Question

The point (3 , k) is a distance of 5 units from (0 , 1). Find the two possible values of k.

Answer 1

d =√(x2-x1)^2 + (y2-y1)^2
......._____________
5 =√(3-0)^2 + (k-1)^2
.......__________
5 =√3^2 + (k-1)^2

25 = 9 + (k-1)^2

16 = k^2 - 2k + 1

k^2 -2k - 15 = 0

(k - 5)(k + 3) = 0

k - 5 = 0.....or.......(k + 3) = 0

k = 5.........or.......k = -3