The sides of a triangle are OA, OB, AB and have equations `2x-y=0, 3x+y=0, x-3y+10=0`, respectively. Find the equation of the three medians of the triangle and verify that they are concurrent.

Answer

To solve the problem, we will find the equations of the three medians of the triangle formed by the lines OA, OB, and AB, and then verify that they are concurrent.

### Step 1: Find the coordinates of the vertices of the triangle

1. **Find point O**:
- The intersection of the lines \(2x - y = 0\) and \(3x + y = 0\).
- From \(2x - y = 0\), we have \(y = 2x\).
- Substitute \(y = 2x\) into \(3x + y = 0\):
\[ 3x + 2x = 0 \implies 5x = 0 \implies x = 0 \implies y = 0 \]
- Thus, \(O(0, 0)\).

2. **Find point A**:
- The intersection of lines OA and AB.
- From \(2x - y = 0\), we have \(y = 2x\).
- Substitute \(y = 2x\) into \(x - 3y + 10 = 0\):
\[ x - 3(2x) + 10 = 0 \implies x - 6x + 10 = 0 \implies -5x + 10 = 0 \implies 5x = 10 \implies x = 2 \]
- Substitute \(x = 2\) back into \(y = 2x\):
\[ y = 2(2) = 4 \]
- Thus, \(A(2, 4)\).

3. **Find point B**:
- The intersection of lines AB and OB.
- From \(3x + y = 0\), we have \(y = -3x\).
- Substitute \(y = -3x\) into \(x - 3y + 10 = 0\):
\[ x - 3(-3x) + 10 = 0 \implies x + 9x + 10 = 0 \implies 10x + 10 = 0 \implies 10x = -10 \implies x = -1 \]
- Substitute \(x = -1\) back into \(y = -3x\):
\[ y = -3(-1) = 3 \]
- Thus, \(B(-1, 3)\).

### Step 2: Find the midpoints of the sides

1. **Midpoint D of AB**:
\[ D\left(\frac{2 + (-1)}{2}, \frac{4 + 3}{2}\right) = D\left(\frac{1}{2}, \frac{7}{2}\right) \]

2. **Midpoint E of OA**:
\[ E\left(\frac{0 + 2}{2}, \frac{0 + 4}{2}\right) = E\left(1, 2\right) \]

3. **Midpoint F of OB**:
\[ F\left(\frac{0 + (-1)}{2}, \frac{0 + 3}{2}\right) = F\left(-\frac{1}{2}, \frac{3}{2}\right) \]

### Step 3: Find the equations of the medians

1. **Median OD** (from O to D):
- Points: \(O(0, 0)\) and \(D\left(\frac{1}{2}, \frac{7}{2}\right)\).
- Slope \(m_{OD} = \frac{\frac{7}{2} - 0}{\frac{1}{2} - 0} = 7\).
- Equation:
\[ y - 0 = 7(x - 0) \implies y = 7x \]

2. **Median AE** (from A to E):
- Points: \(A(2, 4)\) and \(E(1, 2)\).
- Slope \(m_{AE} = \frac{2 - 4}{1 - 2} = \frac{-2}{-1} = 2\).
- Equation:
\[ y - 4 = 2(x - 2) \implies y - 4 = 2x - 4 \implies y = 2x \]

3. **Median BF** (from B to F):
- Points: \(B(-1, 3)\) and \(F\left(-\frac{1}{2}, \frac{3}{2}\right)\).
- Slope \(m_{BF} = \frac{\frac{3}{2} - 3}{-\frac{1}{2} - (-1)} = \frac{\frac{3}{2} - \frac{6}{2}}{\frac{1}{2}} = \frac{-\frac{3}{2}}{\frac{1}{2}} = -3\).
- Equation:
\[ y - 3 = -3\left(x + 1\right) \implies y - 3 = -3x - 3 \implies y = -3x \]

### Step 4: Verify concurrency of the medians

To verify that the medians are concurrent, we need to check if they intersect at a common point.

1. Set \(y = 7x\) equal to \(y = 2x\):
\[ 7x = 2x \implies 5x = 0 \implies x = 0 \implies y = 0 \implies (0, 0) \]

2. Set \(y = 7x\) equal to \(y = -3x\):
\[ 7x = -3x \implies 10x = 0 \implies x = 0 \implies y = 0 \implies (0, 0) \]

3. Set \(y = 2x\) equal to \(y = -3x\):
\[ 2x = -3x \implies 5x = 0 \implies x = 0 \implies y = 0 \implies (0, 0) \]

Since all three medians intersect at the point \(O(0, 0)\), they are concurrent.

### Summary of the Medians
- Median OD: \(y = 7x\)
- Median AE: \(y = 2x\)
- Median BF: \(y = -3x\)