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:
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
The vertices of a triangle ABC are A(-3,-4), B(16,-2) and C(13,-10).
Using the distance formula, we get
[tex]AB=\sqrt{(16-(-3))^2+(-2-(-4))^2}[/tex]
[tex]AB=\sqrt{(19)^2+(2)^2}[/tex]
[tex]AB=\sqrt{361+4}[/tex]
[tex]AB=\sqrt{365}[/tex]
Similarly,
[tex]BC=\sqrt{\left(13-16\right)^2+\left(-10-\left(-2\right)\right)^2}[/tex]
[tex]BC=\sqrt{73}[/tex]
And,
[tex]AC=\sqrt{\left(13-\left(-3\right)\right)^2+\left(-10-\left(-4\right)\right)^2}[/tex]
[tex]AC=\sqrt{292}[/tex]
Now, add the square of two smaller sides.
[tex]BC^2+AC^2=(\sqrt{73})^2+(\sqrt{292})^2[/tex]
[tex]BC^2+AC^2=73+292[/tex]
[tex]BC^2+AC^2=365[/tex]
[tex]BC^2+AC^2=(AB)^2[/tex]
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.
