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The line 3x2y=24 meets xaxis at A and yaxis at B. The perpendicular bisector of AB meets the line through (0,1) and parallel to xaxis 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)
Given : 3x2y=24 A(8,0) and B(0,12)
mid point of AB(4,6) and slope of AB=32
slope of perndicular =23
equation of perpendicular bisector is
(y+6)=23(x4) 2x+3y+10=0
and the line parallel to xaxis passing through (0,1) is y=1 solving with 2x+3y+10=0
C(72,1)

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