Question

1. rotate \\( \triangle f g h 90 ^ { \circ } \\) counterclockwise around the origin.

1. suppose \\( \triangle f g h \\) is transformed by the following composite transformation rule:

first \\( ( x, y ) \
ightarrow ( x + 2, y - 3 ) \\),
then \\( \left( x ^ { \prime }, y ^ { \prime } \
ight) \
ightarrow ( - x,- y ) \\)

Explanation:

Step1: Rotation rule

The rule for a \(90^{\circ}\) counter - clockwise rotation about the origin is \((x,y)\to(-y,x)\).
For point \(F(4,2)\):
Substitute \(x = 4\) and \(y=2\) into the rule \((x,y)\to(-y,x)\).
We get \((-2,4)\).
For point \(G(4,6)\):
Substitute \(x = 4\) and \(y = 6\) into the rule \((x,y)\to(-y,x)\).
We get \((-6,4)\).
For point \(H(6,2)\):
Substitute \(x = 6\) and \(y=2\) into the rule \((x,y)\to(-y,x)\).
We get \((-2,6)\).

Step2: Composite transformation rule

First transformation: \((x,y)\to(x + 2,y-3)\)
For point \(F(4,2)\):
\(x'=4 + 2=6\), \(y'=2-3=-1\), so the point is \((6,-1)\)
Then second transformation: \((x',y')\to(-x',-y')\)
Substitute \(x'=6\) and \(y'=-1\) into \((x',y')\to(-x',-y')\), we get \((-6,1)\)

For point \(G(4,6)\):
First transformation: \(x'=4 + 2=6\), \(y'=6-3 = 3\), so the point is \((6,3)\)
Second transformation: Substitute \(x'=6\) and \(y'=3\) into \((x',y')\to(-x',-y')\), we get \((-6,-3)\)

For point \(H(6,2)\):
First transformation: \(x'=6+2 = 8\), \(y'=2-3=-1\), so the point is \((8,-1)\)
Second transformation: Substitute \(x'=8\) and \(y'=-1\) into \((x',y')\to(-x',-y')\), we get \((-8,1)\)

Answer:

For the \(90^{\circ}\) counter - clockwise rotation:

• \(F(4,2)\) maps to \((-2,4)\)
• \(G(4,6)\) maps to \((-6,4)\)
• \(H(6,2)\) maps to \((-2,6)\)

For the composite transformation:

• \(F(4,2)\) maps to \((-6,1)\)
• \(G(4,6)\) maps to \((-6,-3)\)
• \(H(6,2)\) maps to \((-8,1)\)