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9.Straight Line
hard
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 $\mathrm{g}+\mathrm{c}+\mathrm{h}-\mathrm{f}$ equals
A
$14$
B
$6$
C
$8$
D
$29$
(JEE MAIN-2024)
Solution
Cocus of point $\mathrm{P}(\mathrm{x}, \mathrm{y})$ whose distance from
Gives
$\mathrm{X}+2 \mathrm{y}+7=0 \& 2 \mathrm{x}-\mathrm{y}+8=0$ 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
$(x-3 y+1)(3 x+y+15)=0$
$3 x^2-3 y^2-8 x y+18 x-44 y+15=0$
$x^2-y^2-\frac{8}{3}xy+6x-\frac{44}{3} y+5=0$
$x^2-y^2+2 h x y+2 g x 2+2 f y+c=0$
$h=\frac{4}{3}, g=3, f=-\frac{22}{3}, c=5$
$g+c+h-f=3+5-\frac{4}{3}+\frac{22}{3}=8+6=14$
Standard 11
Mathematics
