median in analytic geometry

**booper563** · December 11th 2008, 09:53 AM

Alrighty I was doing good until here.

The question says.

Prove that the median, AM, from the right angle to the hypotenuse is one half the length of the hypotenuse, BC.
The points are
A (-3,1) B(-3,7) and C (5,1)

I have lots of questions like this and I'm not sure what to do or what formula to use. And whats exactly the median and AM mean?

Deatails would be very helpful.
Thanks@

**masters** · December 11th 2008, 10:37 AM

Originally Posted by booper563

Alrighty I was doing good until here.

The question says.

Prove that the median, AM, from the right angle to the hypotenuse is one half the length of the hypotenuse, BC.
The points are
A (-3,1) B(-3,7) and C (5,1)

I have lots of questions like this and I'm not sure what to do or what formula to use. And whats exactly the median and AM mean?

Deatails would be very helpful.
Thanks@

Hello booper,

Let's see. From the looks of things you have a $\overline{AB}\perp \overline{AC}$ .

I can tell that $\overline{AB}$ is vertical since the x-coordinates are the same. I can also tell that $\overline{AC}$ is horizontal since the y-coordinates are the same.

The right angle is at A.

The hypotenuse is $\overline{BC}$

Use the distance formula: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

to find the length of $\overline{BC}$

Now find the midpoint M of $\overline{BC}$ using the midpoint formula:

$M_{\overline{BC}}=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

$\overline{AM}$ is the median drawn from the right angle to the midpoint of the hypotenuse M

Next use the distance formula again to find the length of $\overline{AM}$ and see if it is half of $\overline{BC}$

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