Question

# The line 3xâˆ’2y=24 meets xâˆ’axis at A and y axis at B. The perpendicular bisector of AB meets the line through (0,âˆ’1) and parallel to xâˆ’axis at C. Then C is

A
(72,1)
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B
(152,1)
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C
(112,1)
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D
(132,1)
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Solution

## The correct option is A (−72,−1)Slope of the line joining of AB is 32 and it's mid-point is M(4,−6).The perpendicular bisector of this line will have the slope −23 and pass through M(4,−6)................(using m1.m2= -1 product of slopes of two perpendicular lines)∴ The equation of the perpendicular bisector of AB is 2x+3y+10=0→(1).The equation of the other line given in the problem is y=−1→(2).Lines (1) and (2) intersect at (−72,−1).

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