A (1,-2), B(-2,3) and C (2,-5) are the vertives of `Delta` ABC. Find the equation of the
(i) side AC
(ii) altiude from A
(iii) median from B
Perpendicular bisector of side AB.

Answer

To solve the problem step by step, we will find the equations for the side AC, the altitude from A, the median from B, and the perpendicular bisector of side AB for triangle ABC with vertices A(1, -2), B(-2, 3), and C(2, -5).

### Step 1: Find the equation of side AC

1. **Identify the coordinates of points A and C**:
- A(1, -2)
- C(2, -5)

2. **Use the two-point form of the equation of a line**:
\[ y - y_1 = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1) \]
Here, \( (x_1, y_1) = (1, -2) \) and \( (x_2, y_2) = (2, -5) \).

3. **Calculate the slope**:
\[ \text{slope} = \frac{-5 - (-2)}{2 - 1} = \frac{-3}{1} = -3 \]

4. **Substitute into the line equation**:
\[ y + 2 = -3(x - 1) \]
\[ y + 2 = -3x + 3 \]
\[ 3x + y = 1 \]

**Equation of side AC**: \( 3x + y = 1 \)

### Step 2: Find the equation of the altitude from A

1. **Find the slope of side BC**:
- B(-2, 3) and C(2, -5)
\[ \text{slope of BC} = \frac{-5 - 3}{2 - (-2)} = \frac{-8}{4} = -2 \]

2. **Find the slope of the altitude from A** (perpendicular to BC):
\[ \text{slope of altitude} = \frac{1}{2} \]

3. **Use point-slope form** with point A(1, -2):
\[ y - (-2) = \frac{1}{2}(x - 1) \]
\[ y + 2 = \frac{1}{2}x - \frac{1}{2} \]
\[ 2y + 4 = x - 1 \]
\[ x - 2y - 5 = 0 \]

**Equation of the altitude from A**: \( x - 2y - 5 = 0 \)

### Step 3: Find the equation of the median from B

1. **Find the midpoint D of AC**:
\[ D\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = D\left(\frac{1 + 2}{2}, \frac{-2 - 5}{2}\right) = D\left(\frac{3}{2}, -\frac{7}{2}\right) \]

2. **Use point-slope form with point B(-2, 3) and point D**:
\[ \text{slope of BD} = \frac{-\frac{7}{2} - 3}{\frac{3}{2} - (-2)} = \frac{-\frac{7}{2} - \frac{6}{2}}{\frac{3}{2} + \frac{4}{2}} = \frac{-\frac{13}{2}}{\frac{7}{2}} = -\frac{13}{7} \]

3. **Equation of line BD**:
\[ y - 3 = -\frac{13}{7}(x + 2) \]
\[ 7y - 21 = -13x - 26 \]
\[ 13x + 7y + 5 = 0 \]

**Equation of the median from B**: \( 13x + 7y + 5 = 0 \)

### Step 4: Find the perpendicular bisector of side AB

1. **Find the midpoint E of AB**:
\[ E\left(\frac{1 + (-2)}{2}, \frac{-2 + 3}{2}\right) = E\left(-\frac{1}{2}, \frac{1}{2}\right) \]

2. **Find the slope of AB**:
\[ \text{slope of AB} = \frac{3 - (-2)}{-2 - 1} = \frac{5}{-3} = -\frac{5}{3} \]

3. **Find the slope of the perpendicular bisector**:
\[ \text{slope of perpendicular bisector} = \frac{3}{5} \]

4. **Equation of the perpendicular bisector**:
\[ y - \frac{1}{2} = \frac{3}{5}\left(x + \frac{1}{2}\right) \]
\[ 5y - 5 \cdot \frac{1}{2} = 3x + \frac{3}{2} \]
\[ 5y - \frac{5}{2} = 3x + \frac{3}{2} \]
\[ 3x - 5y + 4 = 0 \]

**Equation of the perpendicular bisector of side AB**: \( 3x - 5y + 4 = 0 \)

### Summary of Results
1. **Equation of side AC**: \( 3x + y = 1 \)
2. **Equation of altitude from A**: \( x - 2y - 5 = 0 \)
3. **Equation of median from B**: \( 13x + 7y + 5 = 0 \)
4. **Equation of perpendicular bisector of side AB**: \( 3x - 5y + 4 = 0 \)