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If x+2y+3=0,x+2y7=0 and 2xy4=0 form three sides of a square, the equation of the fourth side nearer the point (1, - 1) is

a
2x−y−6=0
b
2x−y+6=0
c
2x−y−14=0
d
2x−y+14=0

detailed solution

Correct option is B

Equation of the fourth side is 2x−y−k=0such that 4−k5=±3+75⇒k=−6 or 14So the required equation is 2x−y+6=0, it isnearer to (1, –1).

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