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Question

A straight line L1:xa+yb=1 intersects the x-axis and y-axis at P and Q respectively and a straight line L2 perpendicular to L1 cuts the x-axis and y-axis at R and S respectively. The locus of the point of intersection of the lines PS and QR is

A
x2+y2+axby=0
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B
x2+y2axby=0
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C
x2y2axby=0
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D
x2+y2+ax+by=0
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Solution

The correct option is B x2+y2axby=0
Given equation of the line L1 be
xa+yb=1(i)
It cuts the coordinates axes at P(a,0) and Q(0,b).
The equation of line L2 is
xbya=λ, where λ is parameter
xbλyaλ=1

It cuts the coordinate axes at
R(bλ,0) and S(0,aλ)
Equation of line PS and QR are
xayaλ=1 and xbλ+yb=1

Let T(h,k) be the point of intersection of PS and QR
Then,
hakaλ=1 and hbλ+kb=1ha1=kaλ and hbλ=1kbhak=1λ and bkh=1λhak=bkhh2+k2ahbk=0

Hence, the locus of T(h,k) is
x2+y2axby=0

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