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

- 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 \(3y=x+22\).

- the coordinates of \(C\),

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

- 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 \(D(16,6)\).

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

Here you can see a complete sketch of the rectangle.