Question

exercise #3: triangle efg is shown below with vertices at e(1,2), f(-3,0), and g(4,3).
(a) rotate △efg by 90° clockwise about the origin. label the image △efg. show where each vertex gets mapped below.
(b) based on (a), give an algebraic rule for a 90° clockwise rotation about the origin.
(c) rotate △efg by 90° counterclockwise about the origin. label the image △efg. show where each vertex gets mapped below.
(d) based on (c), give an algebraic rule for a 90° counterclockwise rotation about the origin.
(e) use tracing paper to verify that △efg is congruent to both △efg and △efg.

Explanation:

(a)

Step1: Apply rotation formula for \(90^{\circ}\) clockwise

The formula for a \(90^{\circ}\) clockwise rotation about the origin is \((x,y)\to(y, - x)\)
For point \(E(1,2)\):
Substitute \(x = 1\) and \(y=2\) into \((x,y)\to(y, - x)\)
\(E'(2,-1)\)
For point \(F(-3,6)\):
Substitute \(x=-3\) and \(y = 6\) into \((x,y)\to(y, - x)\)
\(F'(6,3)\)
For point \(G(4,3)\):
Substitute \(x = 4\) and \(y=3\) into \((x,y)\to(y, - x)\)
\(G'(3,-4)\)

From part (a), when we rotate a point \((x,y)\) \(90^{\circ}\) clockwise about the origin, we observe the transformation rule.
If we start with a general point \((x,y)\) and after rotation we get \((y,-x)\)

Step1: Apply rotation formula for \(90^{\circ}\) counter - clockwise

The formula for a \(90^{\circ}\) counter - clockwise rotation about the origin is \((x,y)\to(-y,x)\)
For point \(E(1,2)\):
Substitute \(x = 1\) and \(y = 2\) into \((x,y)\to(-y,x)\)
\(E''(-2,1)\)
For point \(F(-3,6)\):
Substitute \(x=-3\) and \(y = 6\) into \((x,y)\to(-y,x)\)
\(F''(-6,-3)\)
For point \(G(4,3)\):
Substitute \(x = 4\) and \(y=3\) into \((x,y)\to(-y,x)\)
\(G''(-3,4)\)

Answer:

\(E'(2,-1)\), \(F'(6,3)\), \(G'(3,-4)\)

(b)