Question

extra practice level up your skills!
graph each figure with the given vertices and its image after the indicated composition of transformations.

1. \\(\overline{xy}\\): \\(x(-4, 3)\\), \\(y(-2, 1)\\)

translation: along \\(<1, 0>\\)
rotation: \\(90^\circ\\) clockwise about origin

Explanation:

Step1: Apply Translation

The translation vector is $\langle 1, 0
angle$, which means we add 1 to the x - coordinate and 0 to the y - coordinate of each vertex.
For point $X(-4, 3)$:
New $x$-coordinate: $-4 + 1=-3$
New $y$-coordinate: $3+0 = 3$
So, the translated point $X'$ is $(-3, 3)$
For point $Y(-2, 1)$:
New $x$-coordinate: $-2 + 1=-1$
New $y$-coordinate: $1+0 = 1$
So, the translated point $Y'$ is $(-1, 1)$

Step2: Apply Rotation

The rule for a $90^{\circ}$ clockwise rotation about the origin is $(x,y)\to(y, - x)$
For point $X'(-3, 3)$:
Using the rotation rule, $x=-3,y = 3$
New $x$-coordinate: $3$
New $y$-coordinate: $-(-3)=3$? Wait, no. Wait, the rule for $90^{\circ}$ clockwise rotation about the origin is $(x,y)\to(y,-x)$. So for $(x,y)=(-3,3)$, the new point $X''$ is $(3,3)$? Wait, no:
If the original point is $(x,y)$, after $90^{\circ}$ clockwise rotation about the origin, the coordinates become $(y, - x)$
So for $X'(-3,3)$:
$x=-3,y = 3$
New $x = y=3$
New $y=-x=-(-3)=3$? Wait, no, $-x$ when $x=-3$ is $-(-3)=3$? Wait, no, let's re - derive the rotation rule.
A $90^{\circ}$ clockwise rotation about the origin:
We can think of it as a rotation matrix. The rotation matrix for a $90^{\circ}$ clockwise rotation is $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$. So if we have a vector $\begin{pmatrix}x\\y\end{pmatrix}$, after rotation, it becomes $\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}y\\-x\end{pmatrix}$
So for $X'(-3,3)$:
$x=-3,y = 3$
After rotation, the coordinates are $(y,-x)=(3, - (-3))=(3,3)$? Wait, no, $-x$ when $x = - 3$ is $-(-3)=3$? Wait, no, $-x$ is the negative of $x$. If $x=-3$, then $-x = 3$. So the new point $X''$ is $(3,3)$
For $Y'(-1,1)$:
Using the rotation rule $(x,y)\to(y,-x)$
$x=-1,y = 1$
New $x = y = 1$
New $y=-x=-(-1)=1$? Wait, no, $-x$ when $x=-1$ is $-(-1) = 1$. So the new point $Y''$ is $(1,1)$? Wait, that can't be right. Wait, maybe I made a mistake in the rotation rule. Wait, the correct rule for $90^{\circ}$ clockwise rotation about the origin is $(x,y)\to(y, - x)$
Wait, let's take a simple point, say $(1,0)$. A $90^{\circ}$ clockwise rotation about the origin should take it to $(0, - 1)$. Using the rule $(x,y)\to(y,-x)$, for $(1,0)$, we get $(0,-1)$, which is correct.
Another example: $(0,1)$ rotated $90^{\circ}$ clockwise about the origin should be $(1,0)$. Using the rule $(x,y)\to(y,-x)$, for $(0,1)$ we get $(1,0)$, which is correct.
Another example: $(1,1)$ rotated $90^{\circ}$ clockwise about the origin should be $(1, - 1)$. Using the rule $(x,y)\to(y,-x)$, for $(1,1)$ we get $(1,-1)$, which is correct.
So going back to $X'(-3,3)$:
$x=-3,y = 3$
Applying the rule $(y,-x)=(3,-(-3))=(3,3)$
For $Y'(-1,1)$:
$x=-1,y = 1$
Applying the rule $(y,-x)=(1,-(-1))=(1,1)$

Wait, but let's check the translation again. The translation is along $\langle1,0
angle$, which is a horizontal translation 1 unit to the right. So $X(-4,3)$ moves to $(-4 + 1,3)=(-3,3)$, $Y(-2,1)$ moves to $(-2 + 1,1)=(-1,1)$. That part is correct.

Now, for the $90^{\circ}$ clockwise rotation about the origin:
The formula is: If a point $(x,y)$ is rotated $90^{\circ}$ clockwise about the origin, the new coordinates $(x',y')$ are given by $x'=y$ and $y'=-x$
So for $X'(-3,3)$:
$x'=3$, $y'=-(-3)=3$? Wait, no, $y'=-x$, where $x=-3$, so $y'=-(-3)=3$. So $X''=(3,3)$
For $Y'(-1,1)$:
$x' = 1$, $y'=-(-1)=1$. So $Y''=(1,1)$

Now, to graph:
- First, plot the original segment $\overline{XY}$ with $X(-4,3)$ and $Y(-2,1)$
- Then plot the translated segment $\overline{X'Y'}$ with $X'(-3,3)$ and $Y'(-1,1)$
- Then plot the rotated segment $\overline{X''Y''}$ with $X''(3,3)$ and $Y''(1,1)$

Answer:

The original points are $X(-4,3),Y(-2,1)$; after translation, $X'(-3,3),Y'(-1,1)$; after rotation, $X''(3,3),Y''(1,1)$. To graph, plot these points and draw the segments.