JEE Main 2024 — Mathematics Coordinate Geometry

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

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

$$\Rightarrow {x}^{2}+4{y}^{2}+49+4xy+28y+14x=4{x}^{2}+{y}^{2}+64-4xy-16y+32x$$⇒x2+4y2+49+4xy+28y+14x=4x2+y2+64−4xy−16y+32x

$$\Rightarrow -3{x}^{2}+3{y}^{2}-15+8xy+44y-18x=0$$⇒−3x2+3y2−15+8xy+44y−18x=0

$$\Rightarrow 3{x}^{2}-3{y}^{2}-8xy+18x-44y+15=0$$⇒3x2−3y2−8xy+18x−44y+15=0

$$\Rightarrow {x}^{2}-{y}^{2}-\frac{8}{3}xy+6x-\frac{44}{3}y+5=0$$⇒x2−y2−38​xy+6x−344​y+5=0

Now, on comparing ${x}^{2}-{y}^{2}+2hxy+2gx+2fy+c=0$x2−y2+2hxy+2gx+2fy+c=0 with above equation we get,

$$\Rightarrow h=\frac{-4}{3},g=3,f=-\frac{22}{3},c=5$$⇒h=3−4​,g=3,f=−322​,c=5

$$\Rightarrow g+c+h-f=3+5-\frac{4}{3}+\frac{22}{3}$$⇒g+c+h−f=3+5−34​+322​

$$\Rightarrow g+c+h-f=14$$⇒g+c+h−f=14