Straight Lines Question 11
Question 11 - 2024 (30 Jan Shift 2)
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 $\mathrm{x}+2 \mathrm{y}+7=0$ and $2 \mathrm{x}-\mathrm{y}+8=0$, then the value of $\mathrm{g}+\mathrm{c}+\mathrm{h}-\mathrm{f}$ equals
(1) 14
(2) 6
(3) 8
(4) 29
Show Answer
Answer (1)
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
$(x-3 y+1)(3 x+y+15)=0$
$3 x^{2}-3 y^{2}-8 x y+18 x-44 y+15=0$
$x^{2}-y^{2}-\frac{8}{3} x y+6 x-\frac{44}{3} y+5=0$
$x^{2}-y^{2}+2 h x y+2 g x 2+2 f y+c=0$
$h=\frac{4}{3}, g=3, f=-\frac{22}{3}, c=5$
$g+c+h-f=3+5-\frac{4}{3}+\frac{22}{3}=8+6=14$
