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 Dx+3y+1=0 Equation of the line passing through A and B is y−5=5−3−4−2(x+4)⇒x+3y=11
and |AB|=2√10
Let P be (x,y).
So the perpendicular distance from P to line x+3y=11 d=|x+3y−11|√10
So area of △PAB=12⋅|x+3y−11|√10⋅2√10⇒12=|x+3y−11| ∴ 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(3−5)+2(5−y)−4(y−3)|=12⇒|−2x−6y+22|=24⇒x+3y=23
or x+3y+1=0