Question

Find the distance between (-2,4) & (5, 1) on the CCS. Please explain how you did it please.

Answer 1

A(x1, y1), B (x2,y2)

d=V(y2-y1)^2+(x2-x1)^2

d^2= (1-4)^2+[5-(-2)]^2=(-3)^2+ (5+2)^2=9+49=58

d=V58