Can we find the coordinates of this tilted rectangle?

Can we find the coordinates of this tilted rectangle?

Question

In the rectangle $ABCD$, $A$ and $B$ are the points $(4,2)$ and $(2,8)$ respectively.

Given that the equation $AC$ is $y=x-2$, find

1. the equation of $BC$,

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1. the coordinates of $C$,

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1. the coordinates of $D$,

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1. the area of the rectangle $ABCD$.

This question needs an exact method that goes beyond accurate drawing.

Suggestion

In the rectangle $ABCD$, $A$ and $B$ are the points $(4,2)$ and $(2,8)$ respectively.

This question needs an exact method that goes beyond accurate drawing.

Given that the equation $AC$ is $y=x-2$, find

1. the equation of $BC$,

If we know that $BC$ is perpendicular to $AB$, what can we say about the gradients of these lines?

We know the equations of $BC$ and $AC$ now…

How do we calculate the area of a rectangle?

Solution

In the rectangle $ABCD$, $A$ and $B$ are the points $(4,2)$ and $(2,8)$ respectively.

This question needs an exact method that goes beyond accurate drawing.

Given that the equation $AC$ is $y=x-2$, find

1. the equation of $BC$,

The gradient of $AB$ is $\dfrac{8-2}{2-4} = - 3$. The gradient of the perpendicular line $BC$ is thus $\dfrac {1}{3}$.

Since we know that the line passes through $B(2,8)$, the equation of $BC$ is given by $y-8 = \frac{1}{3}(x-2)$ which we can rearrange as $y = \dfrac{1}{3}x+\dfrac{22}{3}$ or $3y=x+22$.

We know that

$BC$

and

$AC$

meet at

$C$

. We can calculate their intersection point by setting their equations equal to each other (solving their equations simultaneously). We then have

$$\begin{align*} &&x-2&= \frac{1}{3}x+\frac{22}{3}&&\quad\\ \iff\quad&& \frac{2}{3}x&=\frac{28}{3}\\ \iff\quad&& x&=14. \end{align*}$$

The corresponding $y$ value is then $y=14-2=12$, so $C$ has coordinates $(14,12)$.

Going from $B$ to $C$ is the same translation as going from $A$ to $D$. (In terms of vectors, $\overrightarrow{BC}=\overrightarrow{AD}$.) To get from $B (2,8)$ to $C (14,12)$, we move $12$ units to the right and $4$ units up.

So if we start at $A (4,2)$ and move $12$ units to the right and $4$ units up, we arrive at $D(16,6)$.

The length of the side

$AB$

can be calculated using Pythagoras’ theorem, giving

$$\begin{align*} AB &= \sqrt{(x_A-x_B)^2+(y_A-y_B)^2}\\ &= \sqrt{(2)^2+(-6)^2}\\ &= \sqrt{40}\\ &= 2\sqrt{10}. \end{align*}$$

In the same way we can find

$BC$

:

$$\begin{align*} BC &= \sqrt{(x_B-x_C)^2+(y_B-y_C)^2} \\ &= \sqrt{(-12)^2+(-4)^2} \\ &=\sqrt{160}\\ &= 4\sqrt{10}. \end{align*}$$

If we multiply the two side lengths of our rectangle, we find its area. We therefore have

$$\begin{align*} \text{area} &=AB \times BC \\ &= 2\sqrt{10} \times 4\sqrt{10} \\ &= 8\times10 \\ &= 80. \end{align*}$$

Here you can see a complete sketch of the rectangle.