Question

Triangle ABC has vertices A(- 3, - 4), B(16, - 2) and C(13, - 10) . Show algebraically that ABC is a right angled triangle.​

Answer 1

Given:

The vertices of a triangle ABC are A(-3,-4), B(16,-2) and C(13,-10).

To show:

That the triangle ABC is a right angled triangle.​

Solution:

Distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

The vertices of a triangle ABC are A(-3,-4), B(16,-2) and C(13,-10).

Using the distance formula, we get

$AB=\sqrt{(16-(-3))^2+(-2-(-4))^2}$

$AB=\sqrt{(19)^2+(2)^2}$

$AB=\sqrt{361+4}$

$AB=\sqrt{365}$

Similarly,

$BC=\sqrt{\left(13-16\right)^2+\left(-10-\left(-2\right)\right)^2}$

$BC=\sqrt{73}$

And,

$AC=\sqrt{\left(13-\left(-3\right)\right)^2+\left(-10-\left(-4\right)\right)^2}$

$AC=\sqrt{292}$

Now, add the square of two smaller sides.

$BC^2+AC^2=(\sqrt{73})^2+(\sqrt{292})^2$

$BC^2+AC^2=73+292$

$BC^2+AC^2=365$

$BC^2+AC^2=(AB)^2$

Since the sum of the square of two smaller sides is equal to the square of the largest side, therefore the given triangle is a right angle triangle by using Pythagoras theorem.

Hence proved.