Question

find a point on y axis which is equidistant from(6,5) and (_4,3)​

Answer 1

Step-by-step explanation:

It lies on Y-axis and is equidistant

let point be (0,y)

(x2)^2 +(y2-y)^2 = (x1)^2+ (y1-y)^2

36 + 16 -8y +y^2 = 25 + 9 - 6y +y^2

2y = 14

y = 7

so point is (0,7)