I am looking for an answer to this:

Given that ABCD is a rectangle with $A(2,1)$ and $y=5x+3$, $x-5y+3=0$ are the equations of its its diagonals, find the coordinates of the other edges of the rectangle.

Here is what I've done so far:

Solving simultaneous equations, I found the point $M(-\frac{1}{2},\frac{1}{2})$ which is the point of intersection of the two diagonals. Then, since $MA=MC$,

$x_M=\frac{x_A +x_B}{2} \Rightarrow -\frac{1}{2}=\frac{2 +x_B}{2} \Rightarrow x_C=-3$


$y_M=\frac{y_A +y_B}{2} \Rightarrow \frac{1}{2}=\frac{1 +y_B}{2}\Rightarrow y_C=0$

Therefore, $C(-3,0)$.

I am looking for a solution of how to find the coordinates of the other two edges.


Hints, suggestions:

The two points that you found are on $x-5y+3=0$. So the other two points, $B$ and $D$ are on the other line: $y=5x+3$.

Moreover, $B$ and $D$ are symmetric with respect to $M$, and $BM = AM = CM = DM$.

Knowing the distances $BM$ and $DM$, and the fact that $B$ and $D$ lie on $y = 5x+3$ should be enough to find these two points.

  • $\begingroup$ In other words: The four vertices are on a circle around the intersection of the diagonal $\endgroup$ – Hagen von Eitzen Feb 25 '16 at 22:10
  • $\begingroup$ I got it! I'll try and work this out. Thank you very much for your help! $\endgroup$ – P.D. Feb 25 '16 at 22:18
  • $\begingroup$ @P.D. You are very welcome! $\endgroup$ – GaussTheBauss Feb 25 '16 at 22:19

As the direction vectors $(5,1)$ and $(1,5)$ that we can read off from the diagonal equations obviously have the same lengths, we can find the directions of the angular bisectors between the diagonals by adding/subtracting, i.e., $(6,6)$ and $(4,-4)$, or simplified $(1,1)$ and $(1,-1)$. These directions are parallel to the sides of the rectangle. Thus by intersecting the line $A+t(1,1)$ or $A+t(1,-1)$ with the other diagonal, we can find $B$ and $D$.


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