Let `L_(1)` be a straight line passing through the origin and `L_(2)` be the straight line `x + y = 1`. If the intercepts made by the circle `x^(2)+y^(2)-x+3y=0` on `L_(1)` and `L_(2)` are equal, then which of the following equations can represent `L_(1)`?

Options

A. `x+y=0`
B. `x-y=0`
C. `x+7y=0`
D. `x-7y=0`

Answer

To solve the problem, we need to find the equations of the line \( L_1 \) that passes through the origin and has equal intercepts with the line \( L_2: x + y = 1 \) and the circle defined by the equation \( x^2 + y^2 - x + 3y = 0 \).

### Step 1: Find the center and radius of the circle
The equation of the circle can be rewritten in standard form. We start with:
\[ x^2 + y^2 - x + 3y = 0 \]
We complete the square for \( x \) and \( y \).

For \( x \):
\[ x^2 - x = (x - \frac{1}{2})^2 - \frac{1}{4} \]

For \( y \):
\[ y^2 + 3y = (y + \frac{3}{2})^2 - \frac{9}{4} \]

Substituting these back into the equation gives:
\[ (x - \frac{1}{2})^2 + (y + \frac{3}{2})^2 - \frac{1}{4} - \frac{9}{4} = 0 \]
\[ (x - \frac{1}{2})^2 + (y + \frac{3}{2})^2 = \frac{10}{4} = \frac{5}{2} \]
Thus, the center of the circle is \( C\left(\frac{1}{2}, -\frac{3}{2}\right) \) and the radius \( r = \sqrt{\frac{5}{2}} \).

### Step 2: Find the distance from the center to the line \( L_2 \)
The line \( L_2: x + y - 1 = 0 \) can be used to find the distance from the center \( C\left(\frac{1}{2}, -\frac{3}{2}\right) \) to the line using the formula:
\[ \text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
where \( A = 1, B = 1, C = -1 \) and \( (x_1, y_1) = \left(\frac{1}{2}, -\frac{3}{2}\right) \).

Calculating the distance:
\[ \text{Distance} = \frac{|1 \cdot \frac{1}{2} + 1 \cdot \left(-\frac{3}{2}\right) - 1|}{\sqrt{1^2 + 1^2}} = \frac{|\frac{1}{2} - \frac{3}{2} - 1|}{\sqrt{2}} = \frac{|-2|}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

### Step 3: Find the equation of line \( L_1 \)
Let the equation of line \( L_1 \) be \( y = mx \). The distance from the center \( C\left(\frac{1}{2}, -\frac{3}{2}\right) \) to this line is given by:
\[ \text{Distance} = \frac{|m \cdot \frac{1}{2} - 1 \cdot \left(-\frac{3}{2}\right)|}{\sqrt{m^2 + 1}} = \frac{|m \cdot \frac{1}{2} + \frac{3}{2}|}{\sqrt{m^2 + 1}} \]

### Step 4: Set the distances equal
Since the intercepts are equal, we set the distances equal:
\[ \frac{|m \cdot \frac{1}{2} + \frac{3}{2}|}{\sqrt{m^2 + 1}} = \sqrt{2} \]

Squaring both sides:
\[ \left(m \cdot \frac{1}{2} + \frac{3}{2}\right)^2 = 2(m^2 + 1) \]

Expanding and simplifying:
\[ \frac{1}{4}m^2 + \frac{3}{2}m + \frac{9}{4} = 2m^2 + 2 \]
\[ 0 = 2m^2 - \frac{1}{4}m - \frac{1}{4} \]
Multiplying through by 4 to eliminate fractions:
\[ 0 = 8m^2 - m - 1 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula:
\[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 8 \cdot (-1)}}{2 \cdot 8} \]
\[ = \frac{1 \pm \sqrt{1 + 32}}{16} = \frac{1 \pm \sqrt{33}}{16} \]

### Conclusion
The lines represented by \( L_1 \) can be expressed as:
1. \( y = \frac{1 + \sqrt{33}}{16}x \)
2. \( y = \frac{1 - \sqrt{33}}{16}x \)

Thus, the equations of the lines \( L_1 \) that can represent the required conditions are:
- \( x - y = 0 \) (for \( m = 1 \))
- \( x + 7y = 0 \) (for \( m = -\frac{1}{7} \))