MHB Show that set of points form right-angled triangle

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To determine if the three given points form a right-angled triangle, the slopes of segments AB and AC can be analyzed. The slopes calculated show that their product is -1, indicating that the segments are perpendicular. Additionally, using the Pythagorean theorem, the distances between the points confirm that the sum of the squares of the two shorter sides equals the square of the longest side. Therefore, the three points indeed represent the vertices of a right triangle. This conclusion is supported by both slope analysis and vector calculations.
Yazan975
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View attachment 8415

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

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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?
 

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Yazan975 said:
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 $a^2+ b^2= c^2$?
 
To follow up (which is what we're hoping you will do when given help), we find:

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

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

Hence:

$$m_{\overline{AB}}\cdot m_{\overline{AC}}=-1$$

And so we may conclude that the 3 given points must be the vertices of a right triangle.
 
Alternatively, use vectors:

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