# verify that a triangle contains a right-angle

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.

• A variation on the answers below is to use the law of cosines $c^2=a^2+b^2-2ab \cos C$; if you can pick $a,b,c$ from your three lengths such that $a^2+b^2=c^2$, then automatically $\cos C=0$ and therefore $C=90^\circ$. (The advantage of this approach is that it's also applicable if it's not a right triangle.) This is also equivalent to JasonC's remark regarding the dot product. Sep 13, 2016 at 13:35

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

• could you please show me how to do this as my answers are in surds so i'm abit confused thanks. Sep 13, 2016 at 13:28
• And what if i then wanted to find the area of the triangle PQR once i have all this information about the triangle ? thanks. Sep 13, 2016 at 13:44
• In general, use the formula: Area = $\sqrt{s(s-a)(s-b)(s-c)}$, where $s:=\frac{a+b+c}{2}$ is called the semiperimeter. Here. because the triangle is right-angled, the formula$\frac{1}{2} \times b \times h$, where $b$ denotes base and $h$ denotes height, is more useful. So, area $=\frac{1}{2}\times 2\sqrt{10}\times 5\sqrt{10}$
– JDF
Sep 13, 2016 at 14:02

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.

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.

• So i can use either solutions ? this one or the one below ? thanks. Sep 13, 2016 at 13:39
• assuming 'the one below' is Pythagoras, then yes.
– JMP
Sep 13, 2016 at 13:40
• And what if i then wanted to find the area of the triangle PQR once i have all this information about the triangle ? thanks. Sep 13, 2016 at 13:50
• it's |QR|*|QP|/2
– JMP
Sep 13, 2016 at 13:51
• how did you know which is the height and base from that information thanks. Sep 13, 2016 at 13:53