Question

The distance between the point (1,2) and (x,-1) is 5units. Find the possible value of x

Answer 1

Step-by-step explanation:

1-x/2+1=5. ( x1-x2/y1-y2)

1-x/3=5

1-x=5×3

1-x=15

x=-(15-1)

x=14