Question

unit 4 : hw #2
4 - 3 rotations
directions: give each rule for our
180^{circ}(x,y)
directions: graph and label each figure and its image under
give the coordinates of the image

1. rhombus abcd with vertices a(4,4), b(6,7),

c(5,3), and d(1,2): 180^{circ}

1. trape
2. triangle fgh with vertices f(-7,8), g(-1,1),

and h(-8,4): 270^{circ} counterclockwise

1. so
2. trapezoid qrst with vertices q(2,1), r(2,5),

s(4,5), and t(8,1): 90^{circ} clockwise
6.

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 \(F(-7,8)\):
Substitute \(x=-7\) and \(y = 8\) into the rule \((x,y)\to(y,-x)\).
We get \(F'=(8,7)\).

Step2: Apply the rule to point \(G\)

For point \(G(-1,1)\), substitute \(x=-1\) and \(y = 1\) into \((x,y)\to(y,-x)\).
We get \(G'=(1,1)\).

Step3: Apply the rule to point \(H\)

For point \(H(-8,4)\), substitute \(x=-8\) and \(y = 4\) into \((x,y)\to(y,-x)\).
We get \(H'=(4,8)\).

The rule for a \(90^{\circ}\) clockwise rotation about the origin is \((x,y)\to(y,-x)\) (same as \(270^{\circ}\) counter - clockwise).

Step4: Apply the rule to point \(Q\)

For point \(Q(2,1)\), substitute \(x = 2\) and \(y=1\) into \((x,y)\to(y,-x)\).
We get \(Q'(1,-2)\).

Step5: Apply the rule to point \(R\)

For point \(R(2,5)\), substitute \(x = 2\) and \(y = 5\) into \((x,y)\to(y,-x)\).
We get \(R'(5,-2)\).

Step6: Apply the rule to point \(S\)

For point \(S(4,5)\), substitute \(x = 4\) and \(y = 5\) into \((x,y)\to(y,-x)\).
We get \(S'(5,-4)\).

Step7: Apply the rule to point \(T\)

For point \(T(8,1)\), substitute \(x = 8\) and \(y = 1\) into \((x,y)\to(y,-x)\).
We get \(T'(1,-8)\).

Answer:

For the \(270^{\circ}\) counter - clockwise rotation of \(\triangle FGH\):
\(F'(8,7)\), \(G'(1,1)\), \(H'(4,8)\)
For the \(90^{\circ}\) clockwise rotation of trapezoid \(QRST\):
\(Q'(1,-2)\), \(R'(5,-2)\), \(S'(5,-4)\), \(T'(1,-8)\)