If the line `y=mx` meets the lines `x+2y-1=0` and `2x-y+3=0` at the same point, then m is equal to

Options

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

Answer

To solve the problem, we need to find the value of \( m \) such that the line \( y = mx \) meets the lines \( x + 2y - 1 = 0 \) and \( 2x - y + 3 = 0 \) at the same point. This means that the three lines are concurrent.

### Step-by-step Solution:

1. **Write the equations of the lines in standard form**:
- The first line is \( y = mx \).
- The second line is \( x + 2y - 1 = 0 \).
- The third line is \( 2x - y + 3 = 0 \).

2. **Convert the first line to standard form**:
\[ mx - y = 0 \]

3. **Set up the determinant for the three lines**:
The lines are concurrent if the determinant of the coefficients is zero:
\[ \begin{vmatrix} 1 & 2 & -1 \\ 2 & -1 & 3 \\ m & -1 & 0 \end{vmatrix} = 0 \]

4. **Calculate the determinant**:
Expanding the determinant:
\[ = 1 \cdot \begin{vmatrix} -1 & 3 \\ -1 & 0 \end{vmatrix} - 2 \cdot \begin{vmatrix} 2 & 3 \\ m & 0 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 2 & -1 \\ m & -1 \end{vmatrix} \]
Calculating each of the smaller 2x2 determinants:
- First determinant:
\[ (-1)(0) - (3)(-1) = 3 \]
- Second determinant:
\[ (2)(0) - (3)(m) = -3m \]
- Third determinant:
\[ (2)(-1) - (-1)(m) = -2 + m \]

5. **Substituting back into the determinant**:
\[ 1 \cdot 3 - 2 \cdot (-3m) + (-1)(-2 + m) = 0 \]
Simplifying:
\[ 3 + 6m + 2 - m = 0 \]
\[ 5 + 5m = 0 \]

6. **Solving for \( m \)**:
\[ 5m = -5 \implies m = -1 \]

### Final Answer:
Thus, the value of \( m \) is \( -1 \).