Question

Triangle ABC has vertices located at A( 0, 2), B (2, 5), and C (−1, 7).

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)

Answer 1

To solve the problem regarding triangle ABC with vertices at A(0, 2), B(2, 5), and C(−1, 7), we will move step-by-step through each part: finding side lengths, slopes, and classifying the triangle.

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

To find the lengths of the sides, we can use the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

**Length of side AB:**
- A(0, 2) and B(2, 5)
\[
AB = \sqrt{(2 - 0)^2 + (5 - 2)^2} = \sqrt{(2)^2 + (3)^2} = \sqrt{4 + 9} = \sqrt{13}
\]

**Length of side BC:**
- B(2, 5) and C(−1, 7)
\[
BC = \sqrt{(-1 - 2)^2 + (7 - 5)^2} = \sqrt{(-3)^2 + (2)^2} = \sqrt{9 + 4} = \sqrt{13}
\]

**Length of side CA:**
- C(−1, 7) and A(0, 2)
\[
CA = \sqrt{(0 - (-1))^2 + (2 - 7)^2} = \sqrt{(1)^2 + (-5)^2} = \sqrt{1 + 25} = \sqrt{26}
\]

Thus, the lengths of the sides are:
- \( AB = \sqrt{13} \)
- \( BC = \sqrt{13} \)
- \( CA = \sqrt{26} \)

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

The slope of a line segment between two points (x1, y1) and (x2, y2) is given by the formula:
\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]

**Slope of AB:**
- Points A(0, 2) and B(2, 5)
\[
m_{AB} = \frac{5 - 2}{2 - 0} = \frac{3}{2}
\]

**Slope of BC:**
- Points B(2, 5) and C(−1, 7)
\[
m_{BC} = \frac{7 - 5}{-1 - 2} = \frac{2}{-3} = -\frac{2}{3}
\]

**Slope of CA:**
- Points C(−1, 7) and A(0, 2)
\[
m_{CA} = \frac{2 - 7}{0 - (-1)} = \frac{-5}{1} = -5
\]

Thus, the slopes of the sides are:
- \( m_{AB} = \frac{3}{2} \)
- \( m_{BC} = -\frac{2}{3} \)
- \( m_{CA} = -5 \)

### Part C: Classify the triangle.

To classify triangle ABC, we can check the lengths of the sides obtained in Part A.

1. **Identifying congruent sides**:
- \( AB = BC = \sqrt{13} \)
- \( CA = \sqrt{26} \)

Since two sides of the triangle are equal (\(AB\) and \(BC\)), triangle ABC is classified as an **isosceles triangle**.

2. **Checking for right angle**:
- We can also check the squares of the side lengths to apply the Pythagorean theorem:
- \( AB^2 + BC^2 = ( \sqrt{13} )^2 + ( \sqrt{13} )^2 = 13 + 13 = 26 \)
- \( CA^2 = ( \sqrt{26} )^2 = 26 \)

Thus, \( AB^2 + BC^2 = CA^2 \), confirming that triangle ABC is also a **right triangle**.

### Conclusion
Triangle ABC is classified as an **isosceles right triangle** because it has two equal sides and satisfies the Pythagorean theorem conditions.