Question

Use a coordinate proof to show that triangle ABC is scalene.
What are the lengths of sides AB,BC and AC?
ABC coordinates: A(-5,2) B(-3,5) C(-2,-1)

A. AB=3.6 BC=6.1 and AC = 4.2
B. AB =10.6 BC = 8.1 AC =7.1
C. AB = 8.5 BC 6.4 AC = 7.1
D. AB = 10.6 BC = 6.4 AC = 7.1

Answer 1

To determine if triangle ABC is scalene, we need to calculate the lengths of its sides AB, BC, and AC using the coordinates provided for points A, B, and C.

**Coordinates:**
- A(-5, 2)
- B(-3, 5)
- C(-2, -1)

**Distance Formula:**
The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

**Calculating Lengths of Each Side:**

1. **Length of AB**:
\[
AB = \sqrt{((-3) - (-5))^2 + (5 - 2)^2} = \sqrt{(-3 + 5)^2 + (3)^2}
\]
\[
= \sqrt{(2)^2 + (3)^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.6
\]

2. **Length of BC**:
\[
BC = \sqrt{((-2) - (-3))^2 + (-1 - 5)^2} = \sqrt{(-2 + 3)^2 + (-6)^2}
\]
\[
= \sqrt{(1)^2 + (-6)^2} = \sqrt{1 + 36} = \sqrt{37} \approx 6.1
\]

3. **Length of AC**:
\[
AC = \sqrt{((-2) - (-5))^2 + (-1 - 2)^2} = \sqrt{(-2 + 5)^2 + (-3)^2}
\]
\[
= \sqrt{(3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} \approx 4.2
\]

**Summary of Lengths:**
- \( AB \approx 3.6 \)
- \( BC \approx 6.1 \)
- \( AC \approx 4.2 \)

Since all three side lengths are different, triangle ABC is indeed a scalene triangle.

**Answer:**
A. AB=3.6, BC=6.1, and AC=4.2.