The centroid of the triangle formed by the pair of straight lines `12x^2- 20xy+7y^2=0` and the line `2x-3y + 4 = 0 `is

Options

A. `((8)/(3),(8)/(3))`
B. `((4)/(3),(4)/(3))`
C. `(2,2)`
D. `(1,1)`

Answer

To find the centroid of the triangle formed by the pair of straight lines given by the equation \(12x^2 - 20xy + 7y^2 = 0\) and the line \(2x - 3y + 4 = 0\), we will follow these steps:

### Step 1: Factor the Pair of Straight Lines
We start with the equation of the pair of straight lines:
\[ 12x^2 - 20xy + 7y^2 = 0 \]
We need to factor this quadratic equation. We will use the middle term splitting method.

1. Multiply the coefficient of \(x^2\) (which is 12) by the coefficient of \(y^2\) (which is 7):
\[ 12 \times 7 = 84 \]
2. We need two numbers that multiply to 84 and add up to -20. These numbers are -14 and -6.
3. Rewrite the equation:
\[ 12x^2 - 14xy - 6xy + 7y^2 = 0 \]
4. Group the terms:
\[ (12x^2 - 14xy) + (-6xy + 7y^2) = 0 \]
5. Factor by grouping:
\[ 2x(6x - 7y) - y(6x - 7y) = 0 \]
6. Factor out the common term:
\[ (6x - 7y)(2x - y) = 0 \]
Thus, the two lines are:
\[ 6x - 7y = 0 \quad \text{and} \quad 2x - y = 0 \]

### Step 2: Find the Intersection Points
Next, we need to find the intersection points of these lines with the third line \(2x - 3y + 4 = 0\).

#### Intersection of \(6x - 7y = 0\) and \(2x - 3y + 4 = 0\):
1. From \(6x - 7y = 0\), we can express \(x\) in terms of \(y\):
\[ x = \frac{7}{6}y \]
2. Substitute \(x\) into the second line:
\[ 2\left(\frac{7}{6}y\right) - 3y + 4 = 0 \]
\[ \frac{14}{6}y - 3y + 4 = 0 \]
\[ \frac{14}{6}y - \frac{18}{6}y + 4 = 0 \]
\[ -\frac{4}{6}y + 4 = 0 \]
\[ -\frac{2}{3}y + 4 = 0 \implies y = 6 \]
3. Substitute \(y = 6\) back to find \(x\):
\[ x = \frac{7}{6} \cdot 6 = 7 \]
Thus, one intersection point is \((7, 6)\).

#### Intersection of \(2x - y = 0\) and \(2x - 3y + 4 = 0\):
1. From \(2x - y = 0\), we have:
\[ y = 2x \]
2. Substitute \(y\) into the second line:
\[ 2x - 3(2x) + 4 = 0 \]
\[ 2x - 6x + 4 = 0 \]
\[ -4x + 4 = 0 \implies x = 1 \]
3. Substitute \(x = 1\) back to find \(y\):
\[ y = 2 \cdot 1 = 2 \]
Thus, the second intersection point is \((1, 2)\).

### Step 3: Identify the Vertices of the Triangle
Now we have the three vertices of the triangle:
1. \(A(0, 0)\) (origin)
2. \(B(7, 6)\)
3. \(C(1, 2)\)

### Step 4: Calculate the Centroid
The centroid \(G\) of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by:
\[ G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \]
Substituting the coordinates of the vertices:
\[ G\left(\frac{0 + 7 + 1}{3}, \frac{0 + 6 + 2}{3}\right) = G\left(\frac{8}{3}, \frac{8}{3}\right) \]

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