Rectangle $A B C D$ has diagonal $B D$ with endpoints $B(4,2)$ and $D(12,8)$. Diagonal $A C$ lies on the line with equation $x+2 y-18=0$. Determine the area of $A B C D$.
Suppose $C$ has coordinates $(a, b)$ and is the vertex of the rectangle to the right of $B$ and on the line $x+2 y-18=0$ as pictured below.

![](https://cdn.mathpix.com/cropped/2024_04_17_a164435ebfb99ea5f1c4g-17.jpg?height=667&width=1399&top_left_y=325&top_left_x=409)

By finding the coordinates of $C$, we will be able to compute the lengths of $B C$ and $C D$ and use them to find the area of the rectangle.

First, we find the equation of the line through $B$ and $D$.

Since this line passes through the points $(4,2)$ and $(12,8)$, it has slope $\frac{8-2}{12-4}=\frac{6}{8}=\frac{3}{4}$.

Thus, its equation is of the form $y=\frac{3}{4} x+k$ for some $k$.

Substituting $(x, y)=(4,2)$, we have $2=\frac{3}{4}(4)+k$ so $k=-1$, which means the equation of the line is $y=\frac{3}{4} x-1$.

To find the coordinates of $C$, we will use that the point of intersection of the diagonals of a rectangle is equidistant from all four of its vertices.

The intersection of the diagonals is the intersection of the lines with equations $y=\frac{3}{4} x-1$ and $x+2 y-18=0$.

Substituting, we have $x+2\left(\frac{3}{4} x-1\right)=18$ or $x+\frac{3}{2} x-2=18$. Multiplying through by 2 and rearranging, we have $2 x+3 x=40$, so $x=8$ from which we get $y=\frac{3}{4}(8)-1=5$.

The coordinates of the centre of the rectangle are $\left(\frac{4+12}{2}, \frac{2+8}{2}\right)=(8,5)$.

The distance from this point to point $B$ is

$$
\sqrt{(8-4)^{2}+(5-2)^{2}}=\sqrt{16+9}=\sqrt{25}=5 \text {. }
$$

Therefore, $C$ is the point on the line $x+2 y=18$ that is 5 units away from $(8,5)$.

The condition on the distance gives us $\sqrt{(a-8)^{2}+(b-5)^{2}}=5$ or $(a-8)^{2}+(b-5)^{2}=25$.

Since $(a, b)$ is on $x+2 y=18$, we also have that $a+2 b=18$ which can be rearranged to get $a-8=10-2 b=2(5-b)$.

Substituting $a-8=2(5-b)$ into $(a-8)^{2}+(b-5)^{2}=5$ gives

$$
[2(5-b)]^{2}+(b-5)^{2}=25
$$

which simplifies to $5(b-5)^{2}=25$ or $(b-5)^{2}=5$.

Thus, $(b-5)= \pm 5$, so $b=5 \pm \sqrt{5}$. The point $C$ is to the right of the centre of the rectangle, and since the line connecting $A$ and $C$ has negative slope, this means $b$ is less than 5 .

Therefore, $b=5-\sqrt{5}$. Substituting this into $a+2 b=18$, we have $a=18-2(5-\sqrt{5})=8+2 \sqrt{5}$.

Using the formula for the distance between two points in the plane, we have

$$
\begin{aligned}
B C & =\sqrt{(4-a)^{2}+(2-b)^{2}} \\
& =\sqrt{(4-(8+2 \sqrt{5}))^{2}+(2-(5-\sqrt{5}))} \\
& =\sqrt{(-4-2 \sqrt{5})^{2}+(-3+\sqrt{5})^{2}} \\
& =\sqrt{16+16 \sqrt{5}+20+9-6 \sqrt{5}+5} \\
& =\sqrt{50+10 \sqrt{5}} .
\end{aligned}
$$

Similarly, we compute the length of $C D$

$$
\begin{aligned}
C D & =\sqrt{(a-12)^{2}+(b-8)^{2}} \\
& =\sqrt{(8+2 \sqrt{5}-12)^{2}+(5-\sqrt{5}-8)^{2}} \\
& =\sqrt{(-4+2 \sqrt{5})^{2}+(-3-\sqrt{5})^{2}} \\
& =\sqrt{16-16 \sqrt{5}+20+9+6 \sqrt{5}+5} \\
& =\sqrt{50-10 \sqrt{5}} .
\end{aligned}
$$

Finally, we compute the area of the rectangle

$$
\begin{aligned}
B C \times C D & =\sqrt{50+10 \sqrt{5}} \times \sqrt{50-10 \sqrt{5}} \\
& =\sqrt{(50+10 \sqrt{5})(50-10 \sqrt{5})} \\
& =\sqrt{50^{2}-(10 \sqrt{5})^{2}} \\
& =\sqrt{2500-500} \\
& =\sqrt{2000} .
\end{aligned}
$$

So the area of the rectangle is $\sqrt{2000}$, and since $2000=20^{2} \times 5$, this simplifies to $20 \sqrt{5}$.

ANSWER: $20 \sqrt{5}$