Two sides of a triangle given by `2x+3y-5=0` and `5x-4y+10=0` intersect at A. Centroid of the triangle is at the origin. Find the equation of the median through A.

Answer

To find the equation of the median through point A of the triangle formed by the lines \(2x + 3y - 5 = 0\) and \(5x - 4y + 10 = 0\), we can follow these steps:

### Step 1: Find the coordinates of point A (intersection of the two lines)

To find the intersection point A, we solve the equations of the two lines simultaneously:

1. The first line is:
\[ 2x + 3y - 5 = 0 \quad \text{(Equation 1)} \]
Rearranging gives:
\[ 2x + 3y = 5 \quad \text{(1)} \]

2. The second line is:
\[ 5x - 4y + 10 = 0 \quad \text{(Equation 2)} \]
Rearranging gives:
\[ 5x - 4y = -10 \quad \text{(2)} \]

Now, we can multiply Equation (1) by 5 and Equation (2) by 2 to eliminate \(x\):
\[ 10x + 15y = 25 \quad \text{(3)} \]
\[ 10x - 8y = -20 \quad \text{(4)} \]

Now, subtract Equation (4) from Equation (3):
\[ (10x + 15y) - (10x - 8y) = 25 - (-20) \]
This simplifies to:
\[ 23y = 45 \implies y = \frac{45}{23} \]

Substituting \(y\) back into Equation (1) to find \(x\):
\[ 2x + 3\left(\frac{45}{23}\right) = 5 \]
\[ 2x + \frac{135}{23} = 5 \]
\[ 2x = 5 - \frac{135}{23} = \frac{115}{23} - \frac{135}{23} = -\frac{20}{23} \]
\[ x = -\frac{10}{23} \]

Thus, the coordinates of point A are:
\[ A\left(-\frac{10}{23}, \frac{45}{23}\right) \]

### Step 2: Use the centroid condition

The centroid \(G\) of triangle ABC is given to be at the origin \((0, 0)\). The centroid formula is:
\[ G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) = (0, 0) \]
This gives us two equations:
1. \(x_1 + x_2 + x_3 = 0\)
2. \(y_1 + y_2 + y_3 = 0\)

Substituting \(x_1 = -\frac{10}{23}\) and \(y_1 = \frac{45}{23}\):
\[ -\frac{10}{23} + x_2 + x_3 = 0 \implies x_2 + x_3 = \frac{10}{23} \]
\[ \frac{45}{23} + y_2 + y_3 = 0 \implies y_2 + y_3 = -\frac{45}{23} \]

### Step 3: Find the coordinates of point D (midpoint of BC)

Let \(D\) be the midpoint of line segment \(BC\). The coordinates of \(D\) can be expressed as:
\[ D\left(\frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}\right) \]

Substituting the values we found:
\[ D\left(\frac{\frac{10}{23}}{2}, \frac{-\frac{45}{23}}{2}\right) = D\left(\frac{5}{23}, -\frac{45}{46}\right) \]

### Step 4: Find the equation of the median AD

The slope \(m\) of line segment \(AD\) is given by:
\[ m = \frac{y_D - y_A}{x_D - x_A} = \frac{-\frac{45}{46} - \frac{45}{23}}{\frac{5}{23} - \left(-\frac{10}{23}\right)} \]

Calculating \(y_D - y_A\):
\[ -\frac{45}{46} - \frac{45}{23} = -\frac{45 \cdot 23 + 45 \cdot 46}{46 \cdot 23} = -\frac{45(23 + 46)}{46 \cdot 23} = -\frac{45 \cdot 69}{46 \cdot 23} \]

Calculating \(x_D - x_A\):
\[ \frac{5}{23} + \frac{10}{23} = \frac{15}{23} \]

Thus, the slope \(m\) becomes:
\[ m = \frac{-\frac{45 \cdot 69}{46 \cdot 23}}{\frac{15}{23}} = -\frac{45 \cdot 69}{46 \cdot 15} \]

### Step 5: Form the equation of line AD using point-slope form

Using point-slope form \(y - y_1 = m(x - x_1)\):
\[ y - \frac{45}{23} = m\left(x + \frac{10}{23}\right) \]

Substituting the slope \(m\) and simplifying gives the equation of the median.

### Final Answer

The equation of the median through point A is:
\[ 207x - 46y = 0 \]