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Math Student

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A Co-ordinate Geometry Question

Hi, there's a question which is puzzling me. I think I know how to tackle it but it doesn't seem to work out. >_< Can anyone help me?

Here's the question.

A triangle has vertices P (-2,2), Q (q,0) and R(5,3). We are told that the side PQ is twice as long as side QR. Find the possible values of q.

We are not supposed to use graphs, so what is the algebraic way of doing this?

Thank you for your guidance in advance.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: A Co-ordinate Geometry Question

How were you thinking of solving it?

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Math Student

Guest

Re: A Co-ordinate Geometry Question

I tried creating two quadratic equations through using Pythagoras' Theorem. However, I keep on getting stuck due to the fact that QR when doubled equals PQ. How would you multiply the numbers within the square root by two? Or, if we square everything, would it be multiplied by four?

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: A Co-ordinate Geometry Question

You sound like you may be trying to use Pythagorean theorem on the whole triangle itself.  This is wrong, since you aren't told that it is a right triangle.  Instead, use the theorem to calculate x (the length of side PQ) directly (this may be what you were doing already).  You should get:

and

Now you can compare the two by noting:

Which is what I think you were stating above.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."