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Question

# A line cuts the x-axis at A(5,0) and the y-axis at B(0,−3). A variable line PQ is drawn perpendicular to AB cutting the x--axis at P and the y-axis at Q. If AQ and BP meet at R, then the locus of R is

A
x2+y25x+3y=0
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B
x2+y2+5x+3y=0
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C
x2+y2+5x3y=0
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D
x2+y25x3y=0
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Solution

## The correct option is A x2+y2−5x+3y=0Line AB is x5+y−3=1⇒3x−5y=15any perpendicular line to AB5x+3y=λ So P(λ5,0),Q(0,λ3)AQ is x5+yλ/3=1⇒3yλ=1−x5⇒1λ=13y(1−x5)...(1)and BP is xλ/5−y3=1⇒5xλ=1+1λ=15(1+y3)....(2)13y(1−x5)=15x(1+y3)⇒5(1−x5)=3y(1+y3)⇒5x−x2=3y+y2⇒x2+y2−5x+3y=0

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