Question: Plot the points $P(5, 1)$, $Q(0, 6)$, and $R(-1, 1)$ on a coordinate plane. Where must the point $S$ be located so that the quadrilateral $PQRS$ is a square? Find the area of this square.
My try at it: Finding the area is easy. We can determine the distance $PQ$ between $P$ & $Q$ and then multiply it by itself. However, how do we find the point $S$?
The solution given in the website:
I think that the person has just figured out the point from the diagram and not by using some Math (though, technically he did use Math to draw the figure and everything, but I am referring to something like a formula). I have the following questions:
1) Is there a way to figure out the unknown point $S$ using some formula?
2) If yes, can we do it by knowing only the distance formula and the mid-point formula?
3) Regardless of the answer to the above question; is there a more difficult/advanced way of doing it?
My try at the questions:
1) I thought of using the distance formula but that cancelled everything and didn't give me the coordinates.
2) I think that the diagonals of a square meet at the same point. Using this fact, we can calculate the mid-point of $RP$, say $(x1, y1)$ and then (taking $(x, y)$ to be the coordinates of the point $S$), use $(\frac{(x - 0)}{2}, \frac{(y - 6)}{2}) = (x1, y1)$. Does this make sense?
3) I feel like there's always a more advanced way, maybe you can shed some light on it?
P.S. I found something weird. By calculating $(x + 5, y -1) = (5 - 0, 1 - 6)$ (I subtracted the coordinates of $S$ and $R$ on the $LHS$ and of $P$ & $Q$ on the $RHS$), I am getting the correct answer! Am I going anywhere with this?
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$\begingroup$$$ RS\parallel QP \quad \wedge\quad RS= QP\quad \text{(property of square)}\\ \begin{align*} \implies \vec{RS} &= \vec{QP}\\ \vec{OS} &= \vec{OR}+\vec{RS}\\ &= \vec{OR} + \vec{QP}\\ &= (-5,1) + (5-0,1-6)\\ &= (0,-4) \end{align*}$$ which is the reason behind the weird method you are mentioning.
$\endgroup$ 12 $\begingroup$1) Systematically use the distance formula with each of the co-ordinates $ P,Q $ and $R$.
2) If you are using mid-point formula, then it should be $\left(\frac{x+0}{2},\frac{y+6}{2}\right) = \left(\frac{-5+5}{2},\frac{1+1}{2}\right) = (0,1).$
$\endgroup$ 5 $\begingroup$I don't mean to be critical, but a lot of the above is very over-complicated. Here is the easy and quick way to do this: take advantage of the fact that the relative spacings of the PAIRS of corners in any parallelogram are the same. I.e., whatever you do to get from the top left corner to get to the bottom left corner, do that same move to get from the top RIGHT corner to the bottom RIGHT corner, etc. You can literally do it in seconds, with no formulas. For instance, let's say I have these three corners: (0,3) (6, 1), and (5, -2). When you draw this up on a graph, it is obvious that to get from (6, 1), to (5, -2), you just do "down 3 and left 1". Do the same thing from (0, 3) and you get (-1,0). Done. No need for slopes, intersections, formulas, etc.
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