The vertices of a triangle are A (0, 5), B (-1, -2) and C (11, 7). Write down the equations of BC and the perpendicular from A to BC and hence find the co-ordinates of the foot of the perpendicular.

Answer

To solve the problem step by step, we will find the equations of line BC and the perpendicular line from point A to line BC, and then we will find the coordinates of the foot of the perpendicular.

### Step 1: Find the equation of line BC

1. **Identify the coordinates of points B and C:**
- B = (-1, -2)
- C = (11, 7)

2. **Calculate the slope (m) of line BC using the formula:**
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
Here, \( (x_1, y_1) = (-1, -2) \) and \( (x_2, y_2) = (11, 7) \).
\[ m = \frac{7 - (-2)}{11 - (-1)} = \frac{7 + 2}{11 + 1} = \frac{9}{12} = \frac{3}{4} \]

3. **Use the point-slope form of the line equation:**
\[ y - y_1 = m(x - x_1) \]
Using point B (-1, -2):
\[ y - (-2) = \frac{3}{4}(x - (-1)) \]
\[ y + 2 = \frac{3}{4}(x + 1) \]

4. **Rearrange to get the standard form:**
\[ y + 2 = \frac{3}{4}x + \frac{3}{4} \]
\[ 4y + 8 = 3x + 3 \]
\[ 3x - 4y = 5 \]
This is the equation of line BC.

### Step 2: Find the equation of the perpendicular line from A to BC

1. **Identify the coordinates of point A:**
- A = (0, 5)

2. **Find the slope of the perpendicular line (AD):**
The slope of line BC is \( \frac{3}{4} \), so the slope of the perpendicular line (m') is:
\[ m' = -\frac{1}{m} = -\frac{1}{\frac{3}{4}} = -\frac{4}{3} \]

3. **Use the point-slope form of the line equation for line AD:**
\[ y - y_1 = m'(x - x_1) \]
Using point A (0, 5):
\[ y - 5 = -\frac{4}{3}(x - 0) \]
\[ y - 5 = -\frac{4}{3}x \]
Rearranging gives:
\[ 4x + 3y = 15 \]
This is the equation of line AD.

### Step 3: Find the coordinates of the foot of the perpendicular (intersection of BC and AD)

1. **Set the equations equal to find the intersection:**
We have:
- Line BC: \( 3x - 4y = 5 \)
- Line AD: \( 4x + 3y = 15 \)

2. **Multiply the equations to eliminate y:**
Multiply the first equation by 3 and the second by 4:
\[ 9x - 12y = 15 \quad \text{(1)} \]
\[ 16x + 12y = 60 \quad \text{(2)} \]

3. **Add equations (1) and (2):**
\[ 9x - 12y + 16x + 12y = 15 + 60 \]
\[ 25x = 75 \]
\[ x = 3 \]

4. **Substitute x back to find y:**
Substitute \( x = 3 \) into the equation of line AD:
\[ 4(3) + 3y = 15 \]
\[ 12 + 3y = 15 \]
\[ 3y = 3 \quad \Rightarrow \quad y = 1 \]

5. **Coordinates of the foot of the perpendicular:**
The foot of the perpendicular D is at \( (3, 1) \).

### Final Answer:
The coordinates of the foot of the perpendicular from A to line BC are \( D(3, 1) \).