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

moralesml

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Feb 27, 2019
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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

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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

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Feb 27, 2019
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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!
 

mmm4444bot

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Hello. Why have you typed "please disregard" in your thread title?

If you're trying to say that your question has been answered, please click the green Mark 'Solved' button, at the top of your thread.

😎
 

Otis

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Apr 22, 2015
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Did your son draw a picture? If not, I'd start there.

😎
 
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