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Q.

Let L1:3x+4y−1=0 and L2:5x−12y+2=0 be two given lines. Let image of every point on L1 with respect to a line L lies on L2 then possible equation of ‘L’ can be

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a

14x+112y−23=0

b

64x–8y–3=0

c

11x – 4y = 0

d

52y –45x = 7

answer is A, B.

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Detailed Solution

‘L’ must be angular bisector of L1 and L2 

3x+4y−15=±5x−12y+213

14x+112y−23=0 or 64x−8y−3=0

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