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Question

# A Iine cuts the x-axis at A(7,0) and y-axis at B(0,âˆ’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, then the locus of R is-

A
x(x7)+y(y+5)=0
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B
x(x7)y(y+5)=0
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C
x(x+7)+y(y5)=0
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D
x(x7)+y(y5)=0
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Solution

## The correct option is A x(x−7)+y(y+5)=0The given lines cuts the x−axis at A(7,0) and the y−axis is B(0,−5).Hence, the equation of the line AB is x7+y−5=1............(1)Hence, the equation becomes 5x−7y=35We know that,the equation of the line perpendicular to AB is 7x+5y=μ.Hence, it wil meet x−axis at P(μ7,0)andy−axisatQ(0,μ5)The equations of lines AQ and BP are x7+5yμ=1and7xμ−y5=1Let R(h,k) be the point of intersection of lines AQ and BP.Then,h7+5kμ=17hμ−k5=115k(1−h7)=17h(1+k5)so,h(7−h)=k(5+k)h2+k2−7h+5k=0Hence, the locus of R is x2+y2−7x+5y=0

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