Prove that M(­2,­1), A(1,6), T(8,3) and H(5,­4) are the vertices of a square.

Question

Prove that M(­2,­1), A(1,6), T(8,3) and H(5,­4) are the vertices of a square.

Answer 1

MA has slope (6-1)/(1-2) = -5

AT has slope (3-6)/(8-1) =-3/7

TH has slope (4-3)/(5-8) = - 1/3

HM has slope (4-1)/(5-3) = 3/2

Just not a square. Not even close, not even a parallelogram. Squares have opposite sides with the same slope. All 4 sides of the same length: But AM, MT, HT & HA all have different lengths: sqr26, sqr40, sqr10 and sqr20

squares also have 4 sides all of equal length.

Just plot the 4 points on a graph. Even a rough sketch shows nothing like a square.

Answer 2

Prove opposite sides are parallel (have same slope) and and congruent (have the same "distance".)