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.
- 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.
- 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.
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