Question

suppose we perform a translation, then a reflection, then a rotation, then another translation on the point a.
a gets translated to ____, which gets reflected to __, which gets rotated to __, which finally gets translated again to ____
pre - image (t_{(-3,5)}) (r_{x - axis}) (r_{270}) (t_{(1,3)})
a(6, -1)
notation
which rigid motion would be performed first?
(r_{x - axis}circ r_{270}) (t_{(-5,11)}circ r_{y = x}) (r_{y=-x}circ r_{y - axis}) (r_{x - axis}circ r_{x - axis})
the __________________ is performed first.
for the last box, where would the point image of the point (-4, 2) be?

Explanation:

Step1: First translation of point A

For a translation $T_{(-3,5)}$ on point $A(6, - 1)$, we add the translation values to the coordinates of the point. The $x$-coordinate becomes $6+( - 3)=3$ and the $y$-coordinate becomes $-1 + 5 = 4$. So the point after the first translation is $(3,4)$.

Step2: Reflection across x - axis

The rule for reflecting a point $(x,y)$ across the $x$-axis is $(x,-y)$. For the point $(3,4)$, after reflection across the $x$-axis, it becomes $(3,-4)$.

Step3: Rotation of 270 degrees counter - clockwise

The rule for a $270^{\circ}$ counter - clockwise rotation about the origin $(x,y)\to(y,-x)$. For the point $(3,-4)$, after rotation, it becomes $(-4,-3)$.

Step4: Second translation

For a translation $T_{(1,3)}$ on the point $(-4,-3)$, we add the translation values to the coordinates. The $x$-coordinate becomes $-4 + 1=-3$ and the $y$-coordinate becomes $-3+3 = 0$. So the final point is $(-3,0)$.

For the notation part, in function composition $f\circ g$ means apply $g$ first. So for $r_{x - axis}\circ R_{270}$, $R_{270}$ is performed first; for $T_{(-5,11)}\circ r_{y = x}$, $r_{y = x}$ is performed first; for $r_{y=-x}\circ r_{y - axis}$, $r_{y - axis}$ is performed first; for $r_{x - axis}\circ r_{x - axis}$, $r_{x - axis}$ is performed first.

For the point $(-4,2)$ with $r_{x - axis}\circ r_{x - axis}$, first reflection across the $x$-axis of $(-4,2)$ gives $(-4,-2)$, and then reflecting $(-4,-2)$ across the $x$-axis again gives $(-4,2)$ (since reflecting a point $(x,-y)$ across the $x$-axis gives $(x,y)$).

Answer:

1. First translation of $A(6,-1)$: $(3,4)$
2. Reflection of $(3,4)$ across $x$-axis: $(3,-4)$
3. Rotation of $(3,-4)$ by $270^{\circ}$: $(-4,-3)$
4. Second translation of $(-4,-3)$: $(-3,0)$
5. In $r_{x - axis}\circ R_{270}$, $R_{270}$ is performed first; in $T_{(-5,11)}\circ r_{y = x}$, $r_{y = x}$ is performed first; in $r_{y=-x}\circ r_{y - axis}$, $r_{y - axis}$ is performed first; in $r_{x - axis}\circ r_{x - axis}$, $r_{x - axis}$ is performed first
6. Image of $(-4,2)$ for $r_{x - axis}\circ r_{x - axis}$: $(-4,2)$