Answer :
### Part A: Find the length of each side of the triangle (4 points)
To determine the lengths of the sides of the triangle, we use the distance formula:
[tex]\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
1. Length of side [tex]\( AB \)[/tex]:
[tex]\[ A = (0, 2) \quad \text{and} \quad B = (2, 5) \][/tex]
[tex]\[ AB = \sqrt{(2 - 0)^2 + (5 - 2)^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.605551275463989 \][/tex]
2. Length of side [tex]\( BC \)[/tex]:
[tex]\[ B = (2, 5) \quad \text{and} \quad C = (-1, 7) \][/tex]
[tex]\[ BC = \sqrt{(-1 - 2)^2 + (7 - 5)^2} = \sqrt{(-3)^2 + (2)^2} = \sqrt{9 + 4} = \sqrt{13} \approx 3.605551275463989 \][/tex]
3. Length of side [tex]\( CA \)[/tex]:
[tex]\[ C = (-1, 7) \quad \text{and} \quad A = (0, 2) \][/tex]
[tex]\[ CA = \sqrt{(0 - (-1))^2 + (2 - 7)^2} = \sqrt{1^2 + (-5)^2} = \sqrt{1 + 25} = \sqrt{26} \approx 5.0990195135927845 \][/tex]
The lengths of the sides are:
- [tex]\( AB \approx 3.606 \)[/tex]
- [tex]\( BC \approx 3.606 \)[/tex]
- [tex]\( CA \approx 5.099 \)[/tex]
### Part B: Find the slope of each side of the triangle (3 points)
To find the slopes of the sides of the triangle, we use the slope formula:
[tex]\[ \text{Slope} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
1. Slope of side [tex]\( AB \)[/tex]:
[tex]\[ A = (0, 2) \quad \text{and} \quad B = (2, 5) \][/tex]
[tex]\[ \text{Slope of } AB = \frac{5 - 2}{2 - 0} = \frac{3}{2} = 1.5 \][/tex]
2. Slope of side [tex]\( BC \)[/tex]:
[tex]\[ B = (2, 5) \quad \text{and} \quad C = (-1, 7) \][/tex]
[tex]\[ \text{Slope of } BC = \frac{7 - 5}{-1 - 2} = \frac{2}{-3} = -\frac{2}{3} \approx -0.6666666666666666 \][/tex]
3. Slope of side [tex]\( CA \)[/tex]:
[tex]\[ C = (-1, 7) \quad \text{and} \quad A = (0, 2) \][/tex]
[tex]\[ \text{Slope of } CA = \frac{2 - 7}{0 - (-1)} = \frac{-5}{1} = -5.0 \][/tex]
The slopes of the sides are:
- [tex]\( \text{Slope of } AB = 1.5 \)[/tex]
- [tex]\( \text{Slope of } BC \approx -0.667 \)[/tex]
- [tex]\( \text{Slope of } CA = -5.0 \)[/tex]
### Part C: Classify the triangle (3 points)
To classify the triangle, we compare the lengths of the sides:
1. If all three sides are of the same length, the triangle is equilateral.
2. If exactly two sides are of the same length, the triangle is isosceles.
3. If all three sides are of different lengths, the triangle is scalene.
Comparing the side lengths:
- [tex]\( AB \approx 3.606 \)[/tex]
- [tex]\( BC \approx 3.606 \)[/tex]
- [tex]\( CA \approx 5.099 \)[/tex]
Since [tex]\( AB \)[/tex] and [tex]\( BC \)[/tex] are equal while [tex]\( CA \)[/tex] is different, we classify the triangle as:
- Isosceles, because it has exactly two sides of equal length.
In summary, the triangle ABC is classified as Isosceles. This classification is based on the fact that two of its sides, [tex]\( AB \)[/tex] and [tex]\( BC \)[/tex], have the same length, while the third side [tex]\( CA \)[/tex] is different.
To determine the lengths of the sides of the triangle, we use the distance formula:
[tex]\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
1. Length of side [tex]\( AB \)[/tex]:
[tex]\[ A = (0, 2) \quad \text{and} \quad B = (2, 5) \][/tex]
[tex]\[ AB = \sqrt{(2 - 0)^2 + (5 - 2)^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.605551275463989 \][/tex]
2. Length of side [tex]\( BC \)[/tex]:
[tex]\[ B = (2, 5) \quad \text{and} \quad C = (-1, 7) \][/tex]
[tex]\[ BC = \sqrt{(-1 - 2)^2 + (7 - 5)^2} = \sqrt{(-3)^2 + (2)^2} = \sqrt{9 + 4} = \sqrt{13} \approx 3.605551275463989 \][/tex]
3. Length of side [tex]\( CA \)[/tex]:
[tex]\[ C = (-1, 7) \quad \text{and} \quad A = (0, 2) \][/tex]
[tex]\[ CA = \sqrt{(0 - (-1))^2 + (2 - 7)^2} = \sqrt{1^2 + (-5)^2} = \sqrt{1 + 25} = \sqrt{26} \approx 5.0990195135927845 \][/tex]
The lengths of the sides are:
- [tex]\( AB \approx 3.606 \)[/tex]
- [tex]\( BC \approx 3.606 \)[/tex]
- [tex]\( CA \approx 5.099 \)[/tex]
### Part B: Find the slope of each side of the triangle (3 points)
To find the slopes of the sides of the triangle, we use the slope formula:
[tex]\[ \text{Slope} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
1. Slope of side [tex]\( AB \)[/tex]:
[tex]\[ A = (0, 2) \quad \text{and} \quad B = (2, 5) \][/tex]
[tex]\[ \text{Slope of } AB = \frac{5 - 2}{2 - 0} = \frac{3}{2} = 1.5 \][/tex]
2. Slope of side [tex]\( BC \)[/tex]:
[tex]\[ B = (2, 5) \quad \text{and} \quad C = (-1, 7) \][/tex]
[tex]\[ \text{Slope of } BC = \frac{7 - 5}{-1 - 2} = \frac{2}{-3} = -\frac{2}{3} \approx -0.6666666666666666 \][/tex]
3. Slope of side [tex]\( CA \)[/tex]:
[tex]\[ C = (-1, 7) \quad \text{and} \quad A = (0, 2) \][/tex]
[tex]\[ \text{Slope of } CA = \frac{2 - 7}{0 - (-1)} = \frac{-5}{1} = -5.0 \][/tex]
The slopes of the sides are:
- [tex]\( \text{Slope of } AB = 1.5 \)[/tex]
- [tex]\( \text{Slope of } BC \approx -0.667 \)[/tex]
- [tex]\( \text{Slope of } CA = -5.0 \)[/tex]
### Part C: Classify the triangle (3 points)
To classify the triangle, we compare the lengths of the sides:
1. If all three sides are of the same length, the triangle is equilateral.
2. If exactly two sides are of the same length, the triangle is isosceles.
3. If all three sides are of different lengths, the triangle is scalene.
Comparing the side lengths:
- [tex]\( AB \approx 3.606 \)[/tex]
- [tex]\( BC \approx 3.606 \)[/tex]
- [tex]\( CA \approx 5.099 \)[/tex]
Since [tex]\( AB \)[/tex] and [tex]\( BC \)[/tex] are equal while [tex]\( CA \)[/tex] is different, we classify the triangle as:
- Isosceles, because it has exactly two sides of equal length.
In summary, the triangle ABC is classified as Isosceles. This classification is based on the fact that two of its sides, [tex]\( AB \)[/tex] and [tex]\( BC \)[/tex], have the same length, while the third side [tex]\( CA \)[/tex] is different.
