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Question

The area of the quadrilateral formed by the lines 4x3ya=0,3x4y+a=0, 4x3y3a=0 and 3x4y+2a=0 is

A
2a211 sq. units
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B
2a29 sq. units
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C
a27 sq. units
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D
2a27 sq. units
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Solution

The correct option is D 2a27 sq. units
Given lines
L1:4x3ya=0L2:3x4y+a=0L3:4x3y3a=0L4:3x4y+2a=0
We know the distance between two parallel lines is
p1=∣ ∣3a(a)42+32∣ ∣=2a5unitsp2=∣ ∣a2a32+42∣ ∣=a5units
Slope of the lines are
m1=34,m2=43
Now,
tanθ=m1m21+m1m2tanθ=∣ ∣ ∣ ∣34431+1∣ ∣ ∣ ∣=724sinθ=725

Now, the area of the parallelogram
=p1p2sinθ=25×2a27×25=2a27 sq. units

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