Answer:
See the graph below
Explanation:
Given:
Quadrilaterla EFGH on a coordinate plane
To find:
to graph the quadrilateral after a rotation of 90° counterclockwise around the origin.
The rule for a 90° counterclockwise around the origin is given as:
[tex]\begin{gathered} (x,\text{ y\rparen}\rightarrow\text{ \lparen-y, x\rparen} \\ Switch\text{ x anf y and }negate\text{ the y while keeping x constant} \end{gathered}[/tex]
We will apply the rules to the coordinates of EFGH:
E = (3, 3)
F = (6, 3)
G = (6, 6)
H = (3, 6)
Applying the rule:
[tex]\begin{gathered} (3,\text{ 3\rparen }\rightarrow\text{ \lparen3, 3\rparen }\rightarrow\text{ \lparen-3, 3\rparen} \\ E^{\prime}\text{ = \lparen-3, 3\rparen} \\ \\ (6,\text{ 3\rparen }\rightarrow(3,\text{ 6\rparen }\rightarrow(-3,\text{ 6\rparen} \\ F^{\prime}\text{ = \lparen-3, 6\rparen} \\ \\ (6,\text{ 6\rparen }\rightarrow\text{ \lparen6, 6\rparen }\rightarrow(-6,\text{ 6\rparen} \\ G^{\prime}\text{ = \lparen-6, 6\rparen} \\ \\ (3,\text{ 6\rparen }\rightarrow(6,\text{ 3\rparen }\rightarrow(-6,\text{ 3\rparen} \\ H^{\prime}\text{ = \lparen-6, 3\rparen} \end{gathered}[/tex]
Plotting the graph:
