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Question

A diagonal of rhombus ABCD is member of both the families of lines
(x+y1)+λ1(2x+3y2)=0 and
(xy+2)+λ2(2x3y+5)=0
where λ1 and λ2R and one of the vertex of rhombus is (3,2). If area of the rhombus is 125 square units, then the length of the longer diagonal of the rhombus is

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Solution

Intersection point of x+y1=0 and 2x+3y2=0 is (1,0)
Intersection point of xy+2=0 and 2x3y+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+2y1=0.

Vertex (3,2) does not lie on x+2y1=0.
Finding distance of (3,2) from x+2y1=0:
Distance =∣ ∣3+2×2112+22∣ ∣=65

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=2×65=125

Given, area =125
12×d1×d2=125
d2=10

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