The line x+y=1 cuts the coordinate axes at P and Q and a line perpendicular to it meet the axes - Maths - Conic Sections

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 B) x2+y2-2x-2y=0 C) x2+y2-x-y=0 D) x2+y2+x+y=0

Utsav · Accepted Answer

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 ⊥to PQ is
x1-y1=λ∴Coordinates of R is (λ,0) and coordinates
of S is (0,-λ)Now by intercept form,equations of PS and QR
arePS :- x1+y-λ =1 ⇒λ=yx-1 (1)QR:- xλ+y1=1
⇒λ=-xy-1 (2)Now in order to find the locus of the point
of intersection of these lines ,eliminate the variable
λEquating R H S of (1) and (2) as L H S is same
yx-1=-xy-1⇒y(y-1)=-x(x-1)⇒y2-y=-x2+x⇒x2+y2-x-y =0
∴The locus of the point of intersection of the lines PS and
QR is x2+y2-x-y =0 Hence option (C) is correct

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 B) x2+y2-2x-2y=0 C) x2+y2-x-y=0 D) x2+y2+x+y=0

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 B) x2+y2-2x-2y=0 C) x2+y2-x-y=0 D) x2+y2+x+y=0