The coordinates of the points \(A\), \(B\), \(C\) are \((-2,1)\), \((2,7)\), \((5,5)\) respectively. Prove that these points form three corners of a rectangle…

To prove that the points form three corners of a rectangle, we need to show that the two lines \(AB\) and \(BC\) meet at right angles.

If this is true, then it must be possible to form a rectangle with these three points (and one other).

Graph with the points A, B and C.

The slope of the line \(AB\) is \(m_{AB}=\dfrac{\Delta y}{\Delta x} = \dfrac{7-1}{2-(-2)} = \dfrac{6}{4} = \dfrac{3}{2}\).

The slope of \(BC\) is \(m_{BC} = \dfrac{5-7}{5-2} = \dfrac{-2}{3} = - \dfrac{2}{3}\).

The product of the two slopes is \(\dfrac{3}{2} \times -\dfrac{2}{3} = -1\), so the lines are perpendicular and thus there exists a rectangle \(ABCD\).

This is the most straightforward way to prove what we need, but it’s possible (with quite a lot more work) to use the cosine rule instead.

The length of \(AB\) is \(\sqrt{(7-1)^2 + (2-(-2))^2} = \sqrt{52}\).

The length of \(BC\) is \(\sqrt{(5-7)^2 + (5-2)^2} = \sqrt{13}\).

The length of \(CA\) is \(\sqrt{(5-(-2))^2 + (5-1)^2} = \sqrt{65}\).

The cosine rule says \[(CA)^2 = (AB)^2 + (BC)^2 - 2(AB)(BC) \cos(B).\] Therefore

\[65 = 13 + 52 - 2 \times \sqrt{13} \times \sqrt{52} \times \cos(B)\]

which gives that

\[\cos(B) = 0\] and hence that

\[B = 90^\circ.\]

…and that \(AB=2BC\).

The simplest approach here is to consider the similar triangles which appear in the diagram.

The points A, B and C with vertical lines drawn upward from A and C, and a horizontal line drawn in both directions from B. In this way two right-angled triangles are formed with hypotenuses AB and BC respectively.

\(AB\) is the hypotenuse of a right-angled triangle, with the other sides length \(6\) and \(4\).

\(BC\) is the hypotenuse of a right-angled triangle, with the other sides length \(3\) and \(2\).

Therefore the two triangles are similar, with a scale factor of \(2\), so we must have \(AB\) = \(2BC\).

We could instead calculate the lengths of the sides using Pythagoras’ theorem.

We have \[AB = \sqrt{(2-(-2))^2 + (7-1)^2} = \sqrt{4^2+6^2} = \sqrt{52} \qquad \text{and} \qquad BC = \sqrt{3^2 + (-2)^2} = \sqrt{13}\] and indeed \[AB = \sqrt{52} = \sqrt{4\times 13} = 2\sqrt{13} = 2 BC.\]

If \(D\) is the fourth corner of the rectangle, calculate the distance of \(C\) from the diagonal \(BD\).

There are a few ways to do this problem.

We could calculate \(D\)’s coordinates, but we can solve the problem without doing this.

This first method shown is more efficient than other methods, which require a greater amount of algebraic manipulation.

The distance of \(C\) from the diagonal \(BD\) will be the shortest possible distance, which is the perpendicular distance, shown as \(h\) in the diagram below.

The rectangle ABCD with the diagonal BD, and a line drawn from C which meets BD at a right angle, at a point P. The length of BC is d, and the length of PC is h.
By Pythagoras, \(BD\) is \(\sqrt{5}d\) where \(d\) is the length of \(BC\). Using similar triangles, we also see that \[\begin{align*} \frac{h}{BC} &= \frac{CD}{BD} \\ \iff\quad \frac{h}{d} &= \frac{2d}{\sqrt{5}d} \\ \iff\quad h &= \frac{2d}{\sqrt{5}}. \end{align*}\] Using Pythagoras, we can find the length of \(BC\). We have \[d = \sqrt{3^2 + (-2)^2} = \sqrt{13},\] and therefore \[\begin{align*} h &= \dfrac{2\sqrt{13}}{\sqrt{5}} \\ &= \dfrac{2}{5} \sqrt{65}. \end{align*}\]

Or even more simply, we could consider area. The area of the rectangle is \(d \times 2d = BD \times h = \sqrt{5}d \times h\), and so \(h = \dfrac{2}{5} \sqrt{65}\).

We can instead calculate the position of \(D\), and then calculate the equations of \(BD\) and \(CP\) to find the distance \(CP\).

We know that \(AD\) is parallel to \(BC\), and so we know the horizontal and vertical distance of \(D\) from \(A\), as shown in the diagram below.

The rectangle ABCD showing D as 2 units below and 3 units to the right of A, just as C is 2 units below and 3 units to the right of B

The coordinates of \(D\) are thus \(\quad (-2 + 3, 1 -2) \quad = \quad (1,-1).\)

The gradient of the line \(BD\) is \(m = \dfrac{7-(-1)}{2-1} = \dfrac{8}{1} = 8\).

We can now calculate the equation of the straight line \(BD\), from \[\begin{equation} y + 1 = 8(x - 1).\label{eq:1} \end{equation}\] We can also calculate the line through \(C\) perpendicular to \(BD\) (so its gradient is \(-1/8\)) as \[\begin{equation} y - 5 = -\dfrac{1}{8}(x - 5).\label{eq:2} \end{equation}\] The point of intersection is found by solving the two equations simultaneously. If we calculate \(\eqref{eq:1}\) - \(\eqref{eq:2}\), we find \[\begin{align*} 6 &= 8(x-1) + \dfrac{1}{8}(x - 5) \\ 48 &= 64(x - 1) + (x - 5) \\ &= 64x - 64 + x - 5 \\ 117 &= 65x \\ \text{so} \qquad x &= \dfrac{117}{65} = \dfrac{9}{5}, \\ \text{and also} \qquad y &= 8\left(\dfrac{9}{5} - 1\right) = \dfrac{27}{5}. \end{align*}\]

We can now work out the length \(CP\), as \[\sqrt{\left(5 - \frac{9}{5}\right)^2 + \left(5 - \frac{27}{5}\right)^2} = \frac{2}{5} \sqrt{65}.\]

As we can see, this method required substantially more work than the first two, and with lots of algebraic manipulation there was plenty of scope for errors.