JEE Main 2024 — Straight Lines Question with Solution

Let $P(x,y)$ be the point whose distance from $x+2y+7=0$ and $2x-y+8=0$ is equal.

$\Rightarrow \frac{x+2y+7}{\sqrt{5}}=\pm \frac{2x-y+8}{\sqrt{5}}$

$\Rightarrow x^2+4y^2+49+4xy+28y+14x=4x^2+y^2+64-4xy-16y+32x$

$\Rightarrow -3x^2+3y^2-15+8xy+44y-18x=0$

$\Rightarrow 3x^2-3y^2-8xy+18x-44y+15=0$

$\Rightarrow x^2-y^2-\frac{8}{3}xy+6x-\frac{44}{3}y+5=0$

Now, on comparing $x^2-y^2+2hxy+2gx+2fy+c=0$ with above equation we get,

$\Rightarrow h=\frac{-4}{3}, g=3, f=-\frac{22}{3}, c=5$

$\Rightarrow g+c+h-f=3+5-\frac{4}{3}+\frac{22}{3}$

$\Rightarrow g+c+h-f=14$