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

Options

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

Answer

To solve the problem step-by-step, we will find the value of \( m \) such that 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.

### Step 1: Identify the equations
We have the line:
\[ x + y = 1 \]
And the curve:
\[ x^2 + y^2 + x - 2y - m = 0 \]

### Step 2: Substitute \( y \) from the line equation into the curve equation
From the line equation, 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 3: Simplify the equation
Expanding \( (1 - x)^2 \):
\[ x^2 + (1 - 2x + x^2) + x - 2 + 2x - m = 0 \]
Combining like terms:
\[ 2x^2 + 2x - 1 - m = 0 \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ 2x^2 + 2x + (-1 - m) = 0 \]

### Step 5: Find the condition for perpendicular lines
For the lines joining the origin to the points of intersection to be perpendicular, the product of the slopes of the lines must be \(-1\). The slopes can be found from the roots of the quadratic equation:
\[ 2x^2 + 2x + (-1 - m) = 0 \]
The slopes (roots) are given by:
\[ x_1, x_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{4 - 8(-1 - m)}}{4} = \frac{-2 \pm \sqrt{4 + 8 + 8m}}{4} \]
This simplifies to:
\[ x_1, x_2 = \frac{-2 \pm \sqrt{12 + 8m}}{4} \]

### Step 6: Condition for perpendicularity
The condition for the slopes \( m_1 \) and \( m_2 \) to be perpendicular is:
\[ m_1 \cdot m_2 = -1 \]
Using the product of roots from the quadratic equation:
\[ x_1 x_2 = \frac{c}{a} = \frac{-1 - m}{2} \]
Setting the product equal to \(-1\):
\[ \frac{-1 - m}{2} = -1 \]
Multiplying both sides by 2:
\[ -1 - m = -2 \]
Solving for \( m \):
\[ m = 1 \]

### Conclusion
Thus, the value of \( m \) is:
\[ \boxed{1} \]