# points on an unknown line: "A special right triangle has a hypotenuse that is 13 units long...."

#### moralesml

##### New member
Hello! I am a mom trying (proactively) to help my son.... Here's the problem I'm looking at

A special right triangle has a hypotenuse that is 13 units long, one side of length 5 units and another side of x units. The unknown side has one endpoint at (1,3). If the side with length x lines on a vertical line, what is a midpoint of that side of triangle?

These are practice problems with multiple choice answers, but it's been awhile.... Can someone help me figure this out?

Thank you!

#### MarkFL

##### Super Moderator
Staff member
Hello, and welcome to FMH!

Normally we don't generally provide complete solutions, but since you are a mom helping your son, I'm going to do so.

First we need to figure out the length of the unknown side. The Pythagorean theorem tells us we may state:

$$\displaystyle x^2+5^2=13^2$$

$$\displaystyle x^2=13^2-5^2=169-25=144=12^2$$

Now, since $$x$$ represent a length of a line, it cannot be negative, so we find:

$$\displaystyle x=12$$

If one endpoint of the side 12 units long is at (1,3), and this side lies along a vertical line, then the other endpoint could be at:

$$\displaystyle (1,3\pm12)=(1,3(1\pm4))$$

Notice this represent two possible points...one above and one below.

And so the mid-point of that side would be:

$$\displaystyle (x,y)=\left(\frac{1+1}{2},\frac{3+3(1\pm4)}{2}\right)=\left(1,3(1\pm2)\right)$$

Does that make sense?

#### moralesml

##### New member
Thank you! I was on the right track with the Pythagorean theorem but somehow confused myself. Now, I can help with the question. He ran out just after saying "I need help with this after practice, please." So thank you, thank you!

Staff member