Question

a graph shows the first triangle in a wallpaper pattern. the coordinates are a(-4, -8), b(-7, 4), and c(6, 5). how can the distance formula be used to find the side lengths and classify the triangle? the side lengths are √137, √82, and √13, so this is an isosceles triangle. the side lengths are 3√17, √170, and √269, so this is a scalene triangle. the side lengths are √137, √82, and √13, so this is a scalene triangle. the side lengths are 3√17, √170, and √269, so this is an isosceles triangle.

Explanation:

Step1: Recall distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.

Step2: Calculate length of $AB$

For points $A(-4,-8)$ and $B(-7,4)$:
\[

$$\begin{align*} AB&=\sqrt{(-7+4)^2+(4 + 8)^2}\\ &=\sqrt{(-3)^2+12^2}\\ &=\sqrt{9 + 144}\\ &=\sqrt{153}=3\sqrt{17} \end{align*}$$

\]

Step3: Calculate length of $BC$

For points $B(-7,4)$ and $C(6,5)$:
\[

$$\begin{align*} BC&=\sqrt{(6 + 7)^2+(5 - 4)^2}\\ &=\sqrt{13^2+1^2}\\ &=\sqrt{169+1}\\ &=\sqrt{170} \end{align*}$$

\]

Step4: Calculate length of $AC$

For points $A(-4,-8)$ and $C(6,5)$:
\[

$$\begin{align*} AC&=\sqrt{(6 + 4)^2+(5 + 8)^2}\\ &=\sqrt{10^2+13^2}\\ &=\sqrt{100+169}\\ &=\sqrt{269} \end{align*}$$

\]
Since all side - lengths $3\sqrt{17},\sqrt{170},\sqrt{269}$ are different, it is a scalene triangle.

Answer:

The side lengths are $3\sqrt{17},\sqrt{170}$, and $\sqrt{269}$, so this is a scalene triangle.