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Question
- triangle fgh with vertices f(-7, 8), g(-1, 1), and h(-8, 4); 270° counterclockwise
f(, )
g(, )
h(, )
The rule for a \( 270^\circ \) counterclockwise rotation (or equivalently a \( 90^\circ \) clockwise rotation) about the origin is \((x,y)\to(y, -x)\). We will apply this rule to each vertex of triangle \( FGH \).
Step 1: Rotate point \( F(-7, 8) \)
Using the rotation rule \((x,y)\to(y, -x)\), substitute \( x = -7 \) and \( y = 8 \).
So, \( F'=(8, -(-7))=(8, 7) \)
Step 2: Rotate point \( G(-1, 1) \)
Using the rotation rule \((x,y)\to(y, -x)\), substitute \( x = -1 \) and \( y = 1 \).
So, \( G'=(1, -(-1))=(1, 1) \)
Step 3: Rotate point \( H(-8, 4) \)
Using the rotation rule \((x,y)\to(y, -x)\), substitute \( x = -8 \) and \( y = 4 \).
So, \( H'=(4, -(-8))=(4, 8) \)
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\( F'(8, 7) \)
\( G'(1, 1) \)
\( H'(4, 8) \)