Question

Consider the points A(0, 1) and B(2, 0) and P be a point on the line 4x + 3y + 9 = 0. Coordinates of P such that |PA – PB| is maximum are

(a) \(\left(\frac{-12}5, \frac{17}5\right)\)

(b) \(\left(\frac{-18}5, \frac{13}5\right)\)

(c) \(\left(\frac{31}7, \frac{31}7\right)\)

(d) (0, 0)

Answer 1

|PA - PB| ≤ AB

Thus |PA - PB| is max. if points A,B,P are collinear. Equation of AB is \(y -1 = \frac{0-1}{2-1}(x-0)\) ⇒ \(x+2y-2=0\)

Hence solving x + 2y - 2 = 0 & 4x + 3y + 9 = 0 we get \(\left(-\frac{84}{15}, \frac{13}5\right)\)