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Question

A Iine cuts the x-axis at A(7,0) and 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, 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(x7)+y(y+5)=0
The given lines cuts the xaxis at A(7,0) and the yaxis is B(0,5).
Hence, the equation of the line AB is x7+y5=1............(1)
Hence, the equation becomes 5x7y=35
We know that,
the equation of the line perpendicular to AB is 7x+5y=μ.
Hence, it wil meet xaxis at P(μ7,0)andyaxisatQ(0,μ5)
The equations of lines AQ and BP are x7+5yμ=1and7xμy5=1
Let R(h,k) be the point of intersection of lines AQ and BP.
Then,h7+5kμ=17hμk5=115k(1h7)=17h(1+k5)so,h(7h)=k(5+k)h2+k27h+5k=0
Hence, the locus of R is x2+y27x+5y=0

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