If the straight lines x - 2y = 0 and kx + y = 1 intersect at the point `(1,(1)/(2))`, then what is the value of k ?

Options

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

Answer

To find the value of \( k \) such that the lines \( x - 2y = 0 \) and \( kx + y = 1 \) intersect at the point \( (1, \frac{1}{2}) \), we can follow these steps:

### Step 1: Substitute the point into the first equation
The first equation is \( x - 2y = 0 \). We substitute \( x = 1 \) and \( y = \frac{1}{2} \) into this equation.

\[ 1 - 2\left(\frac{1}{2}\right) = 0 \]

### Step 2: Simplify the equation
Now simplify the left side:

\[ 1 - 1 = 0 \]

This confirms that the point \( (1, \frac{1}{2}) \) lies on the first line.

### Step 3: Substitute the point into the second equation
Now we substitute the same point into the second equation \( kx + y = 1 \):

\[ k(1) + \frac{1}{2} = 1 \]

### Step 4: Solve for \( k \)
Now, we can solve for \( k \):

\[ k + \frac{1}{2} = 1 \]

Subtract \( \frac{1}{2} \) from both sides:

\[ k = 1 - \frac{1}{2} \]

\[ k = \frac{1}{2} \]

### Conclusion
Thus, the value of \( k \) is \( \frac{1}{2} \).