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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=0Equation of line AB is, x5+y−3=1⇒3x−5y=15 Any line perpendicular to AB is, 5x+3y=c So, coordinates of P and Q are (c5,0),(0,c3) respectively. Equation of AQ is, x5+yc/3=1⇒3yc=1−x5⇒1c=13y(1−x5) ...(1) Equation of BP is, xc/5−y3=1⇒1c=15x(1+y3) ...(2) From eqn(1) and (2), we have 13y(1−x5)=15x(1+y3) ⇒5x−x2=3y+y2⇒x2+y2−5x+3y=0

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