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+sqrt3y=0`

Answer

To solve the problem, we need to find the equation 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 given by \( x^2 + y^2 - x + 3y = 0 \).

### Step-by-Step Solution:

1. **Equation of Line \( L_1 \)**:
Since \( L_1 \) passes through the origin, we can express it in the form:
\[ L_1: y = mx \]
where \( m \) is the slope of the line.

2. **Finding the Center of the Circle**:
The given equation of the circle is:
\[ x^2 + y^2 - x + 3y = 0 \]
We can rewrite this in standard form by completing the square.

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

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

Combining these, we get:
\[ \left(x - \frac{1}{2}\right)^2 + \left(y + \frac{3}{2}\right)^2 = \frac{10}{4} = \frac{5}{2} \]
Thus, the center of the circle is:
\[ \left(\frac{1}{2}, -\frac{3}{2}\right) \]

3. **Finding the Distance from the Center to Line \( L_2 \)**:
The line \( L_2: x + y = 1 \) can be rewritten as:
\[ x + y - 1 = 0 \]
The distance \( d_2 \) from the center \( \left(\frac{1}{2}, -\frac{3}{2}\right) \) to line \( L_2 \) is given by the formula:
\[ d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \]
Here, \( a = 1, b = 1, c = -1 \):
\[ d_2 = \frac{|1 \cdot \frac{1}{2} + 1 \cdot \left(-\frac{3}{2}\right) - 1|}{\sqrt{1^2 + 1^2}} = \frac{| \frac{1}{2} - \frac{3}{2} - 1 |}{\sqrt{2}} = \frac{| -2 |}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

4. **Finding the Distance from the Center to Line \( L_1 \)**:
The line \( L_1: mx - y = 0 \) can be expressed as:
\[ mx - y + 0 = 0 \]
The distance \( d_1 \) from the center \( \left(\frac{1}{2}, -\frac{3}{2}\right) \) to line \( L_1 \) is:
\[ d_1 = \frac{|m \cdot \frac{1}{2} - \left(-\frac{3}{2}\right) + 0|}{\sqrt{m^2 + 1}} = \frac{| \frac{m}{2} + \frac{3}{2} |}{\sqrt{m^2 + 1}} \]

5. **Setting Distances Equal**:
Since the intercepts made by the circle on \( L_1 \) and \( L_2 \) are equal, we set \( d_1 = d_2 \):
\[ \frac{| \frac{m}{2} + \frac{3}{2} |}{\sqrt{m^2 + 1}} = \sqrt{2} \]

6. **Squaring Both Sides**:
Squaring both sides gives:
\[ \left( \frac{m}{2} + \frac{3}{2} \right)^2 = 2(m^2 + 1) \]
Expanding and simplifying leads to:
\[ m^2 + 3m + \frac{9}{4} = 2m^2 + 2 \]
Rearranging gives:
\[ 7m^2 - 6m - 1 = 0 \]

7. **Solving the Quadratic Equation**:
Using the quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ m = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 7 \cdot (-1)}}{2 \cdot 7} = \frac{6 \pm \sqrt{36 + 28}}{14} = \frac{6 \pm \sqrt{64}}{14} = \frac{6 \pm 8}{14} \]
This gives:
\[ m = \frac{14}{14} = 1 \quad \text{or} \quad m = \frac{-2}{14} = -\frac{1}{7} \]

8. **Finding the Equations of Lines**:
Thus, the possible equations for \( L_1 \) are:
\[ y = x \quad \text{or} \quad y = -\frac{1}{7}x \]
In standard form, these can be written as:
\[ x - y = 0 \quad \text{or} \quad x + 7y = 0 \]

### Conclusion:
The correct option representing \( L_1 \) is:
\[ \text{Option: } x + 7y = 0 \]