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.
Idk why I am not getting this out... please help..
thanks
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.
Idk why I am not getting this out... please help..
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)
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?