Description : The point P is equidistant from A(1, 3), B(–3, 5) and C(5, –1). Then PB is equal to : -Maths 9th
Last Answer : (b) (2, 2)Let the point be P whose abscissa = ordinate = a. ∴ P ≡ (a, a) Given, PA = PB ⇒ (a + 1)2 + a2 = a2 + (a – 5)2 ⇒ 2a2 + 2a + 1 = 2a2 – 10a + 25 ⇒ 12a = 24 ⇒ a = 2. ∴ The point is (2, 2).