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Three lines x+2y+3=0,x+2y-7=0,a n d2x-y-...

Three lines `x+2y+3=0,x+2y-7=0,a n d2x-y-4=0` form the three sides of two squares. The equation of the four side of the each square is

A

2x-y+6=0

B

2x-y+8=0

C

2x-y-10=0

D

2x-y-14=0

Text Solution

Verified by Experts

The correct Answer is:
A, D
Distance between x+2y+3 = 0 and x+2y-7 = 0 si `10//sqrt(5)`. Let the remaining side parallel to 2x-y-4=0 be `2x-y+lambda = 0`. We have `(|lambda +4|)/(sqrt(5)) = (10)/(sqrt(5)) " or " lambda =6, -14`
Thus, the remaining side is 2x-y+6 = 0 2x-y-14=0.
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