Question

putting it all together! name: teneighlilliamson period 3 transformations, constructions, symmetry! 1. find the image of the point (3, -5) after each transformation: a. r_x - axis b. r_y - axis c. r_origin d. r_y = x e. r_y=-x f. r_90 g. r_180 h. r_270 i. t_-6,7 1. a. what is the minimum number of degrees needed to carry a regular decagon onto itself? b. what are nine more degree measures that will carry the regular decagon onto itself? 3. in the diagram below, parallelogram efgh is mapped onto parallelogram ijkh after a reflection over line l. use the properties of rigid motions to explain why parallelogram efgh is congruent to parallelogram ijkh.

Explanation:

Step1: Point - reflection rules

For a point $(x,y)$ reflected over the $x$-axis, the new point is $(x, - y)$. For the point $(3,-5)$ reflected over the $x$-axis ($r_{x - axis}$), we change the sign of the $y$-coordinate. So, the new point is $(3,5)$.

Step2: $y$-axis reflection

For a point $(x,y)$ reflected over the $y$-axis, the new point is $(-x,y)$. For the point $(3,-5)$ reflected over the $y$-axis ($r_{y - axis}$), we change the sign of the $x$-coordinate. So, the new point is $(-3,-5)$.

Step3: Origin reflection

For a point $(x,y)$ reflected over the origin, the new point is $(-x,-y)$. For the point $(3,-5)$ reflected over the origin ($r_{origin}$), we change the signs of both coordinates. So, the new point is $(-3,5)$.

Step4: Reflection over $y = x$

For a point $(x,y)$ reflected over the line $y=x$, the new point is $(y,x)$. For the point $(3,-5)$ reflected over $y = x$ ($r_{y=x}$), the new point is $(-5,3)$.

Step5: Reflection over $y=-x$

For a point $(x,y)$ reflected over the line $y =-x$, the new point is $(-y,-x)$. For the point $(3,-5)$ reflected over $y=-x$ ($r_{y=-x}$), the new point is $(5,-3)$.

Step6: $90^{\circ}$ rotation

For a $90^{\circ}$ counter - clockwise rotation ($R_{90}$) of a point $(x,y)$ about the origin, the new point is $(-y,x)$. For the point $(3,-5)$, the new point is $(5,3)$.

Step7: $180^{\circ}$ rotation

For a $180^{\circ}$ rotation ($R_{180}$) of a point $(x,y)$ about the origin, the new point is $(-x,-y)$. For the point $(3,-5)$, the new point is $(-3,5)$.

Step8: $270^{\circ}$ rotation

For a $270^{\circ}$ counter - clockwise rotation ($R_{270}$) of a point $(x,y)$ about the origin, the new point is $(y,-x)$. For the point $(3,-5)$, the new point is $(-5,-3)$.

Step9: Translation

For a translation $T_{-6,7}$, we add $- 6$ to the $x$-coordinate and $7$ to the $y$-coordinate of the point $(x,y)$. For the point $(3,-5)$, the new point is $(3-6,-5 + 7)=(-3,2)$.

Step10: Regular decagon rotation

A regular decagon has $n = 10$ sides. The minimum angle of rotation to carry a regular polygon onto itself is given by $\frac{360^{\circ}}{n}$. For a decagon, $\frac{360^{\circ}}{10}=36^{\circ}$.

Step11: Other rotation angles

Other angles that will carry a regular decagon onto itself are multiples of the minimum angle. So the nine more angles are $72^{\circ},108^{\circ},144^{\circ},180^{\circ},216^{\circ},252^{\circ},288^{\circ},324^{\circ},360^{\circ}$.

Step12: Rigid - motion congruence

A reflection is a rigid motion. Rigid motions (translations, rotations, and reflections) preserve distance and angle measures. Since parallelogram $EFGH$ is mapped onto parallelogram $IJKH$ by a reflection (a rigid motion), the corresponding sides and angles of the two parallelograms are congruent. So, parallelogram $EFGH$ is congruent to parallelogram $IJKH$.

Answer:

a. $(3,5)$
b. $(-3,-5)$
c. $(-3,5)$
d. $(-5,3)$
e. $(5,-3)$
f. $(5,3)$
g. $(-3,5)$
h. $(-5,-3)$
i. $(-3,2)$

1. a. $36^{\circ}$

b. $72^{\circ},108^{\circ},144^{\circ},180^{\circ},216^{\circ},252^{\circ},288^{\circ},324^{\circ},360^{\circ}$

1. A reflection is a rigid motion which preserves distance and angle - measures, so parallelogram $EFGH$ is congruent to parallelogram $IJKH$.