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Question

The locus of P such that the area of ΔPAB=12 sq. units, where A=(2,3) and B(4,5) is/are

A
x+3y=23
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B
x+3y+23=0
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C
x+3y=1
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D
x+3y+1=0
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Solution

The correct option is D x+3y+1=0
Equation of the line passing through A and B is
y5=5342(x+4)x+3y=11
and |AB|=210
Let P be (x,y).
So the perpendicular distance from P to line x+3y=11
d=|x+3y11|10
So area of PAB=12|x+3y11|1021012=|x+3y11|
locus of P will be
x+3y=23 or x+3y+1=0


Alternate solution:
Let P(x,y)
so area of the triangle will be
12|x(35)+2(5y)4(y3)|=12|2x6y+22|=24x+3y=23
or x+3y+1=0

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