Question

question 1 (essay worth 10 points) (mc) 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)

Explanation:

Part A:

Step1: Calculate length of \(AB\)

Use distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). For \(A(0,2)\) and \(B(2,5)\), \(x_1 = 0,y_1=2,x_2 = 2,y_2 = 5\).
\(AB=\sqrt{(2 - 0)^2+(5 - 2)^2}=\sqrt{4 + 9}=\sqrt{13}\)

Step2: Calculate length of \(BC\)

For \(B(2,5)\) and \(C(-1,7)\), \(x_1 = 2,y_1 = 5,x_2=-1,y_2 = 7\).
\(BC=\sqrt{(-1 - 2)^2+(7 - 5)^2}=\sqrt{9+4}=\sqrt{13}\)

Step3: Calculate length of \(AC\)

For \(A(0,2)\) and \(C(-1,7)\), \(x_1 = 0,y_1=2,x_2=-1,y_2 = 7\).
\(AC=\sqrt{(-1 - 0)^2+(7 - 2)^2}=\sqrt{1 + 25}=\sqrt{26}\)

Part B:

Step1: Calculate slope of \(AB\)

Use slope formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\). For \(A(0,2)\) and \(B(2,5)\), \(m_{AB}=\frac{5 - 2}{2 - 0}=\frac{3}{2}\)

Step2: Calculate slope of \(BC\)

For \(B(2,5)\) and \(C(-1,7)\), \(m_{BC}=\frac{7 - 5}{-1 - 2}=-\frac{2}{3}\)

Step3: Calculate slope of \(AC\)

For \(A(0,2)\) and \(C(-1,7)\), \(m_{AC}=\frac{7 - 2}{-1 - 0}=- 5\)

Part C:

Since \(AB = BC=\sqrt{13}\), two sides are equal. Also, \(AB^{2}+BC^{2}=13 + 13=26=AC^{2}\) (by Pythagorean theorem).

Answer:

• Part A: \(AB=\sqrt{13}\), \(BC=\sqrt{13}\), \(AC=\sqrt{26}\)
• Part B: Slope of \(AB=\frac{3}{2}\), slope of \(BC=-\frac{2}{3}\), slope of \(AC=-5\)
• Part C: The triangle is an isosceles right - triangle. Because two sides (\(AB\) and \(BC\)) are equal (\(AB = BC\)) and \(AB^{2}+BC^{2}=AC^{2}\) (satisfies Pythagorean theorem for right - triangle).