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Question

# A line cuts the x - axis at A(7, 0) and the y - axis at B(0, -5). 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 intersect at R, the locus of R is

A

x2+y2+7x5y=0

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B

x2+y27x+5y=0

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C

5x - 7y = 35

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D

none of these

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Solution

## The correct option is B x2+y2−7x+5y=0 Let P(a, 0) and Q(0, b) Slope of PQ =−ba −ba×57=−1⇒ ab=57 Equation of AQ is x7+yb=1 ⇒ b=7y7−x Equation of BP is xa−y5=1 ⇒ a=5x5+y so that 5x5+y×7−x7y=ab=57⇒ x(7−x)=y(5+y) ⇒ x2+y2−7x+5y=0 which is the locus of R(x, y).

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