The point of intersection of two lines given by `2x^(2)-5xy+2y^(2)-3x+3y+1=0` is

Options

A. `(1/2,1/3)`
B. `(-1/7,-1/7)`
C. `(-1/3,1/3)`
D. `(1/3,-1/3)`

Answer

To find the point of intersection of the two lines given by the equation \(2x^2 - 5xy + 2y^2 - 3x + 3y + 1 = 0\), we will follow these steps:

### Step 1: Identify coefficients
We will compare the given equation with the general form of a conic section:
\[ ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \]
From the given equation, we can identify the coefficients:
- \(a = 2\)
- \(b = 2\)
- \(h = -\frac{5}{2}\)
- \(g = -\frac{3}{2}\)
- \(f = \frac{3}{2}\)
- \(c = 1\)

### Step 2: Use the formula for the point of intersection
The point of intersection of the two lines can be found using the formula:
\[ \left( \frac{HF - BG}{AB - H^2}, \frac{TH - AF}{AB - H^2} \right) \]
where:
- \(H = h\)
- \(F = f\)
- \(B = b\)
- \(G = g\)
- \(A = a\)
- \(T = a\)

### Step 3: Substitute the values into the formula
Substituting the identified values into the formula:
- \(H = -\frac{5}{2}\)
- \(F = \frac{3}{2}\)
- \(B = 2\)
- \(G = -\frac{3}{2}\)
- \(A = 2\)
- \(T = 2\)

Now, we calculate:
1. **Numerator for x-coordinate**:
\[ HF - BG = \left(-\frac{5}{2} \cdot \frac{3}{2}\right) - \left(2 \cdot -\frac{3}{2}\right) = -\frac{15}{4} + 3 = -\frac{15}{4} + \frac{12}{4} = -\frac{3}{4} \]

2. **Denominator**:
\[ AB - H^2 = (2)(2) - \left(-\frac{5}{2}\right)^2 = 4 - \frac{25}{4} = \frac{16}{4} - \frac{25}{4} = -\frac{9}{4} \]

3. **x-coordinate**:
\[ x = \frac{-\frac{3}{4}}{-\frac{9}{4}} = \frac{3}{9} = \frac{1}{3} \]

4. **Numerator for y-coordinate**:
\[ TH - AF = (2)(-\frac{5}{2}) - (2)(\frac{3}{2}) = -5 - 3 = -8 \]

5. **y-coordinate**:
\[ y = \frac{-8}{-\frac{9}{4}} = \frac{8 \cdot 4}{9} = \frac{32}{9} \]

### Final Result
Thus, the point of intersection of the two lines is:
\[ \left( \frac{1}{3}, \frac{32}{9} \right) \]