The vertices of a triangle are the points `(2,1)` , `(-2,3)` and `(4,5)`. Find the equation of its sides.

Answer

To find the equations of the sides of the triangle formed by the vertices \( A(2,1) \), \( B(-2,3) \), and \( C(4,5) \), we will calculate the equations for each side: AB, BC, and AC.

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

1. **Calculate the slope of line AB**:
\[ \text{slope} (m_{AB}) = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 1}{-2 - 2} = \frac{2}{-4} = -\frac{1}{2} \]

2. **Use point-slope form to find the equation**:
\[ y - y_1 = m_{AB}(x - x_1) \]
Substituting \( (x_1, y_1) = (2, 1) \):
\[ y - 1 = -\frac{1}{2}(x - 2) \]
Multiplying through by 2 to eliminate the fraction:
\[ 2(y - 1) = -(x - 2) \implies 2y - 2 = -x + 2 \]
Rearranging gives:
\[ x + 2y - 4 = 0 \]

### Step 2: Find the equation of side BC

1. **Calculate the slope of line BC**:
\[ \text{slope} (m_{BC}) = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 3}{4 - (-2)} = \frac{2}{6} = \frac{1}{3} \]

2. **Use point-slope form to find the equation**:
\[ y - y_1 = m_{BC}(x - x_1) \]
Substituting \( (x_1, y_1) = (-2, 3) \):
\[ y - 3 = \frac{1}{3}(x + 2) \]
Multiplying through by 3:
\[ 3(y - 3) = x + 2 \implies 3y - 9 = x + 2 \]
Rearranging gives:
\[ x - 3y + 11 = 0 \]

### Step 3: Find the equation of side AC

1. **Calculate the slope of line AC**:
\[ \text{slope} (m_{AC}) = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 1}{4 - 2} = \frac{4}{2} = 2 \]

2. **Use point-slope form to find the equation**:
\[ y - y_1 = m_{AC}(x - x_1) \]
Substituting \( (x_1, y_1) = (2, 1) \):
\[ y - 1 = 2(x - 2) \]
Rearranging gives:
\[ y - 1 = 2x - 4 \implies 2x - y - 3 = 0 \]

### Final Equations

The equations of the sides of the triangle are:
1. Side AB: \( x + 2y - 4 = 0 \)
2. Side BC: \( x - 3y + 11 = 0 \)
3. Side AC: \( 2x - y - 3 = 0 \)