Given: A(2a, 0), B(0, 2b) and C(0, 0). Prove that the midpoint of the hypotenuse of right ΔABC is equidistant from vertices A, B, and C.

Question

Given: A(2a, 0), B(0, 2b) and C(0, 0). Prove that the midpoint of the hypotenuse of

right ΔABC is equidistant from vertices A, B, and C.

Answer 1

the midpoint of AB is by definition equidistant from A and B

but use the distance formula

from (2a,0) to the midpoint (a,b) = (2a-a)^2 + (0-b)^2 = a^2 + b^2 = d^2

same with from (0,2b) to (a,b) d^2= (0-a)^2 + (2b-b)^2 = a^2 + b^2

same with from the origin to (a,b): d^2 = (a-0)^2 + (b-0)^2 = a^2 + b^2

MA = MB = MC where M = (a,b) the midpoint