\frac{|5p-12q+26|}{\sqrt{(5^2+{12}^2)}}=\frac{|5p-12q+26|}{13}$ …(v)

Distance of $(p,q)$ from equation (iii) is

$d=\frac{|7p+24q-50|}{\sqrt{(7^2+(24{)}^2)}}=\frac{|7p+24q-50|}{25}$ …(vi)

Given that $(p,q)$ is equidistant from the given lines, then

$\frac{|4p+3q+10|}{5}=\frac{|5p-12q+26|}{13}=\frac{|7p+24q-50|}{25}$

By using the given options, if we put $(0,0)$, then

$\frac{|10|}{5}=\frac{|26|}{13}=\frac{|-50|}{25}=2$

So, $(0,0)$ is the point which is equidistant from the given lines.

Hence, Option (C) is correct.