`A(-3, 1), B(4,4) and C(1, -2)` are the vertices of a triangle ABC. Find:
the equation of altitude AE.

Answer

To find the equation of the altitude AE from vertex A to side BC of triangle ABC with vertices A(-3, 1), B(4, 4), and C(1, -2), we can follow these steps:

### Step 1: Find the slope of line segment BC.
The slope \( m \) of a line through points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by the formula:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
Here, \( B(4, 4) \) and \( C(1, -2) \) are the points.

Substituting the coordinates:
\[ m_{BC} = \frac{-2 - 4}{1 - 4} = \frac{-6}{-3} = 2 \]

### Step 2: Determine the slope of the altitude AE.
Since the altitude AE is perpendicular to line BC, the slope of AE \( m_{AE} \) can be found using the property that the product of the slopes of two perpendicular lines is -1:
\[ m_{AE} \cdot m_{BC} = -1 \]
Substituting \( m_{BC} = 2 \):
\[ m_{AE} \cdot 2 = -1 \implies m_{AE} = -\frac{1}{2} \]

### Step 3: Use point-slope form to find the equation of line AE.
The point-slope form of a line is given by:
\[ y - y_1 = m(x - x_1) \]
Using point A(-3, 1) and slope \( m_{AE} = -\frac{1}{2} \):
\[ y - 1 = -\frac{1}{2}(x + 3) \]

### Step 4: Simplify the equation.
Distributing the slope:
\[ y - 1 = -\frac{1}{2}x - \frac{3}{2} \]
Adding 1 to both sides:
\[ y = -\frac{1}{2}x - \frac{3}{2} + 1 \]
Converting 1 to a fraction:
\[ y = -\frac{1}{2}x - \frac{3}{2} + \frac{2}{2} = -\frac{1}{2}x - \frac{1}{2} \]

### Step 5: Rearranging to standard form.
Multiplying through by 2 to eliminate the fraction:
\[ 2y = -x - 1 \]
Rearranging gives:
\[ x + 2y + 1 = 0 \]

### Final Answer:
The equation of the altitude AE is:
\[ \boxed{x + 2y + 1 = 0} \]