Question

1. reasoning a trapezoid has vertices ( a(-6,-2), b(-3,-2), c(-1,-4) ), and ( d(-6,-4) ).

a. rotate the trapezoid ( 180^{circ} ) about the origin. what are the coordinates of the image?
b. describe a way to obtain the same image without using rotations.

Explanation:

Step1: Recall the rule for a \(180^{\circ}\) rotation about the origin

The rule for a \(180^{\circ}\) rotation about the origin \((x,y)\to(-x,-y)\).

Step2: Apply the rule to each vertex

• For \(A(-6,-2)\):

\(x = - 6,y=-2\), then \((-(-6),-(-2))=(6,2)\)

• For \(B(-3,-2)\):

\(x=-3,y = - 2\), then \((-(-3),-(-2))=(3,2)\)

• For \(C(-1,-4)\):

\(x=-1,y=-4\), then \((-(-1),-(-4))=(1,4)\)

• For \(D(-6,-4)\):

\(x=-6,y=-4\), then \((-(-6),-(-4))=(6,4)\)

Step3: For part b

A translation followed by a reflection (or vice - versa).
First, reflect the trapezoid over the \(x\) - axis. The rule for reflection over the \(x\) - axis is \((x,y)\to(x, - y)\).

• \(A(-6,-2)\to(-6,2)\), \(B(-3,-2)\to(-3,2)\), \(C(-1,-4)\to(-1,4)\), \(D(-6,-4)\to(-6,4)\)

Then translate the reflected trapezoid \(12\) units to the right. The rule for translation \(12\) units to the right is \((x,y)\to(x + 12,y)\)

• \((-6,2)\to(6,2)\), \((-3,2)\to(9,2)\) (This is wrong. Let's correct:

Another way: Reflect over the \(y\) - axis \((x,y)\to(-x,y)\)
\(A(-6,-2)\to(6,-2)\), \(B(-3,-2)\to(3,-2)\), \(C(-1,-4)\to(1,-4)\), \(D(-6,-4)\to(6,-4)\)
Then reflect over the \(x\) - axis \((x,y)\to(x,-y)\)
\((6,-2)\to(6,2)\), \((3,-2)\to(3,2)\), \((1,-4)\to(1,4)\), \((6,-4)\to(6,4)\)

Answer:

a. \(A'(6,2)\), \(B'(3,2)\), \(C'(1,4)\), \(D'(6,4)\)
b. Reflect the trapezoid over the \(y\) - axis and then over the \(x\) - axis.