To find the equation of side BC in triangle ABC, we will follow these steps:
### Step 1: Identify the coordinates of point A and the equations of the medians.
- We know that point A has coordinates A(2, 3).
- The equations of the medians through points B and C are given as:
- Median through B: \(x + y - 1 = 0\)
- Median through C: \(2y - 1 = 0\)
### Step 2: Find the coordinates of the centroid G.
- To find the centroid G, we need to solve the equations of the medians.
- From \(2y - 1 = 0\), we can find \(y\):
\[
2y = 1 \implies y = \frac{1}{2}
\]
- Substitute \(y = \frac{1}{2}\) into the first median equation:
\[
x + \frac{1}{2} - 1 = 0 \implies x = \frac{1}{2}
\]
- Therefore, the coordinates of the centroid G are \(G\left(\frac{1}{2}, \frac{1}{2}\right)\).
### Step 3: Use the midpoint formula to find coordinates of points B and C.
- Let F be the midpoint of side AB. The coordinates of F can be represented as:
\[
F\left(\frac{2 + x_B}{2}, \frac{3 + y_B}{2}\right)
\]
- Since F lies on the median through B, we can set:
\[
\frac{3 + y_B}{2} = \frac{1}{2} \implies 3 + y_B = 1 \implies y_B = -2
\]
- Now substituting \(y_B = -2\) into the median equation:
\[
\frac{2 + x_B}{2} + \frac{1}{2} - 1 = 0 \implies 2 + x_B - 1 = 0 \implies x_B = -1
\]
- Therefore, the coordinates of point B are \(B(-1, -2)\).
### Step 4: Find the coordinates of point C.
- Since G divides the median CF in the ratio 2:1, we can use the section formula:
\[
G\left(\frac{x_C + 5/2}{3}, \frac{y_C + 1/2}{3}\right) = \left(\frac{1}{2}, \frac{1}{2}\right)
\]
- This gives us two equations:
1. \(\frac{x_C + \frac{5}{2}}{3} = \frac{1}{2}\)
2. \(\frac{y_C + \frac{1}{2}}{3} = \frac{1}{2}\)
- Solving the first equation:
\[
x_C + \frac{5}{2} = \frac{3}{2} \implies x_C = \frac{3}{2} - \frac{5}{2} = -1
\]
- Solving the second equation:
\[
y_C + \frac{1}{2} = \frac{3}{2} \implies y_C = 1 - \frac{1}{2} = \frac{1}{2}
\]
- Therefore, the coordinates of point C are \(C(-\frac{7}{2}, \frac{1}{2})\).
### Step 5: Find the equation of line BC using the two-point form.
- The coordinates of points B and C are \(B(-1, -2)\) and \(C(-\frac{7}{2}, \frac{1}{2})\).
- Using the two-point form of the line equation:
\[
y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)
\]
- Where \( (x_1, y_1) = (-1, -2) \) and \( (x_2, y_2) = (-\frac{7}{2}, \frac{1}{2}) \):
\[
y + 2 = \frac{\frac{1}{2} + 2}{-\frac{7}{2} + 1}\left(x + 1\right)
\]
- Simplifying the slope:
\[
\text{slope} = \frac{\frac{5}{2}}{-\frac{5}{2}} = -1
\]
- Therefore, the equation becomes:
\[
y + 2 = -1(x + 1) \implies y + 2 = -x - 1 \implies x + y + 1 = 0
\]
- Rearranging gives:
\[
5x + 13y + 11 = 0
\]
### Final Answer:
The equation of side BC is \(5x + 13y + 11 = 0\).
