# How do you show that the points (18, 4), (12, 12), and (8, 4) are the vertices of an isosceles triangle?

Aug 28, 2017

See a solution process below:

#### Explanation:

An isosceles triangle has two sides the same length. We can find the length of the three line segments. If two are the same length then the points are the vertices of an isosceles triangle.

Let's call the Points:

$A \left(18 , 4\right)$

$B \left(12 , 12\right)$

$C \left(8 , 4\right)$

The formula for calculating the distance between two points is:

$d = \sqrt{{\left(\textcolor{red}{{x}_{2}} - \textcolor{b l u e}{{x}_{1}}\right)}^{2} + {\left(\textcolor{red}{{y}_{2}} - \textcolor{b l u e}{{y}_{1}}\right)}^{2}}$

${d}_{A - B} = \sqrt{{\left(\textcolor{red}{12} - \textcolor{b l u e}{18}\right)}^{2} + {\left(\textcolor{red}{12} - \textcolor{b l u e}{4}\right)}^{2}}$

${d}_{A - B} = \sqrt{{\left(- 6\right)}^{2} + {8}^{2}}$

${d}_{A - B} = \sqrt{36 + 64}$

${d}_{A - B} = \sqrt{100}$

${d}_{A - B} = 10$

${d}_{A - C} = \sqrt{{\left(\textcolor{red}{8} - \textcolor{b l u e}{18}\right)}^{2} + {\left(\textcolor{red}{4} - \textcolor{b l u e}{4}\right)}^{2}}$

${d}_{A - C} = \sqrt{{\left(- 10\right)}^{2} + {0}^{2}}$

${d}_{A - C} = \sqrt{100 + 0}$

${d}_{A - C} = \sqrt{100}$

${d}_{A - C} = 10$

${d}_{B - C} = \sqrt{{\left(\textcolor{red}{8} - \textcolor{b l u e}{12}\right)}^{2} + {\left(\textcolor{red}{4} - \textcolor{b l u e}{12}\right)}^{2}}$

${d}_{B - C} = \sqrt{{\left(- 4\right)}^{2} + {\left(- 4\right)}^{2}}$

${d}_{B - C} = \sqrt{16 + 16}$

${d}_{B - C} = \sqrt{32}$

${d}_{B - C} = \sqrt{16 \cdot 2}$

${d}_{B - C} = \sqrt{16} \sqrt{2}$

${d}_{B - C} = 4 \sqrt{2}$

Because ${d}_{A - B} = {d}_{A - C} = 10$ these points are the vertices of an isosceles triangle.