# How would you show that a triangle with vertices (13,-2), (9,-8), (5,-2) is isosceles?

May 7, 2018

color(crimson)("The distances AB and BC are " sqrt52 " and therefore isosceles."

#### Explanation:

If we label them A(13,-2) B(9-8) and C(5-2)

The distance between A and B:
$13 \implies 9 = 4$
$- 2 \implies - 8 = 6$
${4}^{2} + {6}^{2} = 52$ so the distance is $\sqrt{52}$

The distance between B and C:
$9 \implies 5 = 4$
$- 8 \implies - 2 = 6$
${4}^{2} + {6}^{2} = 52$ the distance is $\sqrt{52}$

The distance between A and C is 8 as they are both on -2 for $y$

color(crimson)("The distances AB and BC are " sqrt52 " and therefore isosceles."

May 7, 2018

see below

#### Explanation:

An isosceles triangle is one with two equal lengths.

To find the lengths of the sides with coordinates

$\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right)$ we use Pythagoras' theorem

$d = \sqrt{{\left({x}_{2} - {x}_{2}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

from the information given

${d}_{1} = \sqrt{{\left(13 - 9\right)}^{2} + {\left(- 2 - - 8\right)}^{2}}$

$\textcolor{red}{{d}_{1} = \sqrt{{4}^{2} + {6}^{2}} = \sqrt{52} - - \left(1\right)}$

${d}_{2} = \sqrt{{\left(13 - 5\right)}^{2} + {\left(- 2 - - 2\right)}^{2}}$

color(red)(d_2=sqrt(8^2+0^2)=sqrt64=8--(2)

d_3=sqrt((9-5)^2+(-8 --2 )^2

color(red)(d_3=sqrt(4^2+6^2)=sqrt52---(3)

we see ${d}_{1} = {d}_{3}$

since we have two equal sides the triangle with the given vertices is isosceles

May 7, 2018

Below

#### Explanation:

One way of proving that it is an isosceles triangle is by calculating the length of each side since two sides of equal lengths means that it is an isosceles triangle.

Length of (13,-2) & (9,-8)
=$\sqrt{{\left(13 - 9\right)}^{2} + {\left(- 2 + 8\right)}^{2}}$
=$\sqrt{16 + 36}$
=$\sqrt{52}$
=$2 \sqrt{13}$ units

Length of (9,-8) & (5,-2)
=$\sqrt{{\left(9 - 5\right)}^{2} + {\left(- 8 + 2\right)}^{2}}$
=$\sqrt{16 + 36}$
=$\sqrt{52}$
=$2 \sqrt{13}$ units

Length of (13,-2) & (5,-2)
=$\sqrt{{\left(13 - 5\right)}^{2} + {\left(- 2 + 2\right)}^{2}}$
=$\sqrt{64}$
=$8$ units

From the above calculations, you'll notice that length of (13,-2) & (9,-8) and length of (9,-8) & (5,-2) are the same. therefore, you can prove that the triangle is isosceles