Question

1. rotate figure mno 90° clockwise around point o. label the vertices m, n, and o so that point m corresponds to the rotation of point m and n corresponds to n.

m = ____ n = ____

1. rotate figure mno 180° clockwise around point o. label the vertices m, n, and o so that point m corresponds to the rotation of point m and n corresponds to n.

m = ____ n = ____

1. rotate figure mno 270° clockwise around point o. label the vertices m, n, and o so that point m corresponds to the rotation of point m and n corresponds to n.

m = ____ n = ____
2.4 transformation practice (dok 2)

Explanation:

Step1: Recall rotation rules

For a 90 - degree clock - wise rotation around the origin $(x,y)\to(y, - x)$. For a 180 - degree clock - wise rotation around the origin $(x,y)\to(-x,-y)$. For a 270 - degree clock - wise rotation around the origin $(x,y)\to(-y,x)$. Assume point $O$ is the origin $(0,0)$. Let's assume the coordinates of $M$ are $(x_1,y_1)$ and of $N$ are $(x_2,y_2)$.

Step2: 90 - degree clock - wise rotation

If $M(x_1,y_1)$, then $M'=(y_1,-x_1)$. If $N(x_2,y_2)$, then $N'=(y_2,-x_2)$.

Step3: 180 - degree clock - wise rotation

If $M(x_1,y_1)$, then $M''=(-x_1,-y_1)$. If $N(x_2,y_2)$, then $N''=(-x_2,-y_2)$.

Step4: 270 - degree clock - wise rotation

If $M(x_1,y_1)$, then $M'''=(-y_1,x_1)$. If $N(x_2,y_2)$, then $N'''=(-y_2,x_2)$.

However, since we don't have the actual coordinates of $M$ and $N$ from the graph:
Let's assume $M(-3,3)$ and $N(1,1)$ for illustration purposes.

For 90 - degree clock - wise rotation:

$M'=(3,3)$ and $N'=(1, - 1)$

For 180 - degree clock - wise rotation:

$M''=(3,-3)$ and $N''=(-1,-1)$

For 270 - degree clock - wise rotation:

$M'''=(-3,-3)$ and $N'''=(-1,1)$

Answer:

1. Without specific coordinates of $M$ and $N$: In general, if $M(x,y)$ then $M'=(y, - x)$ and if $N(a,b)$ then $N'=(b, - a)$.
2. In general, if $M(x,y)$ then $M''=(-x,-y)$ and if $N(a,b)$ then $N''=(-a,-b)$.
3. In general, if $M(x,y)$ then $M'''=(-y,x)$ and if $N(a,b)$ then $N'''=(-b,a)$.

(If we assume $M(-3,3)$ and $N(1,1)$:

1. $M'=(3,3)$, $N'=(1, - 1)$
2. $M''=(3,-3)$, $N''=(-1,-1)$
3. $M'''=(-3,-3)$, $N'''=(-1,1)$)