Find the other 2 points of a rectangle?

Question

$PQRS$ is a rectangle with vertices $P(-4,-1)$ and $Q(-6,5),$ and $PQ=2(QR).$ Find the coordinates of $R$ and $S$? I'm so stuck please help! There are 2 answers for each point. almagest has the right answers, but i need work for it. I know that the slope of PS and QS has to be 1/3. But i tried making systems and I can't figure out how to solve for the other point. I also know the distance of PQ is 2(radical 10) so the distance of PS and QR have to be 4 (radical 10)

Answer 1

Since the vertices are written consecutively, $\overline{PQ}$ is a side of the rectangle. Its slope is

$m_{\overline{PQ}} = \dfrac{-1 - 5}{-4 - (-6)} = \dfrac{-6}{2} = -3$

Since the sides of a rectangle are perpendicular, the slope of $\overline{QR}$ must be the negative reciprocal of $-3$, which is $1/3$.

If you use the point-slope equation of a line

$y - y_0 = m(x - x_0)$

you can write the equation of the line containing $\overline{QR}$ in the form

$y - 5 = \dfrac{1}{3}(x + 6)$

You want to find a point $R$ on the line such that the distance between $P$ and $Q$ is twice the distance between $Q$ and $R$. The distance between $P$ and $Q$ is

$d(P, Q) = \sqrt{[-6 - (-4)]^2 + [5 - (-1)]^2} = \sqrt{(-2)^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}$

The distance between $Q$ and $R$ is thus $\sqrt{10}$. The distance between $R(x, y)$ and $Q(-6, 5)$ is

$d(Q, R) = \sqrt{(x + 6)^2 + (y - 5)^2} = \sqrt{10}$

Squaring both sides of the equation yields

$(x + 6)^2 + (y - 5)^2 = 10$

You can obtain an equation in $x$ by substituting

$\dfrac{1}{3}(x + 6)$

for $y - 5$, which yields

$(x + 6)^2 + \left[\dfrac{1}{3}(x + 6)\right]^2 = 10$

Solve the quadratic equation for the possible $x$-coordinates of point $R$. You can then obtain the $y$-coordinate of point $R$ using the equation of the line containing $\overline{PR}$. Subtract the coordinates of point $Q$ from those of point $R$. Add these values to point $P$ to find point $R$.

There are two possible solutions.

Answer 2

What happens if you rotate the point $(x,y)$ anticlockwise about the origin by $90^o$? You will find that it goes to $(-y,x)$.

Now think about rotating the point $P$ about $Q$ anticlockwise by $90^o$. Relative to $Q$ the coordinates of $P$ are $(2,-6)$ so when you rotate it, it goes to $(6,2)$. But you are also halving its length (if you want to get $R$) so that is $(3,1)$. That is relative to $Q$. So relative to the usual origin it is $(-3,6)$.

Now the coordinates of $S$ relative to $P$ are the same as those of $R$ relative to $Q$, so $S$ must be $(-1,0)$.

You can get the other possibility in a similar way.