1. **State the problem:** We need to find the reflection of triangle ABC over the vertical line \(\ell\) given by \(x=1\). The vertices of the original triangle are \(A=(3,-4)\), \(B=(8,-5)\), and \(C=(5,-1)\). 2. **Reflect each vertex over the line \(x=1\):** For a point \((x,y)\), its reflection over \(x=1\) is \((2\times1 - x, y) = (2 - x, y)\). - For \(A=(3,-4)\): \[ x' = 2 - 3 = -1, \\ y' = -4 \] So, \(A' = (-1, -4)\). - For \(B=(8,-5)\): \[ x' = 2 - 8 = -6, \\ y' = -5 \] So, \(B' = (-6, -5)\). - For \(C=(5,-1)\): \[ x' = 2 - 5 = -3, \\ y' = -1 \] So, \(C' = (-3, -1)\). 3. **Final answer:** The reflected triangle \(A'B'C'\) has vertices \[ A' = (-1, -4), \\ B' = (-6, -5), \\ C' = (-3, -1) \] This gives the reflection of triangle ABC over the vertical line \(x=1\).