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Question

# The area of the quadrilateral formed by the lines 4xâˆ’3yâˆ’a=0,3xâˆ’4y+a=0, 4xâˆ’3yâˆ’3a=0 and 3xâˆ’4y+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. unitsGiven lines L1:4x−3y−a=0L2:3x−4y+a=0L3:4x−3y−3a=0L4:3x−4y+2a=0 We know the distance between two parallel lines is p1=∣∣ ∣∣−3a−(−a)√42+32∣∣ ∣∣=2a5unitsp2=∣∣ ∣∣a−2a√32+42∣∣ ∣∣=a5units Slope of the lines are m1=34,m2=43 Now, tanθ=∣∣∣m1−m21+m1m2∣∣∣⇒tanθ=∣∣ ∣ ∣ ∣∣34−431+1∣∣ ∣ ∣ ∣∣=724⇒sinθ=725 Now, the area of the parallelogram =p1p2sinθ=25×2a27×25=2a27 sq. units

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