Let's break down each part of the problem for Triangle DEF with vertices at $D(2, 1)$, $E(3, 5)$, and $F(6, 2)$.

Part A: Find the Length of Each Side of the Triangle

The length of a side of a triangle can be found using the distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$:

$\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

1. Length of DE (between points $D(2, 1)$ and $E(3, 5)$): $DE = \sqrt{(3 - 2)^2 + (5 - 1)^2} = \sqrt{1^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17} \approx 4.12$

2. Length of EF (between points $E(3, 5)$ and $F(6, 2)$): $EF = \sqrt{(6 - 3)^2 + (2 - 5)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} \approx 4.24$

3. Length of FD (between points $F(6, 2)$ and $D(2, 1)$): $FD = \sqrt{(6 - 2)^2 + (2 - 1)^2} = \sqrt{4^2 + 1^2} = \sqrt{16 + 1} = \sqrt{17} \approx 4.12$

So, the lengths of the sides are approximately:

• $DE \approx 4.12$
• $EF \approx 4.24$
• $FD \approx 4.12$

Part B: Find the Slope of Each Side of the Triangle

The slope of a line between two points $(x_1, y_1)$ and $(x_2, y_2)$ is calculated using:

$\text{Slope} = \frac{y_2 - y_1}{x_2 - x_1}$

1. Slope of DE (between points $D(2, 1)$ and $E(3, 5)$): $\text{Slope of DE} = \frac{5 - 1}{3 - 2} = \frac{4}{1} = 4$

2. Slope of EF (between points $E(3, 5)$ and $F(6, 2)$): $\text{Slope of EF} = \frac{2 - 5}{6 - 3} = \frac{-3}{3} = -1$

3. Slope of FD (between points $F(6, 2)$ and $D(2, 1)$): $\text{Slope of FD} = \frac{1 - 2}{2 - 6} = \frac{-1}{-4} = \frac{1}{4}$

So, the slopes are:

• $\text{Slope of DE} = 4$
• $\text{Slope of EF} = -1$
• $\text{Slope of FD} = \frac{1}{4}$

Part C: Classify the Triangle

To classify the triangle, we consider the lengths of the sides and the slopes.

1. Length Comparison:
   • $DE \approx 4.12$
   • $EF \approx 4.24$
   • $FD \approx 4.12$

Since two sides $DE$ and $FD$ are equal and $EF$ is slightly longer, the triangle is isosceles.

2. Right Triangle Check:
   • A triangle is a right triangle if the slopes of two sides are negative reciprocals of each other, meaning they form a 90-degree angle.
   • Here, $\text{Slope of DE} = 4$ and $\text{Slope of EF} = -1$ are not negative reciprocals.
   • However, the length calculation suggests it could be close to a right triangle.

To confirm this, use the Pythagorean theorem: $DE^2 + FD^2 = EF^2 \text{?}$ $(\sqrt{17})^2 + (\sqrt{17})^2 = (\sqrt{18})^2$ $17 + 17 = 18 \quad (\text{False, so it’s not a right triangle.})$

Thus, the triangle is isosceles but not a right triangle.

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Do you want any further details or have any questions?

Here are some related questions:

1. How do you find the perimeter of Triangle DEF?
2. Can you explain the significance of the slopes in triangle classification?
3. What would happen if $DE$ and $FD$ had different lengths?
4. How would the classification change if $EF = \sqrt{34}$?
5. What other types of triangles are there besides isosceles?
6. How can you determine the area of Triangle DEF?
7. What is the relationship between slope and angle in a triangle?
8. How could you confirm if the triangle is scalene?

Tip: Always check both side lengths and angles (or slopes) when classifying triangles.