The point of intersection of the lines given by `6x^(2)+xy-40y^(2)-35x-83y+11=0` is

Options

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

Answer

To find the point of intersection of the lines given by the equation \(6x^2 + xy - 40y^2 - 35x - 83y + 11 = 0\), we can follow these steps:

### Step 1: Identify the coefficients
The given equation can be compared with the standard form of a pair of straight lines:
\[ Ax^2 + Bxy + Cy^2 + 2Gx + 2Fy + D = 0 \]
From the equation \(6x^2 + xy - 40y^2 - 35x - 83y + 11 = 0\), we can identify:
- \(A = 6\)
- \(B = 1\)
- \(C = -40\)
- \(G = -\frac{35}{2}\)
- \(F = -\frac{83}{2}\)
- \(D = 11\)

### Step 2: Calculate the point of intersection
The point of intersection of the lines can be found using the formula:
\[ \left( \frac{BF - CD}{B^2 - AC}, \frac{AE - BG}{B^2 - AC} \right) \]
where \(E = 0\) (since there is no \(y^2\) term).

First, we need to compute \(B^2 - AC\):
\[ B^2 - AC = 1^2 - (6)(-40) = 1 + 240 = 241 \]

Now, we can compute the coordinates of the intersection point:
1. **For x-coordinate**:
\[ x = \frac{(1)(-\frac{83}{2}) - (-40)(11)}{241} \]
\[ = \frac{-\frac{83}{2} + 440}{241} \]
\[ = \frac{-\frac{83}{2} + \frac{880}{2}}{241} \]
\[ = \frac{\frac{797}{2}}{241} = \frac{797}{482} \]

2. **For y-coordinate**:
\[ y = \frac{(6)(0) - (1)(-\frac{35}{2})}{241} \]
\[ = \frac{0 + \frac{35}{2}}{241} \]
\[ = \frac{\frac{35}{2}}{241} = \frac{35}{482} \]

### Step 3: Final Result
Thus, the point of intersection of the lines is:
\[ \left( \frac{797}{482}, \frac{35}{482} \right) \]