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

Graph with the rectangle ABCD. A and B are on the left-facing side of the rectangle, C is the top-right vertex, and D is the bottom-right

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 \(3y=x+22\).

  1. the coordinates of \(C\),
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 \((14,12)\).

  1. the coordinates of \(D\),

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 \(D(16,6)\).

  1. the area of the rectangle \(ABCD\).
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.

Graph with the rectangle with the vertex coordinates, diagonal length, long side length, and area marked.