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

Answer

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

### Step 1: Determine the slope of line AC
To find the slope of line AC, we use the formula for the slope between two points (x1, y1) and (x2, y2):

\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \]

Here, the coordinates of points A and C are:
- A(1, 1) → (x1, y1) = (1, 1)
- C(3, 6) → (x2, y2) = (3, 6)

Substituting the values into the slope formula:

\[ \text{slope of AC} = \frac{6 - 1}{3 - 1} = \frac{5}{2} \]

### Step 2: Find the slope of the altitude from point B to line AC
The altitude from point B to line AC will be perpendicular to line AC. The slope of a line that is perpendicular to another line is the negative reciprocal of the slope of the original line.

The slope of AC is \(\frac{5}{2}\). Therefore, the slope of the altitude (let's call it BE) is:

\[ \text{slope of BE} = -\frac{1}{\text{slope of AC}} = -\frac{1}{\frac{5}{2}} = -\frac{2}{5} \]

### Conclusion
Thus, the slope of the altitude to AC from point B is:

\[ \text{slope of altitude BE} = -\frac{2}{5} \]