The vertices of a `DeltaABC` are A (3, 8), B (-1, 2) and C (6, -6). Find :
(i) Slope of BC.

Answer

To find the slope of line segment BC in triangle ABC with vertices A(3, 8), B(-1, 2), and C(6, -6), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the coordinates of points B and C:**
- B has coordinates (-1, 2) which means \( x_1 = -1 \) and \( y_1 = 2 \).
- C has coordinates (6, -6) which means \( x_2 = 6 \) and \( y_2 = -6 \).

2. **Use the slope formula:**
The formula for the slope \( m \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

3. **Substitute the coordinates into the slope formula:**
- Substitute \( y_2 = -6 \), \( y_1 = 2 \), \( x_2 = 6 \), and \( x_1 = -1 \):
\[ m = \frac{-6 - 2}{6 - (-1)} \]

4. **Calculate the differences:**
- Calculate \( y_2 - y_1 \):
\[ -6 - 2 = -8 \]
- Calculate \( x_2 - x_1 \):
\[ 6 - (-1) = 6 + 1 = 7 \]

5. **Substitute the differences back into the slope formula:**
\[ m = \frac{-8}{7} \]

6. **Final answer:**
The slope of line segment BC is:
\[ m = -\frac{8}{7} \]