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+y2−2x−2y=0
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C
x2+y2−x−y=0
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D
x2+y2+x+y=0
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Solution
The correct option is Dx2+y2−x−y=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 x1−y1=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+y−λ=1⟹λ=yx−1
PS:xλ+y1=1⟹λ=−xy−1
Now in order to find the locus of the point of intersection of these lines, eliminate the variable λ
From above two equations
yx−1=−xy−1
⟹y2−y=−x2+x
⟹y2+x2−x−y=0
So, The locus of the point of intersection of the lines PS and QR is y2+x2−x−y=0.