1. ## Coordinate geometry question.

P - (-1, 11), Q - (2,5), R - (-2, 3). Angle PQR = 90 degrees. The line PQ is produced to S, so that QS = PQ. Calculate the coordinates of S.

thanks

2. ## Re: simple coordinate geometry question that is stumping me for some strange reason

Originally Posted by Ife
P - (-1, 11), Q - (2,5), R - (-2, 3). Angle PQR = 90 degrees. The line PQ is produced to S, so that QS = PQ. Calculate the coordinates of S.

thanks
1. Draw a sketch!

2. Let $\displaystyle \overrightarrow{OP} = \vec p = \langle -1, 11 \rangle$

$\displaystyle \overrightarrow{OQ} = \vec q = \langle 2, 5 \rangle$

3. $\displaystyle \overrightarrow{PQ} = \vec q - \vec p = \langle 3,-6 \rangle = \vec d$

4. $\displaystyle \overrightarrow{QS} = \vec q + \vec d = \langle 5, -1 \rangle$

5. Therefore the point S has the coordinates S(5, -1)

3. ## Re: simple coordinate geometry question that is stumping me for some strange reason

another method. Given P,Q and PQ=QS no further info is required.Slope of PQ =-6/3 = slope of QS.S is then 3 units right ofQ and 6 units down from Q (5,-1)

4. ## Re: simple coordinate geometry question that is stumping me for some strange reason

Originally Posted by bjhopper
another method. Given P,Q and PQ=QS no further info is required.Slope of PQ =-6/3 = slope of QS.S is then 3 units right ofQ and 6 units down from Q (5,-1)
thanks, but moving 3 across and 6 down to (5, -1) guarantees that QS = PQ?

5. ## Re: simple coordinate geometry question that is stumping me for some strange reason

Hello, Ife!

Thanks, but moving 3 across and 6 down to (5, -1) guarantees that QS = PQ?

Yes, it does!

Moving from $\displaystyle P(\text{-}1.11)$ to $\displaystyle Q(2,5)$, we move 3 right and 6 down.

Moving from $\displaystyle Q(2,5)$ to $\displaystyle S(5,\text{-}1)$, we move 3 right and 6 down.

The distances are equal . . . aren't they?