InterviewSolution
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InterviewSolution
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A line cuts the x-axis at `A (7, 0)` and the y-axis at `B(0, - 5)` A variable line PQ is drawn perpendicular to AB cutting the x-axis in P and the y-axis in Q. If AQ and BP intersect at R, find the locus of R |
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Answer» The equation of the line `AB` is `(x)/(7)+(y)/(-5)=1`……`(i)` `implies5x-7y=35` Equation of line perpendicular to `AB` is `7x+5y=lambda` ……..`(ii)` It meets `X`-axis at `P(lambda//7,0)` and `Y`-axis at `Q(0,lambda//5)`. The equations of lines `AQ` and `BP` are `(x)/(7)+(5y)/(lambda)=1` and `(7x)/(lambda)-(y)/(5)=1`, respectively. Let `R(h,k)` be their point of intersection of lines `AQ` and `BP`. Then, `(h)/(7)+(5k)/(lambda)=1` and `(7h)/(lambda)-(k)/(5)=1` `implies(1)/(5k)(1-(h)/(7))=(1)/(7h)(1+(k)/(5))` [on eliminating `lambda`] `impliesh(7-h)=k(5+k)` `impliesh^(2)+k^(2)-7h+5k=0` Hence, the locus of a point is `x^(2)+y^(2)-7x+5y=0`. |
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