The point of intersection of lines represented by ` x^(2) -y ^(2) +x+3y -2=0` is

Options

A. `(1,0)`
B. `(0,2)`
C. `((1)/(2),(3)/(2))`
D. `((1)/(2),(1)/(2))`

Answer

To find the point of intersection of the lines represented by the equation \( x^2 - y^2 + x + 3y - 2 = 0 \), we can follow these steps:

### Step 1: Rewrite the equation
The given equation is:
\[ x^2 - y^2 + x + 3y - 2 = 0 \]
This can be rearranged to express it in a more manageable form.

### Step 2: Group the terms
Group the terms involving \(x\) and \(y\):
\[ x^2 + x - y^2 + 3y - 2 = 0 \]

### Step 3: Treat \(y\) as a constant
We can treat the terms involving \(y\) as constants while solving for \(x\). Thus, we can write:
\[ x^2 + x + (3y - y^2 - 2) = 0 \]
Here, we identify \(a = 1\), \(b = 1\), and \(c = 3y - y^2 - 2\).

### Step 4: Use the quadratic formula
To find the values of \(x\), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting our values:
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (3y - y^2 - 2)}}{2 \cdot 1} \]

### Step 5: Simplify the expression
Calculating the discriminant:
\[ x = \frac{-1 \pm \sqrt{1 - 4(3y - y^2 - 2)}}{2} \]
\[ = \frac{-1 \pm \sqrt{1 - 12y + 4y^2 + 8}}{2} \]
\[ = \frac{-1 \pm \sqrt{4y^2 - 12y + 9}}{2} \]
\[ = \frac{-1 \pm \sqrt{(2y - 3)^2}}{2} \]

### Step 6: Solve for \(x\)
This gives us two possible values for \(x\):
\[ x = \frac{-1 + (2y - 3)}{2} \quad \text{and} \quad x = \frac{-1 - (2y - 3)}{2} \]
Calculating these:
1. \(x = \frac{2y - 4}{2} = y - 2\)
2. \(x = \frac{-2y + 2}{2} = -y + 1\)

### Step 7: Set the equations equal to find \(y\)
Now we have two expressions for \(x\):
1. \(x = y - 2\)
2. \(x = -y + 1\)

Setting them equal to each other:
\[ y - 2 = -y + 1 \]
\[ 2y = 3 \implies y = \frac{3}{2} \]

### Step 8: Substitute \(y\) back to find \(x\)
Substituting \(y\) back into one of the equations for \(x\):
\[ x = y - 2 = \frac{3}{2} - 2 = \frac{3}{2} - \frac{4}{2} = -\frac{1}{2} \]

### Final Answer
Thus, the point of intersection of the lines is:
\[ \left(-\frac{1}{2}, \frac{3}{2}\right) \]