Question

graph the image of trapezoid rstu after a reflection over the line x = 4.

Explanation:

Step1: Find the distance between each x - coordinate and the line \(x = 4\)

For a point \((x,y)\) reflected over the line \(x = a\), the formula for the x - coordinate of the reflected point is \(x'=2a - x\), and the y - coordinate remains the same \(y'=y\). Here \(a = 4\).
Let's assume the coordinates of the trapezoid \(RSTU\) are \(R(x_1,y_1)\), \(S(x_2,y_2)\), \(T(x_3,y_3)\), \(U(x_4,y_4)\). From the graph (assuming standard grid where each square is 1 unit), if \(S(- 2,-1)\), \(R(-3,-7)\), \(T(-1,-9)\), \(U(-2,-10)\)
For point \(S(-2,y)\):
The distance between \(x=-2\) and \(x = 4\) is \(d=4-(-2)=6\). The x - coordinate of the reflected point \(S'\) is \(x'=4 + 6=10\), and \(y'=-1\), so \(S'(10,-1)\)
For point \(R(-3,y)\):
The distance between \(x=-3\) and \(x = 4\) is \(d = 4-(-3)=7\). The x - coordinate of the reflected point \(R'\) is \(x'=4 + 7=11\), and \(y'=-7\), so \(R'(11,-7)\)
For point \(T(-1,y)\):
The distance between \(x=-1\) and \(x = 4\) is \(d=4-(-1)=5\). The x - coordinate of the reflected point \(T'\) is \(x'=4 + 5=9\), and \(y'=-9\), so \(T'(9,-9)\)
For point \(U(-2,y)\):
The distance between \(x=-2\) and \(x = 4\) is \(d=4-(-2)=6\). The x - coordinate of the reflected point \(U'\) is \(x'=4 + 6=10\), and \(y'=-10\), so \(U'(10,-10)\)

Step2: Plot the reflected points

Plot the points \(R'(11,-7)\), \(S'(10,-1)\), \(T'(9,-9)\), \(U'(10,-10)\) on the coordinate plane and connect them in the same order as the original trapezoid \(RSTU\) to get the reflected trapezoid.

Answer:

Plot the points \(R'(11,-7)\), \(S'(10,-1)\), \(T'(9,-9)\), \(U'(10,-10)\) and connect them to form the reflected trapezoid.