Posted by Denis Borris on November 05, 2002 at 23:04:48:
In Reply to: Equidistant Midpoint of a Hypoteneuse posted by James on November 05, 2002 at 21:22:07:
: I'm enrolled in a online school and my text is online. The text is very vague in showing me how to do stuff. I am having trouble with a question. If anyone could help it would be much obliged. The question is:
: A right-angled triangle has vertices A(3,4), B(7,-4), C(-5,0). Show that the midpoint of the hypotenuse is equidistant from each vertex.
I did say this was MY way!
Add 5 to the x-points and 4 to the y-points to get rid of the minuses;
so this gives you A(8,8), B(12,0) and C(0,4).
Plot those; easy to calculate AB = AC = 4sqrt(5), and BC = 4sqrt(10).
So we have an isosceles right triangle.
Eary to calculate its height as 2sqrt(10), which is 4sqrt(10) / 2 (half of BC)
So there ya go!
Post a Followup