Question

Let (h, k) be a fixed point, where h > 0, k > 0. A straight line passing through this point cuts the positive direction of the co-ordinate axies at the points P and Q. Find the minimum area of the triangle OPQ, O being the origin.

Answer 1

Equation of any line passing through (h, k) is

(y – h) = m(x – k)

Put y = 0, then x = (k – h/m)

So, the point P is ((k – h/m), 0)

Put x = 0, then y = h – mk

so, the point Q is (0, (h – mk))

Hence, the area of the triangle OPQ

= A = 1/2.OP.OQ

⇒ A = 1/2 .( k – h/m).(h – mk)

Since h > 0, k > 0 and m < 0, so the value of m is m = – h/k

Hence, the min area of the triangle OPQ

= 1/2(k + h/(h/k)) (h + h/k . k)

= 1/2 .2k .2h = 2hk