Triangle ABC have vertices A(0,4), B(1,2), and C(4,6). Is triangle ABC a right triangle?

Nov 14, 2015

We can solve this by using vectors. Let's start by defining
$\vec{A B}$ and $\vec{A C}$ and $\vec{B C}$

$\vec{A B} = \left[{B}_{x} - {A}_{x} , {B}_{y} - {A}_{y}\right] = \left[1 - 0 , 2 - 4\right] = \left[1 , - 2\right]$
$\vec{A C} = \left[{C}_{x} - {A}_{x} , {C}_{y} - {A}_{y}\right] = \left[4 - 0 , 6 - 4\right] = \left[4 , 2\right]$
$\vec{B C} = \left[{C}_{x} - {B}_{x} , {C}_{y} - {B}_{y}\right] = \left[4 - 1 , 6 - 2\right] = \left[3 , 4\right]$

To figure out whether two lines are perpendicular or not (that means that they cause a 90 degree angle between them), we can multiply the two vectors together. If the product is 0, it means that they indeed are perpendicular.

Let's see if whether:
$\vec{A B} \cdot \vec{A C} = 0$ or
$\vec{A B} \cdot \vec{B C} = 0$ or
$\vec{A C} \cdot \vec{B C} = 0$

When we are multiplying vectors (dot multiplying), we multiply the x components together, and add them with the product of the y components.

Let's start with $\vec{A B} \cdot \vec{A C}$
$\vec{A B} \cdot \vec{A C} = \left[1 , - 2\right] \cdot \left[4 , 2\right]$
$= 1 \cdot 4 + \left(- 2\right) \cdot 2$
$= 4 - 4$
$= 0$

Well, that was lucky! Now we know that AB is perpendicular with AC, because $\vec{A B} \cdot \vec{A C} = 0$. We could in theory check if there were other angles that were 90 degrees, but that would be stupid since there will only be one angle that will be 90 degrees.

Hope this helped. If you don't know how to use vectors, I'm sorry to say that I don't know another way of proving this.