Can we show this parallelogram is a rectangle?

Can we show this parallelogram is a rectangle?

Question

The diagram shows a parallelogram $ABCD$ with $A$ and $C$ on the coordinate axes. The equation of $AB$ is $y=2x+1$ and the equation of $BC$ is $2y+x=12$.

1. Show that $ABCD$ is a rectangle.

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1. Find the coordinates of $A$, $B$, $C$ and $D$.

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1. Find the area of $ABCD$.

Solution

Since $ABCD$ is a parallelogram, we just need to check that $AB$ and $BC$ are perpendicular.

The equation of $AB$ is $y=2x+1$, which has gradient $2$, and the equation of $BC$ is $2y+x=12$, or $y=6-\frac{1}{2}x$ and so the gradient of $BC$ is $-\frac{1}{2}$.

Since $2\times -\frac{1}{2}=-1$, $AB$ and $BC$ are perpendicular. Therefore $ABCD$ is a rectangle.

Since $A$ lies on the $y$-axis, it lies at the $y$-intercept of $y=2x+1$, which is $1$. So $A=(0,1)$.

Next, since $C$ lies on the $x$-axis, which is $y=0$, we can substitute this in to the equation of $BC$ to find that $x=12$, so $C=(12,0)$.

To find the coordinates of $B$, we must find where $AB$ and $BC$ meet, so we solve the simultaneous equations $y=2x+1$ and $2y+x=12$. Substituting the first into the second gives $2(2x+1)+x=12,$ which gives us $x=2$, so $y=5$. Thus $B=(2,5)$.

Now let’s find the coordinates of $D$. To get from $B$ to $A$, we go left $2$ and down $4$, so we must do the same to get from $C$ to $D$, and so $D=(10,-4)$.

In terms of vectors, the last sentence says: $\overrightarrow{BA}=\overrightarrow{CD}=\dbinom{-2}{-4}$, so $D=(10,-4)$.

From the sketch, we can see (using Pythagoras’ theorem) that $AB=\sqrt{2^2+4^2}=\sqrt{20}=2\sqrt{5}$.

Likewise, $BC=\sqrt{5^2+10^2}=\sqrt{125}=5\sqrt{5}$.

Therefore the area of $ABCD$ is $AB\times BC=2\sqrt{5}\times 5\sqrt{5}=50$.