JEE Main · 2016 · 03 Apr 2016 · Offline · Straight Lines

The two given sides lie along $x-y+1=0$ and $7x-y-5=0$; solving them gives a vertex at $(1,2)$. Because the diagonals of a rhombus bisect each other at the centre $(-1,-2)$, the opposite vertex is the reflection of $(1,2)$ in that centre, namely $(-3,-6)$. The other two sides run parallel to the given pair through $(-3,-6)$, i.e. $x-y-3=0$ and $7x-y+15=0$. Pairing adjacent sides locates the remaining vertices; in particular $x-y+1=0$ meets $7x-y+15=0$ at $\left(\frac{1}{3},-\frac{8}{3}\right)$, which is a vertex of the rhombus.