Question

graph the image of parallelogram qrst after a reflection over the line ( y=-6 ).

Explanation:

Step1: Find the distance between each point and the line \(y = - 6\)

Let's assume the coordinates of the points \(Q(x_1,y_1)\), \(R(x_2,y_2)\), \(S(x_3,y_3)\), \(T(x_4,y_4)\). The distance \(d\) between a point \((x,y)\) and the line \(y=-6\) is \(d=\vert y + 6\vert\).

Step2: Reflect each point over the line \(y=-6\)

The rule for reflecting a point \((x,y)\) over the line \(y = k\) is \((x,2k - y)\). Here \(k=-6\), so the rule is \((x,-12 - y)\).
If \(Q\) has coordinates \((0,-9)\), then its image \(Q'\) has coordinates \((0,-12+9)=(0,-3)\)
If \(R\) has coordinates \((3,-9)\), then its image \(R'\) has coordinates \((3,-12 + 9)=(3,-3)\)
If \(S\) has coordinates \((6,-8)\), then its image \(S'\) has coordinates \((6,-12+8)=(6,-4)\)
If \(T\) has coordinates \((1,-8)\), then its image \(T'\) has coordinates \((1,-12 + 8)=(1,-4)\)

Step3: Plot the reflected points

Plot the points \(Q'(0,-3)\), \(R'(3,-3)\), \(S'(6,-4)\), \(T'(1,-4)\) and connect them to form the reflected parallelogram.

Answer:

Plot the points \((0,-3)\), \((3,-3)\), \((6,-4)\), \((1,-4)\) and connect them.