Two sides of a rhombus are along the lines $x-y+1=0$ and $7 x-y-5=0$. If its diagonals intersect at $(-1…
Two sides of a rhombus are along the lines $x-y+1=0$ and $7 x-y-5=0$. If its diagonals intersect at $(-1,-2)$, then one of the vertices of this rhombus is
$(3,16)$
$(-1,-12)$
$\left(\frac{1}{3}, \frac{-8}{3}\right)$
$(2,9)$
Solution
Given equation of lines of two sides of rhombus are
$x-y+1=0 \& 7 x-y-5=0$
On solving these two equation, we get $(x, y)=(1,2)=$ point of intersection of two sides.
So equation of diagonal passing through $(1,2)$ and $(-1,-2)$
$\Rightarrow \frac{y-2}{y+2}=\frac{x-1}{1-(-1)} \Rightarrow y=2 x$ is one diagonal so equation of other diagonal $y=-\frac{1}{2} x+c$
$\Rightarrow-2=-\frac{1}{2} \times(-1)+\mathrm{C} \Rightarrow \mathrm{C}=-\frac{5}{2}$
$\Rightarrow$ equation of other diagonal is
$2 \mathrm{y}+\mathrm{x}+5=0$ ... (i)
So solving given equation with equation (i) we get $\left(\frac{1}{3},-\frac{8}{3}\right)$