Question

directions: find the measures of the sides of \\( \triangle def \\), then classify it by its sides.\

1. \\( d(8, -6) \\), \\( e(-1, -3) \\), \\( f(-2, 5) \\)\

\\( de: \because d = \sqrt{} \\)\
\\( ef: \because d = \sqrt{} \\)\
\\( df: \because d = \sqrt{} \\)

Explanation:

Step1: Recall distance formula

The distance 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 DE

Substitute $D(8,-6), E(-1,-3)$:
$d_{DE}=\sqrt{(-1-8)^2+(-3-(-6))^2}=\sqrt{(-9)^2+(3)^2}=\sqrt{81+9}=\sqrt{90}=3\sqrt{10}$

Step3: Calculate length of EF

Substitute $E(-1,-3), F(-2,5)$:
$d_{EF}=\sqrt{(-2-(-1))^2+(5-(-3))^2}=\sqrt{(-1)^2+(8)^2}=\sqrt{1+64}=\sqrt{65}$

Step4: Calculate length of DF

Substitute $D(8,-6), F(-2,5)$:
$d_{DF}=\sqrt{(-2-8)^2+(5-(-6))^2}=\sqrt{(-10)^2+(11)^2}=\sqrt{100+121}=\sqrt{221}$

Step5: Classify the triangle

All side lengths are unique, so it is scalene.

Answer:

$DE = 3\sqrt{10}$, $EF = \sqrt{65}$, $DF = \sqrt{221}$; $\triangle DEF$ is a scalene triangle.