The vertices of a triangle are `A(10,4) ,B(-4,9) and C(-2,-1).`Find the equation of the altitude through A.

Answer

To find the equation of the altitude through point A(10, 4) of triangle ABC with vertices A(10, 4), B(-4, 9), and C(-2, -1), we can follow these steps:

### Step 1: Find the slope of line BC
The coordinates of points B and C are B(-4, 9) and C(-2, -1). The formula for the slope (m) between two points (x1, y1) and (x2, y2) is given by:

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

Substituting the coordinates of B and C:

\[ m_{BC} = \frac{-1 - 9}{-2 - (-4)} = \frac{-10}{-2 + 4} = \frac{-10}{2} = -5 \]

### Step 2: Find the slope of the altitude AD
Since the altitude AD is perpendicular to line BC, we can use the property that the product of the slopes of two perpendicular lines is -1. Therefore:

\[ m_{BC} \cdot m_{AD} = -1 \]

Substituting the slope of BC:

\[ -5 \cdot m_{AD} = -1 \]

Solving for \(m_{AD}\):

\[ m_{AD} = \frac{1}{5} \]

### Step 3: Use the point-slope form to find the equation of line AD
The point-slope form of a line is given by:

\[ y - y_1 = m(x - x_1) \]

Here, we will use point A(10, 4) and the slope \(m_{AD} = \frac{1}{5}\):

\[ y - 4 = \frac{1}{5}(x - 10) \]

### Step 4: Simplify the equation
Now, we will simplify this equation:

\[ y - 4 = \frac{1}{5}x - 2 \]

Adding 4 to both sides:

\[ y = \frac{1}{5}x + 2 \]

### Step 5: Rearranging to standard form
To express this in standard form (Ax + By + C = 0), we can rearrange the equation:

\[ -\frac{1}{5}x + y - 2 = 0 \]

Multiplying through by 5 to eliminate the fraction:

\[ -x + 5y - 10 = 0 \]

Rearranging gives us:

\[ x - 5y + 10 = 0 \]

Thus, the equation of the altitude through point A is:

\[ x - 5y + 10 = 0 \]