To find the equation of the image of the line \( y = x \) with respect to the line mirror \( 2x - y = 1 \), we can follow these steps:
### Step 1: Identify Points on the Line
Choose two points on the line \( y = x \). We can take:
- Point A: \( (-1, -1) \)
- Point B: \( (0, 0) \)
### Step 2: Find the Image of Point A
To find the image of point A with respect to the line \( 2x - y - 1 = 0 \):
1. Write the line in standard form: \( 2x - y = 1 \).
2. The slope of the line \( 2x - y = 1 \) is \( 2 \), and the slope of the perpendicular line is \( -\frac{1}{2} \).
3. The equation of the line passing through point A and perpendicular to the mirror line is:
\[
y + 1 = -\frac{1}{2}(x + 1)
\]
Simplifying this gives:
\[
y = -\frac{1}{2}x - \frac{3}{2}
\]
### Step 3: Find the Intersection Point
To find the intersection point of the perpendicular line with the mirror line:
1. Set the equations equal:
\[
-\frac{1}{2}x - \frac{3}{2} = 2x - 1
\]
2. Rearranging gives:
\[
-\frac{1}{2}x - 2x = -1 + \frac{3}{2}
\]
\[
-\frac{5}{2}x = \frac{1}{2}
\]
\[
x = -\frac{1}{5}
\]
3. Substitute \( x \) back into the equation of the mirror line to find \( y \):
\[
y = 2(-\frac{1}{5}) - 1 = -\frac{2}{5} - 1 = -\frac{7}{5}
\]
4. Thus, the intersection point is \( \left(-\frac{1}{5}, -\frac{7}{5}\right) \).
### Step 4: Find the Image of Point A
Using the intersection point, we can find the image of point A:
1. The midpoint between A and its image \( (h, k) \) is the intersection point:
\[
\left(\frac{-1 + h}{2}, \frac{-1 + k}{2}\right) = \left(-\frac{1}{5}, -\frac{7}{5}\right)
\]
2. This gives us two equations:
\[
\frac{-1 + h}{2} = -\frac{1}{5} \quad \Rightarrow \quad -1 + h = -\frac{2}{5} \quad \Rightarrow \quad h = \frac{3}{5}
\]
\[
\frac{-1 + k}{2} = -\frac{7}{5} \quad \Rightarrow \quad -1 + k = -\frac{14}{5} \quad \Rightarrow \quad k = -\frac{9}{5}
\]
3. Thus, the image of point A is \( \left(\frac{3}{5}, -\frac{9}{5}\right) \).
### Step 5: Find the Image of Point B
Repeat the process for point B:
1. The perpendicular line through B is:
\[
y = -\frac{1}{2}x
\]
2. Set it equal to the mirror line:
\[
-\frac{1}{2}x = 2x - 1
\]
Rearranging gives:
\[
-\frac{5}{2}x = -1 \quad \Rightarrow \quad x = \frac{2}{5}
\]
3. Substitute back to find \( y \):
\[
y = -\frac{1}{2}(\frac{2}{5}) = -\frac{1}{5}
\]
4. The intersection point is \( \left(\frac{2}{5}, -\frac{1}{5}\right) \).
5. The midpoint gives us:
\[
\left(\frac{0 + h'}{2}, \frac{0 + k'}{2}\right) = \left(\frac{2}{5}, -\frac{1}{5}\right)
\]
This leads to:
\[
h' = \frac{4}{5}, \quad k' = -\frac{2}{5}
\]
6. Thus, the image of point B is \( \left(\frac{4}{5}, -\frac{2}{5}\right) \).
### Step 6: Find the Equation of the Image Line
Now we have the images of points A and B:
- Image of A: \( \left(\frac{3}{5}, -\frac{9}{5}\right) \)
- Image of B: \( \left(\frac{4}{5}, -\frac{2}{5}\right) \)
1. The slope \( m \) of the line through these points is:
\[
m = \frac{-\frac{2}{5} + \frac{9}{5}}{\frac{4}{5} - \frac{3}{5}} = \frac{7/5}{1/5} = 7
\]
2. Using point-slope form \( y - y_1 = m(x - x_1) \):
\[
y + \frac{9}{5} = 7\left(x - \frac{3}{5}\right)
\]
Simplifying gives:
\[
y + \frac{9}{5} = 7x - \frac{21}{5}
\]
\[
y = 7x - \frac{30}{5} = 7x - 6
\]
3. Rearranging gives the final equation:
\[
7x - y = 6
\]
### Final Answer
The equation of the image of the line \( y = x \) with respect to the line mirror \( 2x - y = 1 \) is:
\[
\boxed{7x - y = 6}
\]
