Two sides of a rhombus are along the lines, x - y + 1 = 0 and 7x -...

${d}_{1}: y - 2 = \dfrac{- 2 - 2}{- 1 - 1}\left(x - 1\right)<br /><br />$y - 2 = 2 \left(x - 1\right)$<br /><br />$y - 2x = 0$<br /><br />${d}_{2} ⊥ {d}_{1} ⇒ 2y + x = k, \left(- 1, - 2\right) Lies on it $∴ k = -5$

$⇒ 2y + x + 5 = 0$

Now point of intersection of ${d}_{2} and {L}_{1}$
$x - y + 1 = 0 . . . . . . \left(1\right)$

$x + 2y + 5 = 0 . . . . . . . \left(2\right)</p><p>on solving \left(1\right)\text{\text{ and } }$\left(2\right)

$- 3y - 4 = 0$

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

And point$of intersection of{d}_{2} and {L}_{2}$

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

$14x - 2y - 10 = 0 . . . . . . . . \left(4\right)<br />On solving \left(3\right)\text{\text{ and } }$\left(4\right)
And $y = -$\dfrac{8}{3}$<br /><br />$15x - 5 = 0$<br /><br />$⇒ x = \dfrac{1}{3}