Question

triangle def has vertices located at d (2, 1), e (3, 5), and f (6, 2).

part a: find the length of each side of the triangle. show your work. (4 points)

part b: find the slope of each side of the triangle. show your work. (3 points)

part c: classify the triangle. explain your reasoning. (3 points)

Explanation:

Step1: Use distance formula for side lengths

$DE=\sqrt{(3 - 2)^2+(5 - 1)^2}=\sqrt{1 + 16}=\sqrt{17}$
$EF=\sqrt{(6 - 3)^2+(2 - 5)^2}=\sqrt{9 + 9}=3\sqrt{2}$
$DF=\sqrt{(6 - 2)^2+(2 - 1)^2}=\sqrt{16 + 1}=\sqrt{17}$

Step2: Use slope formula for slopes

Slope of $DE=\frac{5 - 1}{3 - 2}=4$
Slope of $EF=\frac{2 - 5}{6 - 3}=-1$
Slope of $DF=\frac{2 - 1}{6 - 2}=\frac{1}{4}$

Step3: Classify triangle

Two sides ($DE = DF$) are equal, so isosceles.

Answer:

Part A: $DE=\sqrt{17}$, $EF = 3\sqrt{2}$, $DF=\sqrt{17}$
Part B: Slope of $DE = 4$, slope of $EF=-1$, slope of $DF=\frac{1}{4}$
Part C: Isosceles triangle as $DE = DF$.