Question

Find the locus of a point which moves so that the perpendiculars drawn from it to the two straight lines $3x-4y+10=0$ and $5x-12y-10=0$ are equal.

Answer 1

Given,

Let $P(a,b)$ be any point

and $3x-4y+10=0$ …….$\left(1\right)$

$5x-12y-10=0$ …….$\left(2\right)$

if perpendicular from line $\left(1\right)$ & $\left(2\right)$ to P are equal then

$\left|\frac{3a-4b+10}{\sqrt{3^2+4^2}}\right|=\left|\frac{5a-12b-10}{\sqrt{(5{)}^2+(12{)}^2}}\right|$

$\frac{3a-4b+10}{5}=\frac{5a-12b-10}{13}$

$39a-52b+130=25a-60b-50$

$14a+8b+180=0$

$7a+4b+90=0$ ………$\left(3\right)$

Hence locus of point P is a straight line $\left(3\right)$.