(b) (4, 1)
Let (a, b) be on 2x – 3y = 5 ⇒ 2a – 3b = 5
It is equidistance from (1, 2) and (3, 4)
(a – 1)2 + (b – 2)2 = (a – 3)2 + (6 – 4)2
a2 – 2a + 1 + b2 – 4b + 4 = a2 – 6a + 9 + b2 – 8b + 16
4a + 4b = 20
2a + 2b = 10
2a – 3b = 5
5b = 5
b = 1
∴ a = 4
∴ The point is (4, 1)