The vertices of a `DeltaABC` are A(1, 1), B(7, 3) and C(3, 6). State the slope of the altitude to BC

Answer

To find the slope of the altitude from point A to line BC in triangle ABC with vertices A(1, 1), B(7, 3), and C(3, 6), we can follow these steps:

### Step 1: Find the slope of line BC
The slope \( m \) of a line passing through two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by the formula:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
For points B(7, 3) and C(3, 6), we can assign:
- \( (x_1, y_1) = (7, 3) \)
- \( (x_2, y_2) = (3, 6) \)

Now, substituting these values into the slope formula:
\[ m_{BC} = \frac{6 - 3}{3 - 7} = \frac{3}{-4} = -\frac{3}{4} \]

### Step 2: Use the property of perpendicular slopes
The altitude from point A to line BC is perpendicular to line BC. The product of the slopes of two perpendicular lines is -1. Let the slope of the altitude from A to BC be \( m_{AL} \).

Thus, we have:
\[ m_{BC} \cdot m_{AL} = -1 \]
Substituting \( m_{BC} = -\frac{3}{4} \):
\[ -\frac{3}{4} \cdot m_{AL} = -1 \]

### Step 3: Solve for the slope of the altitude
To find \( m_{AL} \), we rearrange the equation:
\[ m_{AL} = \frac{-1}{-\frac{3}{4}} = \frac{4}{3} \]

### Conclusion
The slope of the altitude from point A to line BC is:
\[ \boxed{\frac{4}{3}} \]

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