Question

Find thevalueofkforpointsp(2,4)is equidistantfromthe poinf *5,k) and (k,7)

Answer 1

let (2,4) be (x,y) ,

(5,k) be (x1,y1),

(k,7) be (x2, y2)

Therefore, x = 2, y = 4

x1 =5, y1 =k

x2 =k, y2 =7

(2,4) is equidistant

therefore by midpoint formula

x = x1 + x2/2

2 = 5 + k/2

4 - 5 = k

k = -1