Question

graph the image of parallelogram pqrs after a reflection over the line $x = 5$.

Explanation:

Step1: Find the reflection formula

The formula for reflecting a point \((x,y)\) over the line \(x = a\) is \((2a - x,y)\). Here \(a = 5\), so the formula is \((10 - x,y)\).

Step2: Find the coordinates of \(P\), \(Q\), \(R\), \(S\)

Assume \(P(0,-9)\), \(Q(0,-7)\), \(R(2,-2)\), \(S(2,-4)\) (from the graph).

Step3: Apply the reflection formula

• For \(P(0,-9)\): \(x = 0\), \(y=-9\). Using \((10 - x,y)\), we get \(P'(10,-9)\).
• For \(Q(0,-7)\): \(x = 0\), \(y = -7\). Using \((10 - x,y)\), we get \(Q'(10,-7)\).
• For \(R(2,-2)\): \(x = 2\), \(y=-2\). Using \((10 - x,y)\), we get \(R'(8,-2)\).
• For \(S(2,-4)\): \(x = 2\), \(y=-4\). Using \((10 - x,y)\), we get \(S'(8,-4)\).

Answer:

Plot the points \(P'(10,-9)\), \(Q'(10,-7)\), \(R'(8,-2)\), \(S'(8,-4)\) and connect them to form the reflected parallelogram.