Vertices of right triangle

[danimalswim]

I have spent 3 hours on this problem: Find all possible values of k so that (3,5), (5,-1), and (k,8) are the vertices of a right triangle. I tried to do this problem 10 times---3 pages front and back and not one thing worked!!! I started with the distance formula to find k. Then, I put the new ks into a distance formula (to find the length of all the sides to see if it was a right triangle). Finally, I checked to see if it was a right triangle. I ended up with 58=82, 113=181, 58=36, and a lot more. Please help me!!!

 

[tkhunny]

3 hours? Never do that. Take a break and come back.

Construct your fixed line segment. Let's call it AB, Starting at A(3,5) and Ending at B(5,-1)

1) Construct a line perpendicular to AB at A. This line is not parallel to y = 8, so it will intersect y = 8 in one point. This is one solution.

2) Construct a line perpendicular to AB at B. This line is also not parallel to y = 8, so it will intersect y = 8 in one point. This is one solution.

3) Your only remaining challenge is to see where it is possible to construct a right angle and have one Ray include A and the other include B. Is there such an angle on y = 8? There is some Distance Formula in your future or maybe you can recall a theorem about angles inscribed in circles and having rays that include the ends of a diameter.

Let's see what you get.