# Midpoint M

• Dec 14th 2009, 12:42 PM
sologuitar
Midpoint M
Show that the midpoint M of the hypotenuse of a right triangle is equidistant from the vertices of the triangle.

NOTE: The given points on the picture are (0, r) and (s, 0).

See attachment.
• Dec 14th 2009, 01:35 PM
masters
Quote:

Originally Posted by sologuitar
Show that the midpoint M of the hypotenuse of a right triangle is equidistant from the vertices of the triangle.

NOTE: The given points on the picture are (0, r) and (s, 0).

See attachment.

Hi sologuitar,

Here's a way. Maybe someone has another. We'll see.

Draw the segment from M to the origin (0, 0). Find its distance using the distance formula.

You should find it to be $\displaystyle \frac{\sqrt{s^2+r^2}}{2}$

Find the coordinates of the midpoint M on the hypotenuse using the midpoint formula.

You should come up with $\displaystyle \left(\frac{s}{2}, \frac{r}{2}\right)$

Now find the distances from M to the vertices on the hypotenuse and you'll see they're the same as the distance from M to the origin.
• Dec 14th 2009, 04:01 PM
sologuitar
I see...
Quote:

Originally Posted by masters
Hi sologuitar,

Here's a way. Maybe someone has another. We'll see.

Draw the segment from M to the origin (0, 0). Find its distance using the distance formula.

You should find it to be $\displaystyle \frac{\sqrt{s^2+r^2}}{2}$

Find the coordinates of the midpoint M on the hypotenuse using the midpoint formula.

You should come up with $\displaystyle \left(\frac{s}{2}, \frac{r}{2}\right)$

Now find the distances from M to the vertices on the hypotenuse and you'll see they're the same as the distance from M to the origin.

This is more a proof question than a question I would normally see on a math test.