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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 in P and the y-axis in Q. If AQ and BP intersect at R, the locus of R is

A
x2+y2+7x+5y=0
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B
x2+y27x+5y=0
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C
x2+y2+7x5y=0
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D
x2y27x+5y=0
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Solution

## The correct option is B x2+y2−7x+5y=0Equation of line AB is x7+y−5=1⇒5x−7y=35 ...(1)Equation of line perpendicular to AB is 5x+7y=λ ...(2)It meets x-axis at P(λ7,0) and y-axis at Q(0,λ5)The equation of lines AQ and BP are x7+5yλ=1 and 7xλ−y5=1 respectively Let R(h,k) be their point of intersection of lines AQ and BPThen h7+5kλ=1 and 7hλ−k5=1⇒15k(1−h7)=17h(1+k5)⇒h(7−h)=k(5+k)⇒h2+k2−7h+5k=0Hence, the locus of a point is x2+y2−7x+5y=0

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