Question

A (-3, 2), B (3, 2) and C (-3, -2)are the vertices of the right triangle, right angled at A. Show that the mid point of the hypotenuse is equidistant from the vertices.

Answer 1

Mid point of BC

Hence proved

Answer 2

IF the triangle is not stated as rt angled in the question.

Slope of side AB=(2-2)/(3+3)=0

Slope of side BC= (2+2)/(3+3)=2/3

Slope of side CA=(2+2)/(3-3)=undefined

So AB is parallel to X-axis and CA is perpendicular to X-axis. Hence angle A of the triangle ABC is rt angle and BC is its hypotenuse.

So mid point of BC (say it D ) will be

D=((2-2)/2,(3-3)/2)=(0,0)

Length of  AD=√[((-3)^2+2^2]=√13

Length of BD=√[3^2+2^2]=√13

Length of CD=√[3^2+2^2]=√13

So we can say that mid point of the hypotenuse is equidistant from the vertices.