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Question

The line x+y=1 cuts the coordinate axes at P and Q and a line perpendicular to it meet the axes in R and S. The equation to the locus of the point of intersection of the lines PS and QR is?

A
x2+y2=1
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B
x2+y22x2y=0
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C
x2+y2xy=0
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D
x2+y2+x+y=0
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Solution

The correct option is D x2+y2xy=0
Equation of line PQ in intercept form
x1+y1=1

Coordinates of P is (1,0)and coordinates of Q is (0,1)

Any line perpendicular to PQ is x1y1=λ

Coordinates of R is (λ,0)and coordinates of S is (0,λ)

Now, by intercept form, equation of PS and QR are
PS:x1+yλ=1λ=yx1

PS:xλ+y1=1λ=xy1

Now in order to find the locus of the point of intersection of these lines, eliminate the variable λ

From above two equations

yx1=xy1

y2y=x2+x

y2+x2xy=0

So, The locus of the point of intersection of the lines PS and QR is y2+x2xy=0.

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