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Two sides of a rhombus are along the lines, $x -y+ 1 = 0$ and $7x-y-5 =0.$ If its diagonals intersect at $(-1,-2),$ then which one of the following is a vertex of this rhombus?
$\left( {\frac{1}{3}, - \frac{8}{3}} \right)$
$\left( { - \frac{{10}}{3}, - \frac{7}{3}} \right)$
$\left( { - 3, - 9} \right)$
$\;\left( { - 3, - 8} \right)$
Solution
Equation of angle bisector of the lines $x-y+1=0$ and $7 \mathrm{x}-\mathrm{y}-5=0$ is given by
$\frac{x-y+1}{\sqrt{2}}=\pm \frac{7 x-y-5}{5 \sqrt{2}}$
$\Rightarrow 5(\mathrm{x}-\mathrm{y}+1)=7 \mathrm{x}-\mathrm{y}-5$
and
$5(x-y+1)=-7 x+y+5$
$\therefore 2 \mathrm{x}+4 \mathrm{y}-10=0 \Rightarrow \mathrm{x}+2 \mathrm{y}-5=0$ and
$12 x-6 y=0 \Rightarrow 2 x-y=0$
Now equation of diagonals are
$(x+1)+2(y+2)=0 \Rightarrow x+2 y+5=0$ …….$(1)$
and
$2(x+1)-(y+2)=0 \Rightarrow 2 x-y=0$ ….$(2)$
Clearly $\left(\frac{1}{3},-\frac{8}{3}\right)$ lies on $( 1)$
