verify that a triangle contains a right-angle

Question

For the question "The corners of a triangle are the points $P(4,7)$, $Q(-2,5)$ and $R(3, -10)$. Find the length of each side of $\triangle PQR$, giving your answers in terms of surds". I have the answer to the first part as : $|PQ| = 2\sqrt{10}$, $|PR| = \sqrt{290}$, $|RQ| = 5\sqrt{10}$.

The second part asks "Hence, verify that $\triangle PQR$ contains a right-angle". Could you help me with this part showing all the working out so that i understand how to tackle the question better thanks.

Answer 1

To verify that $PQR$ contains a right angle, you can use Pythagoras' Theorem, which says that in a right angled triangle, if the side lengths are $a\leq b\leq c$, then $a^2+b^2=c^2$.

Note that $(2\sqrt{10})^2+(5\sqrt{10})^2 = 40 + 250 = 290 = (\sqrt{290})^2$

Answer 2

Well, it's an application of Pythagoras' Theorem. If the square of your largest edge length (call it $c$) is equal to the sum of the squares of the other two (call them $a$ and $b$), then you've got a right triangle, because the edge lengths satisfy $a^2+b^2=c^2$.

Another way is to check if the dot product of any two of your edges is 0, then you've got a right triangle, but since you've already computed the edge lengths it's simpler just to do the above.

Answer 3

As $|PR|$ is the maximum length, the right-angle, if any, must be $\angle PQR$.

The gradient of $QR$ is $\dfrac{-2-3}{5-(-10)}=\dfrac{-5}{15}=\dfrac{-1}3$.

The gradient of $QP$ is $\dfrac{-2-4}{5-7}=\dfrac{-6}{-2}=3$.

Two lines are perpedicular iff $m _1=\dfrac{-1}{m_2}$, where $m_1,m_2$ are the gradients of the lines.

So $Q$ is a right-angle.