Question

Show that the points A(-1, 2), B(5, 2) and C(2, 5) are the vertices of an isosceles triangle. Findthe area of triangle ABC.

Answer 1

First, we need to find the side lengths for each side.

Then, we need to use the next equation:

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

For AB:

$\begin{gathered} AB=\sqrt[]{(2-2)^2+(5+1)^2} \\ AB=\sqrt{36} \\ AB=6 \end{gathered}$

For BC:

$\begin{gathered} BC=\sqrt[]{(5-2)^2+(2-5)^2} \\ BC=\sqrt[]{18} \\ BC=3\sqrt[]{2} \end{gathered}$

For AC:

$\begin{gathered} AC=\sqrt{(5-2)^2+(2+1)^2} \\ AC=\sqrt[]{18} \\ AC=3\sqrt{2} \end{gathered}$

Now, isosceles triangles have two equal sides.

In this case, BC and AC are equal.

So, the triangle ABC is isosceles.

To find the area of triangle ABC, we need to use the Heron's formula given by:

$A=s(s-AB)(s-BC)(s-AC)$

where

$\begin{gathered} s=\frac{AB+BC+AC}{2} \\ s=\frac{6+3\sqrt[]{2}+3\sqrt[]{2}}{2} \\ s=3+3\sqrt[]{2} \end{gathered}$

Replacing :

$undefined$