Question

question 1 (essay worth 10 points)
(04.01 mc)
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:

Part A: Find the length of each side of the triangle.

Step 1: Recall the distance formula

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

Step 2: Calculate the length of \(DE\)

For points \(D(2, 1)\) and \(E(3, 5)\):
\[

$$\begin{align*} DE&=\sqrt{(3 - 2)^2 + (5 - 1)^2}\\ &=\sqrt{(1)^2 + (4)^2}\\ &=\sqrt{1 + 16}\\ &=\sqrt{17} \end{align*}$$

\]

Step 3: Calculate the length of \(EF\)

For points \(E(3, 5)\) and \(F(6, 2)\):
\[

$$\begin{align*} EF&=\sqrt{(6 - 3)^2 + (2 - 5)^2}\\ &=\sqrt{(3)^2 + (-3)^2}\\ &=\sqrt{9 + 9}\\ &=\sqrt{18}=3\sqrt{2} \end{align*}$$

\]

Step 4: Calculate the length of \(FD\)

For points \(F(6, 2)\) and \(D(2, 1)\):
\[

$$\begin{align*} FD&=\sqrt{(2 - 6)^2 + (1 - 2)^2}\\ &=\sqrt{(-4)^2 + (-1)^2}\\ &=\sqrt{16 + 1}\\ &=\sqrt{17} \end{align*}$$

\]

Part B: Find the slope of each side of the triangle.

Step 1: Recall the slope formula

The slope between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \(m=\frac{y_2 - y_1}{x_2 - x_1}\).

Step 2: Calculate the slope of \(DE\)

For points \(D(2, 1)\) and \(E(3, 5)\):
\[
m_{DE}=\frac{5 - 1}{3 - 2}=\frac{4}{1} = 4
\]

Step 3: Calculate the slope of \(EF\)

For points \(E(3, 5)\) and \(F(6, 2)\):
\[
m_{EF}=\frac{2 - 5}{6 - 3}=\frac{-3}{3}=- 1
\]

Step 4: Calculate the slope of \(FD\)

For points \(F(6, 2)\) and \(D(2, 1)\):
\[
m_{FD}=\frac{1 - 2}{2 - 6}=\frac{-1}{-4}=\frac{1}{4}
\]

Part C: Classify the triangle.

Step 1: Analyze the side lengths

We found that \(DE = \sqrt{17}\) and \(FD=\sqrt{17}\), so two sides of the triangle are equal.

Step 2: Analyze the slopes (to check for right angle)

The product of the slopes of two perpendicular lines is \(- 1\). Let's check the product of slopes of different sides:

• \(m_{DE}\times m_{EF}=4\times(-1)=-4

eq - 1\)

• \(m_{EF}\times m_{FD}=(-1)\times\frac{1}{4}=-\frac{1}{4}

eq - 1\)

• \(m_{FD}\times m_{DE}=\frac{1}{4}\times4 = 1

eq - 1\)

Since two sides are equal and no two sides are perpendicular, the triangle is an isosceles triangle (not right - angled).

Part A Answers:

\(DE=\sqrt{17}\), \(EF = 3\sqrt{2}\), \(FD=\sqrt{17}\)

Part B Answers:

\(m_{DE}=4\), \(m_{EF}=-1\), \(m_{FD}=\frac{1}{4}\)

Part C Answer:

The triangle \(DEF\) is an isosceles triangle because two of its sides (\(DE\) and \(FD\)) have equal length (\(\sqrt{17}\)) and it does not have a right angle (since the product of the slopes of any two sides is not \(- 1\)).

Answer:

Part A: Find the length of each side of the triangle.

Step 1: Recall the distance formula

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

Step 2: Calculate the length of \(DE\)

For points \(D(2, 1)\) and \(E(3, 5)\):
\[

$$\begin{align*} DE&=\sqrt{(3 - 2)^2 + (5 - 1)^2}\\ &=\sqrt{(1)^2 + (4)^2}\\ &=\sqrt{1 + 16}\\ &=\sqrt{17} \end{align*}$$

\]

Step 3: Calculate the length of \(EF\)

For points \(E(3, 5)\) and \(F(6, 2)\):
\[

$$\begin{align*} EF&=\sqrt{(6 - 3)^2 + (2 - 5)^2}\\ &=\sqrt{(3)^2 + (-3)^2}\\ &=\sqrt{9 + 9}\\ &=\sqrt{18}=3\sqrt{2} \end{align*}$$

\]

Step 4: Calculate the length of \(FD\)

For points \(F(6, 2)\) and \(D(2, 1)\):
\[

$$\begin{align*} FD&=\sqrt{(2 - 6)^2 + (1 - 2)^2}\\ &=\sqrt{(-4)^2 + (-1)^2}\\ &=\sqrt{16 + 1}\\ &=\sqrt{17} \end{align*}$$

\]

Part B: Find the slope of each side of the triangle.

Step 1: Recall the slope formula

The slope between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \(m=\frac{y_2 - y_1}{x_2 - x_1}\).

Step 2: Calculate the slope of \(DE\)

For points \(D(2, 1)\) and \(E(3, 5)\):
\[
m_{DE}=\frac{5 - 1}{3 - 2}=\frac{4}{1} = 4
\]

Step 3: Calculate the slope of \(EF\)

For points \(E(3, 5)\) and \(F(6, 2)\):
\[
m_{EF}=\frac{2 - 5}{6 - 3}=\frac{-3}{3}=- 1
\]

Step 4: Calculate the slope of \(FD\)

For points \(F(6, 2)\) and \(D(2, 1)\):
\[
m_{FD}=\frac{1 - 2}{2 - 6}=\frac{-1}{-4}=\frac{1}{4}
\]

Part C: Classify the triangle.

Step 1: Analyze the side lengths

We found that \(DE = \sqrt{17}\) and \(FD=\sqrt{17}\), so two sides of the triangle are equal.

Step 2: Analyze the slopes (to check for right angle)

The product of the slopes of two perpendicular lines is \(- 1\). Let's check the product of slopes of different sides:

• \(m_{DE}\times m_{EF}=4\times(-1)=-4

eq - 1\)

• \(m_{EF}\times m_{FD}=(-1)\times\frac{1}{4}=-\frac{1}{4}

eq - 1\)

• \(m_{FD}\times m_{DE}=\frac{1}{4}\times4 = 1

eq - 1\)

Since two sides are equal and no two sides are perpendicular, the triangle is an isosceles triangle (not right - angled).

Part A Answers:

\(DE=\sqrt{17}\), \(EF = 3\sqrt{2}\), \(FD=\sqrt{17}\)

Part B Answers:

\(m_{DE}=4\), \(m_{EF}=-1\), \(m_{FD}=\frac{1}{4}\)

Part C Answer:

The triangle \(DEF\) is an isosceles triangle because two of its sides (\(DE\) and \(FD\)) have equal length (\(\sqrt{17}\)) and it does not have a right angle (since the product of the slopes of any two sides is not \(- 1\)).