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graph the image of parallelogram pqrs after a reflection over the line …

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 the original points

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

Step3: Apply the reflection formula

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

Step4: Plot the new points

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

Answer:

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