To find the equations of the sides of the triangle with vertices at the points A(0, 1), B(2, 0), and C(-1, -2), we can use the two-point form of the equation of a line. The formula for the line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[
y - y_1 = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1)
\]
We will calculate the equations for the three sides of the triangle: AB, BC, and CA.
### Step 1: Find the equation of side AB
1. Identify the coordinates of points A and B:
- \(A(0, 1)\) and \(B(2, 0)\)
- Here, \(x_1 = 0\), \(y_1 = 1\), \(x_2 = 2\), \(y_2 = 0\)
2. Substitute these values into the line equation:
\[
y - 1 = \frac{0 - 1}{2 - 0} (x - 0)
\]
\[
y - 1 = \frac{-1}{2} x
\]
3. Rearranging gives:
\[
2y - 2 = -x \quad \Rightarrow \quad x + 2y - 2 = 0
\]
Thus, the equation of side AB is:
\[
x + 2y - 2 = 0
\]
### Step 2: Find the equation of side BC
1. Identify the coordinates of points B and C:
- \(B(2, 0)\) and \(C(-1, -2)\)
- Here, \(x_1 = 2\), \(y_1 = 0\), \(x_2 = -1\), \(y_2 = -2\)
2. Substitute these values into the line equation:
\[
y - 0 = \frac{-2 - 0}{-1 - 2} (x - 2)
\]
\[
y = \frac{-2}{-3} (x - 2)
\]
\[
y = \frac{2}{3} (x - 2)
\]
3. Rearranging gives:
\[
3y = 2x - 4 \quad \Rightarrow \quad 2x - 3y - 4 = 0
\]
Thus, the equation of side BC is:
\[
2x - 3y - 4 = 0
\]
### Step 3: Find the equation of side CA
1. Identify the coordinates of points C and A:
- \(C(-1, -2)\) and \(A(0, 1)\)
- Here, \(x_1 = -1\), \(y_1 = -2\), \(x_2 = 0\), \(y_2 = 1\)
2. Substitute these values into the line equation:
\[
y + 2 = \frac{1 - (-2)}{0 - (-1)} (x + 1)
\]
\[
y + 2 = \frac{3}{1} (x + 1)
\]
\[
y + 2 = 3(x + 1)
\]
3. Rearranging gives:
\[
y + 2 = 3x + 3 \quad \Rightarrow \quad 3x - y + 1 = 0
\]
Thus, the equation of side CA is:
\[
3x - y + 1 = 0
\]
### Summary of the equations of the sides of the triangle:
1. Side AB: \(x + 2y - 2 = 0\)
2. Side BC: \(2x - 3y - 4 = 0\)
3. Side CA: \(3x - y + 1 = 0\)
