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)

A( 0, 2), B (2, 5), and C (−1, 7)

The length sides of the triangle are

AB , AC and BC

For AB A( 0, 2), B (2, 5)

| AB | = √ ( 2 - 0)² + ( 5 - 2)²

= √ 2² + 3² = √ 4 + 9

= √13 units

For AC A( 0, 2), C (−1, 7)

| AC | = √ ( - 1 - 0)² + (7-2)²

= √1 + 5² = √ 1 + 25

= √ 26 units

For BC B (2, 5), C (−1, 7)

| BC | = √ ( - 1 - 2)² + (7-5)²

= √ (-3)² + 2² = √ 9 + 4

= √13 units

For AB A( 0, 2), B (2, 5)

Slope = 5 - 2 / 2 - 0 = 3/2

For AC A( 0, 2), C (−1, 7)

Slope = 7-2/ -1-0 = 5/-1 = - 5

For BC B (2, 5), C (−1, 7)

Slope = -1-2/7-5 = -3/2

The triangle is an isosceles triangle since two sides | AB | and | BC | are equal that's they both have √13units.

Hope this helps you