Question

1. a table in the shape of a trapezoid has vertices: a(3,3), b(5,3), c(6,0), and d(2,0). the teacher rotates the table 270° counterclockwise about the origin. what are the new coordinates of the table and what rule did you use? 12) a trapezoid has vertices a(-6,-2), b(-3,-2), c(-1,-4), and d(-6,-4). what are the coordinates of the image after a 180° about the origin and what rule did you use?

Explanation:

Step1: Recall rotation rules

The rule for a \(270^{\circ}\) counter - clockwise rotation about the origin is \((x,y)\to(y, - x)\).
For point \(A(3,3)\):
Substitute \(x = 3\) and \(y = 3\) into the rule \((x,y)\to(y, - x)\).
We get \(A'(3,-3)\).
For point \(B(5,3)\):
Substitute \(x = 5\) and \(y = 3\) into the rule \((x,y)\to(y, - x)\).
We get \(B'(3,-5)\).
For point \(C(6,0)\):
Substitute \(x = 6\) and \(y = 0\) into the rule \((x,y)\to(y, - x)\).
We get \(C'(0,-6)\).
For point \(D(2,0)\):
Substitute \(x = 2\) and \(y = 0\) into the rule \((x,y)\to(y, - x)\).
We get \(D'(0,-2)\).

The rule for a \(180^{\circ}\) rotation about the origin is \((x,y)\to(-x,-y)\).
For point \(A(-6,-2)\):
Substitute \(x=-6\) and \(y = - 2\) into the rule \((x,y)\to(-x,-y)\).
We get \(A'(6,2)\).
For point \(B(-3,-2)\):
Substitute \(x=-3\) and \(y = - 2\) into the rule \((x,y)\to(-x,-y)\).
We get \(B'(3,2)\).
For point \(C(-1,-4)\):
Substitute \(x=-1\) and \(y = - 4\) into the rule \((x,y)\to(-x,-y)\).
We get \(C'(1,4)\).
For point \(D(-6,-4)\):
Substitute \(x=-6\) and \(y = - 4\) into the rule \((x,y)\to(-x,-y)\).
We get \(D'(6,4)\).

Answer:

For problem 11: The new coordinates are \(A'(3,-3)\), \(B'(3,-5)\), \(C'(0,-6)\), \(D'(0,-2)\) and the rule is \((x,y)\to(y, - x)\).
For problem 12: The new coordinates are \(A'(6,2)\), \(B'(3,2)\), \(C'(1,4)\), \(D'(6,4)\) and the rule is \((x,y)\to(-x,-y)\).