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+ax−by=0
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B
x2+y2−ax−by=0
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C
x2−y2−ax−by=0
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D
x2+y2+ax+by=0
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Solution
The correct option is Bx2+y2−ax−by=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 xb−ya=λ, where λ is parameter ⇒xbλ−yaλ=1
It cuts the coordinate axes at R(bλ,0) and S(0,−aλ) Equation of line PS and QR are xa−yaλ=1 and xbλ+yb=1
Let T(h,k) be the point of intersection of PS and QR Then, ha−kaλ=1 and hbλ+kb=1⇒ha−1=kaλ and hbλ=1−kb⇒h−ak=1λ and b−kh=1λ⇒h−ak=b−kh⇒h2+k2−ah−bk=0