# Re: My way...

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Posted by Joel on November 06, 2002 at 00:59:33:

In Reply to: My way... posted by Denis Borris on November 05, 2002 at 23:04:48:

: : 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!

Oops. Guess I missed the point there. Forgot about the 3rd vertex. :(

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