The equation of the image of line `y=x` wire respect to the line mirror `2x-y=1` is

Options

A. `y=7x-5`
B. `y=7x-6`
C. `y=3x-7`
D. `y=6x-5`

Answer

To find the equation of the image of the line \(y = x\) with respect to the mirror line \(2x - y = 1\), we can follow these steps:

### Step 1: Find the point of intersection
First, we need to find the point of intersection of the line \(y = x\) and the mirror line \(2x - y = 1\).

Substituting \(y = x\) into \(2x - y = 1\):
\[ 2x - x = 1 \implies x = 1 \]
Now substituting \(x = 1\) back into \(y = x\):
\[ y = 1 \]
Thus, the point of intersection is \((1, 1)\).

**Hint:** To find the point of intersection, substitute the equation of one line into the other.

### Step 2: Reflect the origin (0, 0) across the mirror line
Next, we need to find the reflection of the point \((0, 0)\) across the line \(2x - y - 1 = 0\).

Using the formula for the reflection of a point \((x_1, y_1)\) across the line \(Ax + By + C = 0\):
\[ \left( H, K \right) = \left( x_1 - \frac{2A(Ax_1 + By_1 + C)}{A^2 + B^2}, y_1 - \frac{2B(Ax_1 + By_1 + C)}{A^2 + B^2} \right) \]
Here, \(A = 2\), \(B = -1\), \(C = -1\), and \((x_1, y_1) = (0, 0)\).

Calculating \(Ax_1 + By_1 + C\):
\[ 2(0) - 1(0) - 1 = -1 \]
Now substituting into the reflection formula:
\[ H = 0 - \frac{2 \cdot 2 \cdot (-1)}{2^2 + (-1)^2} = 0 + \frac{4}{5} = \frac{4}{5} \]
\[ K = 0 - \frac{2 \cdot (-1) \cdot (-1)}{2^2 + (-1)^2} = 0 - \frac{2}{5} = -\frac{2}{5} \]
Thus, the reflection point \(P' = \left(\frac{4}{5}, -\frac{2}{5}\right)\).

**Hint:** Use the reflection formula to find the new coordinates after reflecting across the line.

### Step 3: Find the equation of the line through the points (1, 1) and (4/5, -2/5)
Now we have two points: the point of intersection \((1, 1)\) and the reflection point \(\left(\frac{4}{5}, -\frac{2}{5}\right)\).

Using the two-point form of the equation of a line:
\[ y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \]
Substituting \((x_1, y_1) = (1, 1)\) and \((x_2, y_2) = \left(\frac{4}{5}, -\frac{2}{5}\right)\):
\[ y - 1 = \frac{-\frac{2}{5} - 1}{\frac{4}{5} - 1}(x - 1) \]
Calculating the slope:
\[ \text{slope} = \frac{-\frac{2}{5} - \frac{5}{5}}{\frac{4}{5} - \frac{5}{5}} = \frac{-\frac{7}{5}}{-\frac{1}{5}} = 7 \]
Now substituting back into the line equation:
\[ y - 1 = 7(x - 1) \]
Expanding:
\[ y - 1 = 7x - 7 \implies y = 7x - 6 \]

### Final Answer
The equation of the image of the line \(y = x\) with respect to the line mirror \(2x - y = 1\) is:
\[ \boxed{y = 7x - 6} \]