# Prove $\blacktriangle ABC$ is a right triangle

Am I missing something here? My homework says to prove that the given triangle is a right triangle, but it does not appear to be a right triangle mathematically.

Let $A =(-3, 2)$, $B=(1, 0)$, and $C=(4,6)$.

Prove that $\blacktriangle ABC$ is a right triangle.

I have tried the slope method and it didn't work, I also tried the distance (Pythagorean) method and that didn't work either.

• Work out the squares of the side lengths and show $AB^2+BC^2=AC^2$ - no square roots required! Commented Sep 23, 2015 at 21:52
• It works with both methods. Make a drawing first to see what angle is right.
– A.Γ.
Commented Sep 23, 2015 at 21:54

Let's try the Pythagorean theorem:

Side $AB = [1- -3,0-2] = [4,-2]$, with length $\sqrt{20}$

Side $BC = [4- 1,6-0] = [3,6]$, with length $\sqrt{45}$

Side $AC = [4- -3,6-2] = [7,4]$, with length $\sqrt{65}$

$|AB|^2 + |BC|^2 = (\sqrt{20})^2+(\sqrt{45})^2 = 20 + 45 = 65 = (\sqrt{65})^2 = |AC|^2$

So the triangle must be right

• I would call it by the name Law of cosines rather than Pythagorean theorem (which for me is just the other direction), but it is okay. Commented Sep 23, 2015 at 22:11

I tried to solve it using the slope concept and indeed $A$, $B$ and $C$ are endpoints of a right triangle.

The slope of the line connecting $AB$ is $\frac{-1}{2}$.

The slope of the line connecting $AC$ is $\frac{4}{7}$.

Lastly, the slope of the line connecting $BC$ is 2.

Now since the product of the slope of line connecting $AB$ and the line connecting $BC$ is $\frac{-1}{2}\cdot 2=-1$, the line connecting $AB$ and the line connecting $BC$ are perpendicular by including line connecting $AC$ one can generate a triangle. This triangle a right triangle since the line connecting $AB$ and the line connecting $BC$ are perpendicular .

Translate the triangle by $(-1,0)$. The new coordinates of the vertices become: $$A(-4,2),\quad B(0,0),\quad C(3,6)$$ and since the dot product between $A$ and $C$ is zero, $\widehat{ABC}$ is a right angle.

Add points $D=(-3,0)$ and $E=(4,0)$ and show that $\Delta ABD \sim\Delta BCE$. Then show that $\measuredangle ABC=90^\circ$.

Do you see why the slopes of perpendicular lines are negative reciprocals of each other?

If you calculate again the distances you will see that $$AC^2=AB^2+BC^2$$