A triangle has vertices D(-2,7), E(-4,2), and F(6,-2). a) Show algrebraically that this triangle is a right triangle. b) Find the midpoint of the hypotenuse. c) Show that this midpoint is equidistant from each of the vertices.

To solve this problem, we will approach it step by step, as follows:

### Part (a): Show algebraically that the triangle is a right triangle.
To determine if the triangle is a right triangle, we need to check if any two sides are perpendicular by calculating the slopes of the sides. Two sides are perpendicular if the product of their slopes is \(-1\).

The vertices of the triangle are:
- \( D(-2, 7) \)
- \( E(-4, 2) \)
- \( F(6, -2) \)

#### Step 1: Calculate the slopes of the sides.

1. **Slope of DE**:
\[
\text{slope of } DE = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 7}{-4 - (-2)} = \frac{-5}{-2} = \frac{5}{2}
\]

2. **Slope of DF**:
\[
\text{slope of } DF = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - 7}{6 - (-2)} = \frac{-9}{8}
\]

3. **Slope of EF**:
\[
\text{slope of } EF = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - 2}{6 - (-4)} = \frac{-4}{10} = \frac{-2}{5}
\]

#### Step 2: Check if the product of any two slopes is \(-1\).

- **Product of slopes of DE and EF**:
\[
\frac{5}{2} \times \frac{-2}{5} = -1
\]
Since the product is \(-1\), the sides \( DE \) and \( EF \) are perpendicular. Therefore, \(\triangle DEF\) is a right triangle.

### Part (b): Find the midpoint of the hypotenuse.

Since \( \triangle DEF \) is a right triangle, the hypotenuse is the side opposite the right angle, which is \( DF \).

The formula for the midpoint of a line segment between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:
\[
\text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
\]

Here, the hypotenuse is \( DF \), so the midpoint is:
\[
\text{Midpoint of } DF = \left( \frac{-2 + 6}{2}, \frac{7 + (-2)}{2} \right) = \left( \frac{4}{2}, \frac{5}{2} \right) = (2, 2.5)
\]

### Part (c): Show that the midpoint is equidistant from each of the vertices.

To prove that the midpoint is equidistant from each of the vertices \( D \), \( E \), and \( F \), we need to calculate the distances between the midpoint and each vertex, and show that they are equal.

The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:
\[
\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

1. **Distance from the midpoint to \( D(-2, 7) \)**:
\[
d(M, D) = \sqrt{(2 - (-2))^2 + (2.5 - 7)^2} = \sqrt{(4)^2 + (-4.5)^2} = \sqrt{16 + 20.25} = \sqrt{36.25} = 6.02
\]

2. **Distance from the midpoint to \( E(-4, 2) \)**:
\[
d(M, E) = \sqrt{(2 - (-4))^2 + (2.5 - 2)^2} = \sqrt{(6)^2 + (0.5)^2} = \sqrt{36 + 0.25} = \sqrt{36.25} = 6.02
\]

3. **Distance from the midpoint to \( F(6, -2) \)**:
\[
d(M, F) = \sqrt{(2 - 6)^2 + (2.5 - (-2))^2} = \sqrt{(-4)^2 + (4.5)^2} = \sqrt{16 + 20.25} = \sqrt{36.25} = 6.02
\]

Since the distances are equal, the midpoint is equidistant from all three vertices.

---

### Final Answer:
- \( \triangle DEF \) is a right triangle because the slopes of \( DE \) and \( EF \) are perpendicular.
- The midpoint of the hypotenuse is \( (2, 2.5) \).
- The midpoint is equidistant from each of the vertices, with a distance of \( 6.02 \) units from each.

Would you like more details on any part of this problem?

Here are five related questions to explore:
1. How do you calculate the distance between two points in 3D space?
2. What is the formula for the area of a right triangle given its vertices?
3. How can you find the equation of the line through two points?
4. What are the properties of midpoints in different types of triangles?
5. How can the Pythagorean theorem be used to verify right triangles?

**Tip**: Always double-check your slope calculations when verifying perpendicularity, as this is crucial to identifying right triangles.

A triangle has vertices D(-2,7), E(-4,2), and F(6,-2). a) Show algebraically that this triangle is a right triangle. b) Find the midpoint of the hypotenuse. c) Show that this midpoint is equidistant from each of the vertices.

Learn how to algebraically prove that triangle DEF with vertices D(-2,7), E(-4,2), and F(6,-2) is a right triangle. Discover how to find the midpoint of the hypotenuse and prove that it is equidistant from all vertices. This solution covers key geometry concepts including the slope, midpoint, and distance formulas. Suitable for Grades 9-10 students.

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