If x^2-y^2+2 h x y+2 g x+2 f y+c=0 is the locus of a point, which moves such that it is al…

Question

If $x^2-y^2+2 h x y+2 g x+2 f y+c=0$ is the locus of a point, which moves such that it is always equidistant from the lines $x+2 y+7=0$ and $2 x-y+8=0$, then the value of $g+c+h-f$ equals

Accepted Answer

Cocus of point $$\mathrm{P}(\mathrm{x}, \mathrm{y})$$ whose distance from Gives
$X+2 y+7=0$ \& $2 x-y+8=0$ are equal is $$\frac{x+2 y+7}{\sqrt{5}}= \pm \frac{2 x-y+8}{\sqrt{5}}$$

$(x+2 y+7)^2-(2 x-y+8)^2=0$

Combined equation of lines

$$\begin{aligned}
& (x-3 y+1)(3 x+y+15)=0 \
& 3 x^2-3 y^2-8 x y+18 x-44 y+15=0 \
& x^2-y^2-\frac{8}{3} x y+6 x-\frac{44}{3} y+5=0 \
& x^2-y^2+2 h x y+2 g x 2+2 f y+c=0 \
& h=\frac{4}{3}, g=3, f=-\frac{22}{3}, c=5 \
& g+c+h-f=3+5-\frac{4}{3}+\frac{22}{3}=8+6=14
\end{aligned}$$

Solutions · 1