Question

prove that mid-point of the Hypotenuse of a right angle is equidistant from it's three vertices P(-2, 5), Q(1, 3) & R(-1, 0)

Answer 1

First, start by drawing the corresponding points and form the triangle

Then, we can see that the hypotenuse is formed by vertices P and R.

Continue by finding the midpoint from this segment using the formula

$(x_m,y_m)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

apply to vertices P and R

$\begin{gathered} (x_m,y_m)=(\frac{-2+(-1)}{2},\frac{5+0}{2}) \\ (x_m,y_m)=(-\frac{3}{2},\frac{5}{2}) \\ (x_m,y_m)=(-1.5,2.5) \end{gathered}$

then, find the distance between the three vertices using the distance formula

$d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}$

Since we are talking about the mid-point of the hypotenuse then the distance from points P and R is going to be the same.

$\begin{gathered} d=\sqrt[]{(-1-(-1.5))^2+(0-(2.5))^2} \\ d=\frac{\sqrt[]{26}}{2} \end{gathered}$

Prove for Q.

$\begin{gathered} d=\sqrt[]{(1-(-1.5))^2+(3-(2.5))^2} \\ d=\frac{\sqrt[]{26}}{2} \end{gathered}$