Two sides of a rhombus are along the lines, x - y + 1 = 0 and 7 x - y - 5 = 0 . If its diagonals intersect…

$d_1 :y-2=\frac{-2-2}{-1-1} \left(x-1\right)$

$y-2=2\left(x-1\right)$

$y-2x=0$

$d_2\perp d_1 \Rightarrow 2y+x=k\mathrm{ ,}$ $\left(-1, -2\right)$ Lies on it $\therefore k=-5$

$\Rightarrow 2y+x+5=0$

Now point of intersection  of $d_2$ and $L_1$
$x-y+1=0 ......\left(1\right)$

$x+2y+5=0 .......\left(2\right)$

on solving $\left(1\right) \& \left(2\right)$

$-3y-4=0$

$y= -\frac{4}{3}$

And point of intersection of $d_2$ and $L_2$

$x+2y+5=0 ......\left(3\right)$

$14x-2y-10=0 ........\left(4\right)$
On solving $\left(3\right) \& \left(4\right)$
And $y= -\frac{8}{3}$

$15x-5=0$

$\Rightarrow$ $x=\frac{1}{3}$