A diagonal of rhombus ABCD is a member of both the family of lines (x+y−1)+λ1(2x+3y−2)=0 and (x−y+2)+λ2(2x−3y+5) where λ1,λ2∈R and one of the vertex of the rhombus is (3,2). If area of the rhombus is 125 square units then the length of longer diagonal of the rhombus is ___________

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A diagonal of rhombus ABCD is a member of both the family of lines $(x + y - 1) + λ_{1} (2 x + 3 y - 2) = 0$ and $(x - y + 2) + λ_{2} (2 x - 3 y + 5)$ where $λ_{1}, λ_{2} \in R$ and one of the vertex of the rhombus is (3,2). If area of the rhombus is $12 \sqrt{5}$ square units then the length of longer diagonal of the rhombus is ___________

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answer is 10.

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Detailed Solution

Since diagonal is a member of both the families so it will pass through $(1, 0)$ and $(- 1, 1)$
Equation of diagonal AC is $x + 2 y - 1 = 0$
Since one of the vertex $(3, 2)$ which does not be on AC, so equation of BD is $2 x - y = 4$
Point of intersection of AC and BD is $P (\frac{9}{5}, - \frac{2}{5})$
If vertex B is $(3, 2)$ then vertex D is $(\frac{3}{5}, - \frac{14}{5})$ also $B D = \frac{12 \sqrt{5}}{5} (say d_{1})$
Area of rhombus $= \frac{1}{2} \times d_{1} \times d_{2}$
$\frac{1}{2} \times (\frac{12 \sqrt{5}}{5}) \times d_{2} = 12 \sqrt{5} d_{2} = 10$