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.
Text Solution
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Sol. Equation of the line AB is 
– 
= 1 [A(7, 0), B(0, –5)]
⇒ 5x –7y –35 = 0 … (1)
Equation of line PQ ⊥ AB is 7x + 5y + λ = 0 which meets axes of x and y at points P(– λ /7, 0) and Q (0, – λ /5) respectively.
Equation of AQ is, 
+ 
= 1
⇒ λ x – 35y – 7 λ = 0… (2)
Equation of BP is,
– 
= 1 ⇒ 35 x+ λ y + 5 λ = 0 … (3)
Locus of R, the point of intersection of (2) and (3) can be obtained by eliminating λ from these equations as follows
35x + (5 + y) 
= 0
⇒ 35x (x –7) + 35y (5 + y) = 0
⇒ x 2 + y 2 –7x + 5y = 0
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