5) Find the image of the line after \( l_{1}: y=2 x-1 \) after it has been reffected along the line \( l: y=x+4 \)

To find the image of the line \( l_{1}: y=2x-1 \) after it has been reflected along the line \( l: y=x+4 \), we can follow these steps: 1. **Find the intersection point**: Substitute \( y = 2x - 1 \) into \( y = x + 4 \) to find where the two lines intersect. \[ 2x - 1 = x + 4 \] \[ x = 5 \] \[ y = 2(5) - 1 = 9 \] Thus, the lines intersect at the point \( (5, 9) \). 2. **Calculate the slope of the original line and the line of reflection**: The slope of \( l_{1} \) is 2, and the slope of \( l \) is 1. The slope of the normal to line \( l \) at the point of intersection can be calculated as the negative reciprocal of 1, which is -1. 3. **Use the intersection point and the slopes**: From the intersection point \( (5, 9) \), create a line using the slope of \( l_{1} \): \[ y - 9 = 2(x - 5) \implies y = 2x - 1 \] Now, to find the reflection, we need to find the distance from the intersection point along the normal's slope to locate the reflected point. 4. **Reflect the entire line**: One effective method is to compute the angle between the original line and the normal, then use that angle to find the reflected line. The result yields the equation of the reflected line. After performing these calculations, the image of line \( l_{1} \) after reflection over line \( l \) is \( y = -\frac{1}{2}x + \frac{23}{2} \). Hence, the equation of the reflected line is \( y = -\frac{1}{2}x + \frac{23}{2} \).