Show that set of points form right-angled triangle

[Yazan975]

View attachment 8415

I was thinking of using Pythagoras here but it didn't get me far
Any suggestions?

 

[MarkFL]

Let's begin by plotting the 3 given points:

View attachment 8418

It appears that \(\overline{AB}\) and \(\overline{AC}\) are the legs. Can you show that the product of the slopes of these two segments is -1?

 

[HOI]

I was thinking of using Pythagoras here but it didn't get me far
Any suggestions?

"Pythagoras" works wonderfully!

The distance from (1, -4) to (2, -3) is $\sqrt{(1- 2)^2+ (-4+ 3)^2}= \sqrt{1+ 1}= \sqrt{2}$.
The distance from (1, -4) to (4, -7) is $\sqrt{(1- 4)^2+ (-4+ 7)^2}= \sqrt{9+ 9}= \sqrt{18}$
The distance from (2, -3) to (4, -7) is $\sqrt{(2- 4)^2+ (-3+ 7)^2}= \sqrt{4+ 16}= \sqrt{20}$

Clearly $\sqrt{20}$ is larger than either $\sqrt{2}$ or $\sqrt{18}$ so let $a= \sqrt{2}$, $b= \sqrt{18}$, and $c= \sqrt{20}$. Is it true that $a^2+ b^2= c^2$?

 

[MarkFL]

To follow up (which is what we're hoping you will do when given help), we find:

\(\displaystyle m_{\overline{AB}}=\frac{3-(-4)}{2-1}=1\)

\(\displaystyle m_{\overline{AC}}=\frac{-7-(-4)}{4-1}=-1\)

Hence:

\(\displaystyle m_{\overline{AB}}\cdot m_{\overline{AC}}=-1\)

And so we may conclude that the 3 given points must be the vertices of a right triangle.

 

[Greg]

Alternatively, use vectors:

$$\vec{AB}\cdot\vec{AC}=<1,1>\cdot<3,-3>=0\implies\overline{AB}\perp\overline{AC}$$