If the straight lines joining origin to the points of intersection of the line x+y=1 with the curve `x^2+y^2 +x-2y -m =0 ` are perpendicular to each other , then the value of m should be

Options

A. `-1/2`
B. 0
C. `1/2`
D. 1

Answer

To solve the problem step by step, we need to find the value of \( m \) for which the lines joining the origin to the points of intersection of the line \( x + y = 1 \) with the curve \( x^2 + y^2 + x - 2y - m = 0 \) are perpendicular to each other.

### Step 1: Substitute \( y \) from the line equation into the curve equation

The line equation is \( x + y = 1 \). We can express \( y \) in terms of \( x \):
\[ y = 1 - x \]
Now, substitute \( y \) into the curve equation:
\[ x^2 + (1 - x)^2 + x - 2(1 - x) - m = 0 \]

### Step 2: Expand the equation

Expanding \( (1 - x)^2 \):
\[ x^2 + (1 - 2x + x^2) + x - 2 + 2x - m = 0 \]
Combine like terms:
\[ 2x^2 + 2x - 1 - m = 0 \]

### Step 3: Rearrange the equation

Rearranging gives:
\[ 2x^2 + 2x + (-1 - m) = 0 \]

### Step 4: Identify coefficients for the quadratic equation

The quadratic equation is in the form \( ax^2 + bx + c = 0 \):
- \( a = 2 \)
- \( b = 2 \)
- \( c = -1 - m \)

### Step 5: Use the condition for perpendicular lines

The slopes of the lines from the origin to the intersection points can be found using the quadratic formula. The slopes \( m_1 \) and \( m_2 \) are given by:
\[ m_1, m_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For the lines to be perpendicular, the product of their slopes must equal \(-1\):
\[ m_1 \cdot m_2 = -1 \]

### Step 6: Calculate the product of the slopes

The product of the roots (slopes) of the quadratic equation is given by:
\[ m_1 \cdot m_2 = \frac{c}{a} = \frac{-1 - m}{2} \]
Setting this equal to \(-1\):
\[ \frac{-1 - m}{2} = -1 \]

### Step 7: Solve for \( m \)

Multiplying both sides by 2:
\[ -1 - m = -2 \]
Rearranging gives:
\[ m = 1 \]

### Conclusion

Thus, the value of \( m \) should be:
\[ \boxed{1} \]