Review question

# Can we show we have three corners of a rectangle? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5386

## Solution

The coordinates of the points $A$, $B$, $C$ are $(-2,1)$, $(2,7)$, $(5,5)$ respectively. Prove that these points form three corners of a rectangle…

To prove that the points form three corners of a rectangle, we need to show that the two lines $AB$ and $BC$ meet at right angles.

If this is true, then it must be possible to form a rectangle with these three points (and one other).

The slope of the line $AB$ is $m_{AB}=\dfrac{\Delta y}{\Delta x} = \dfrac{7-1}{2-(-2)} = \dfrac{6}{4} = \dfrac{3}{2}$.

The slope of $BC$ is $m_{BC} = \dfrac{5-7}{5-2} = \dfrac{-2}{3} = - \dfrac{2}{3}$.

The product of the two slopes is $\dfrac{3}{2} \times -\dfrac{2}{3} = -1$, so the lines are perpendicular and thus there exists a rectangle $ABCD$.

…and that $AB=2BC$.

The simplest approach here is to consider the similar triangles which appear in the diagram.

$AB$ is the hypotenuse of a right-angled triangle, with the other sides length $6$ and $4$.

$BC$ is the hypotenuse of a right-angled triangle, with the other sides length $3$ and $2$.

Therefore the two triangles are similar, with a scale factor of $2$, so we must have $AB$ = $2BC$.

If $D$ is the fourth corner of the rectangle, calculate the distance of $C$ from the diagonal $BD$.

There are a few ways to do this problem.

We could calculate $D$’s coordinates, but we can solve the problem without doing this.

This first method shown is more efficient than other methods, which require a greater amount of algebraic manipulation.

The distance of $C$ from the diagonal $BD$ will be the shortest possible distance, which is the perpendicular distance, shown as $h$ in the diagram below.

By Pythagoras, $BD$ is $\sqrt{5}d$ where $d$ is the length of $BC$. Using similar triangles, we also see that \begin{align*} \frac{h}{BC} &= \frac{CD}{BD} \\ \iff\quad \frac{h}{d} &= \frac{2d}{\sqrt{5}d} \\ \iff\quad h &= \frac{2d}{\sqrt{5}}. \end{align*} Using Pythagoras, we can find the length of $BC$. We have $d = \sqrt{3^2 + (-2)^2} = \sqrt{13},$ and therefore \begin{align*} h &= \dfrac{2\sqrt{13}}{\sqrt{5}} \\ &= \dfrac{2}{5} \sqrt{65}. \end{align*}

Or even more simply, we could consider area. The area of the rectangle is $d \times 2d = BD \times h = \sqrt{5}d \times h$, and so $h = \dfrac{2}{5} \sqrt{65}$.