Intersection point of x+y−1=0 and 2x+3y−2=0 is (1,0)
Intersection point of x−y+2=0 and 2x−3y+5=0 is (−1,1)
Since diagonal of the rhombus is a member of both the given families of lines, it will pass through the points (1,0) and (−1,1).
⇒ Equation of one diagonal is x+2y−1=0.
Vertex (3,2) does not lie on x+2y−1=0.
Finding distance of (3,2) from x+2y−1=0:
Distance =∣∣
∣∣3+2×2−1√12+22∣∣
∣∣=6√5
Let length of one diagonal be d1 and length of the other be d2.
We know that diagonals of a rhombus bisect each other.
∴d1=2×6√5=12√5
Given, area =12√5
⇒12×d1×d2=12√5
⇒d2=10