Review question

# Can we find a rectangle from two vertices and a diagonal? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5967

## Solution

The points $A(1,2)$ and $B(-2,1)$ are two vertices of a rectangle $ABCD$. The diagonal $CA$ produced passes through the point $(2,9)$. Calculate the coordinates of the vertices $C$ and $D$.

Let’s draw a diagram to help us visualise the question.

The point $A$ has coordinates $(1,2)$ and $B$ has coordinates $(-2,1)$, so the gradient of $AB$ is $\frac{\text{change in y}}{\text{change in x}}=\frac{2-1}{1-(-2)}=\frac{1}{3}.$

Now we know that the line $BC$ will be perpendicular to $AB$, and so will have gradient $-3$. As it passes through $B(-2,1)$, it has equation $y-1=-3(x+2)$ (using the general formula $y-y_1=m(x-x_1)$), which we can rearrange to obtain $y=-3x-5.$

In order to find $C$, we must find where $BC$ intersects the line $AC$, whose equation we can find since we know two points on it, $(1,2)$ and $(2,9)$. The gradient of this line is then $\frac{9-2}{2-1}=7,$ and so it has equation $y-2=7(x-1)$, or $y=7x-5$.

So to find $C$ we must find where $AC$ meets $BC$, i.e., where $y=7x-5$ and $y=-3x-5$ meet. We see that $7x-5=-3x-5,$ from which we obtain $x=0$, and so $C$ is the point $(0,-5)$.

Finally, we need to find the coordinates of $D$. To get from $B$ to $A$, we move right $3$, up $1$. So if we go right $3$, up $1$ from $C$, we get to $D$, which therefore has coordinates $(3, -4)$.

We can think of “to get from $B$ to $A$” as the vector $\overrightarrow{BA}=\dbinom{3}{1}$, so the final sentence could be rewritten as:

As $\overrightarrow{BA}=\dbinom{3}{1}$ and $\overrightarrow{CD}=\overrightarrow{BA}$, we see that $D$ has coordinates $(3,-4)$.