Review question

# Can we show this triangle is right-angled and find its area? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5442

## Solution

Three points have co-ordinates $P\,(-2, -3)$, $Q\,(2, 0)$, $R\,(8, -8)$.

1. Prove that $\widehat{PQR}=90^\circ$.

Two lines are perpendicular if their gradients multiply to $-1$. Consider the figure

The horizontal and vertical distances are shown in red. The gradient of the line $PQ$ is $\dfrac{3}{4}$, and the gradient of the line $QR$ is $-\dfrac{8}{6}=-\dfrac{4}{3}$. As these gradients multiply to $-1$, the angle $\widehat{PQR}$ is $90^\circ$.

1. Calculate the area of $\Delta PQR$.

The area of a triangle is $\dfrac{base \times height}{2}$. In our case the base and height are the sides $a$ and $b$ as shown in the following diagram.

By considering the red triangles drawn in part (i) we can use Pythagoras’ Theorem to compute the length of $a$ and $b$. Using this we get $a^2 = 3^2 + 4^2 \quad \text{ and } \quad b^2 = 6^2 + 8^2.$ This yields $a = \sqrt{3^2 + 4^2} = \sqrt{25}=5,$ and $b = \sqrt{6^2 + 8^2} = \sqrt{100} = 10.$ Therefore, the area of $\Delta PQR$ is $\dfrac{5 \times 10}{2} = 25$.

1. Calculate the length of $PR$ and hence, or otherwise, find the perpendicular distance of $Q$ from $PR$.

We can use Pythagoras’ Theorem once more to compute the length of $PR$.

We know that $a = 5$ and $b = 10$, so by Pythagoras’ Theorem, $c^2 = 5^2 + 10^2$ and $c = \sqrt{5^2 + 10^2} = \sqrt{125}= 5\sqrt{5} (\approx 11.18)$.

The perpendicular distance of $Q$ from $PR$ is shown in the following diagram, where we call it $h$.

Notice that $h$ is the height of the triangle $\Delta PQR$ with base $PR$ as $h$ is perpendicular to $PR$.

By the formula for the area of a triangle given in part (ii), and knowing that the area of the triangle is $25$, we have $\dfrac{c \times h}{2} = 25$.

Hence $\dfrac{5\sqrt{5} \times h}{2} = 25$ and therefore $h = \frac{50}{5\sqrt{5}} = \frac{10}{\sqrt{5}} = \sqrt{\frac{100}{5}} = \sqrt{20} = 2\sqrt{5} (\approx 4.47),$ and so the exact perpendicular distance of $Q$ from $PR$ is $2\sqrt{5}$.