Find the coordinates of the centroid of the triangle whose sides are `12x^2-20xy+7y^2=0 and 2x-3y+4=0`

Answer

To find the coordinates of the centroid of the triangle formed by the given lines, we can follow these steps:

### Step 1: Identify the equations of the lines
We have two equations:
1. \( 12x^2 - 20xy + 7y^2 = 0 \)
2. \( 2x - 3y + 4 = 0 \)

### Step 2: Factor the first equation
The first equation represents a pair of straight lines. We can factor it as follows:
\[ 12x^2 - 20xy + 7y^2 = 0 \]
We can rewrite it as:
\[ 12x^2 - 14xy - 6xy + 7y^2 = 0 \]
Grouping terms, we have:
\[ (12x^2 - 14xy) + (-6xy + 7y^2) = 0 \]
Factoring out common terms:
\[ 2x(6x - 7y) - y(6x - 7y) = 0 \]
This gives us:
\[ (2x - y)(6x - 7y) = 0 \]
Thus, we have two lines:
1. \( 2x - y = 0 \) (or \( y = 2x \))
2. \( 6x - 7y = 0 \) (or \( y = \frac{6}{7}x \))

### Step 3: Solve for intersection points
Now we will find the intersection points of these lines with the third line \( 2x - 3y + 4 = 0 \) (or \( y = \frac{2}{3}x + \frac{4}{3} \)).

#### Finding intersection of \( 2x - y = 0 \) and \( 2x - 3y + 4 = 0 \):
Substituting \( y = 2x \) into \( 2x - 3y + 4 = 0 \):
\[ 2x - 3(2x) + 4 = 0 \\ 2x - 6x + 4 = 0 \\ -4x + 4 = 0 \\ 4x = 4 \\ x = 1 \]
Now, substituting \( x = 1 \) back into \( y = 2x \):
\[ y = 2(1) = 2 \]
So, the first intersection point is \( (1, 2) \).

#### Finding intersection of \( 6x - 7y = 0 \) and \( 2x - 3y + 4 = 0 \):
Substituting \( y = \frac{6}{7}x \) into \( 2x - 3y + 4 = 0 \):
\[ 2x - 3\left(\frac{6}{7}x\right) + 4 = 0 \\ 2x - \frac{18}{7}x + 4 = 0 \\ \left(2 - \frac{18}{7}\right)x + 4 = 0 \\ \left(\frac{14}{7} - \frac{18}{7}\right)x + 4 = 0 \\ -\frac{4}{7}x + 4 = 0 \\ \frac{4}{7}x = 4 \\ x = 7 \]
Now substituting \( x = 7 \) back into \( y = \frac{6}{7}x \):
\[ y = \frac{6}{7}(7) = 6 \]
So, the second intersection point is \( (7, 6) \).

#### Finding the origin point:
The third vertex of the triangle is the origin, which is \( (0, 0) \).

### Step 4: Coordinates of the vertices
Now we have the vertices of the triangle:
1. \( A(1, 2) \)
2. \( B(7, 6) \)
3. \( O(0, 0) \)

### Step 5: Calculate the centroid
The coordinates of the centroid \( (x_c, y_c) \) of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is given by:
\[ x_c = \frac{x_1 + x_2 + x_3}{3}, \quad y_c = \frac{y_1 + y_2 + y_3}{3} \]
Substituting the coordinates:
\[ x_c = \frac{1 + 7 + 0}{3} = \frac{8}{3}, \quad y_c = \frac{2 + 6 + 0}{3} = \frac{8}{3} \]

### Final Answer
The coordinates of the centroid of the triangle are:
\[ \left( \frac{8}{3}, \frac{8}{3} \right) \]