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, then find the locus of R.

The given line cuts the x-axis at A (7, 0) and the y-axis at B (0, -5).

Hence, the equation of the line AB is x/7 + y/-5 = 1 …… (1)

Hence, the equation becomes 5x-7y = 35.

We know that the equation of the line perpendicular to AB is

7x+5y = μ.

Hence, it will meet x-axis at P (μ/7, 0) and y-axis at Q (0, μ/5).

The equations of lines AQ and BP are x/7 + 5y/μ = 1 and 7x/μ – y/5 =1 respectively.

Let R(h, k) be the point of intersection of lines AQ and BP.

Then, h/7 + 5k/μ = 1

And 7h/μ – k/5 = 1

This gives 1/5k (1-h/7) = 1/7h (1+k/5)

So, h(7-h) = k(5+k)

This gives h² + k² – 7h + 5k = 0

Hence, the locus of the point is x²+y²-7x+5y = 0.