Worksheet for graphing rotations of geometric figures on a coordinate plane.
Graph paper with six coordinate grids, each showing a polygon and instructions to rotate it by a specified angle about the origin.
JPG
495×640
41 KB
Free · Personal Use
Quality Assured by Worksheets Library Team
Reviewed for educational accuracy and age-appropriateness
ID: #691507
⭐
Show Answer Key & Explanations
Step-by-step solution for: Rotations - Kuta Software
▼
Show Answer Key & Explanations
Step-by-step solution for: Rotations - Kuta Software
To solve these rotation problems, we need to apply specific rules to the coordinates $(x, y)$ of each vertex of the figures. Here are the rules for rotations about the origin $(0,0)$:
1. Rotation $180^\circ$: The rule is $(x, y) \rightarrow (-x, -y)$. This flips the sign of both coordinates.
2. Rotation $90^\circ$ Counterclockwise: The rule is $(x, y) \rightarrow (-y, x)$. Swap the numbers and make the new first number negative.
3. Rotation $90^\circ$ Clockwise: The rule is $(x, y) \rightarrow (y, -x)$. Swap the numbers and make the new second number negative.
Let's solve each problem step-by-step by identifying the original points and applying the correct rule.
Original Points:
Looking at the graph, the vertices of the quadrilateral are:
* $P = (-2, 4)$
* $Q = (-5, 2)$
* $R = (-2, -2)$
* $S = (2, 0)$
Apply Rule $(x, y) \rightarrow (-x, -y)$:
* $P(-2, 4) \rightarrow P'(2, -4)$
* $Q(-5, 2) \rightarrow Q'(5, -2)$
* $R(-2, -2) \rightarrow R'(2, 2)$
* $S(2, 0) \rightarrow S'(-2, 0)$
New Coordinates: $P'(2, -4), Q'(5, -2), R'(2, 2), S'(-2, 0)$
---
Original Points:
The vertices of the triangle are:
* $P = (1, -1)$
* $Q = (3, -2)$
* $R = (4, -3)$
Apply Rule $(x, y) \rightarrow (-x, -y)$:
* $P(1, -1) \rightarrow P'(-1, 1)$
* $Q(3, -2) \rightarrow Q'(-3, 2)$
* $R(4, -3) \rightarrow R'(-4, 3)$
New Coordinates: $P'(-1, 1), Q'(-3, 2), R'(-4, 3)$
---
Original Points:
The vertices of the triangle are:
* $K = (-4, 1)$
* $L = (-3, -1)$
* $M = (-5, -4)$
Apply Rule $(x, y) \rightarrow (-y, x)$:
* $K(-4, 1) \rightarrow K'(-1, -4)$ *(Swap to $1, -4$, then negate first: $-1, -4$)*
* $L(-3, -1) \rightarrow L'(1, -3)$ *(Swap to $-1, -3$, then negate first: $1, -3$)*
* $M(-5, -4) \rightarrow M'(4, -5)$ *(Swap to $-4, -5$, then negate first: $4, -5$)*
New Coordinates: $K'(-1, -4), L'(1, -3), M'(4, -5)$
---
Original Points:
The vertices of the triangle are:
* $J = (1, 3)$
* $K = (3, 3)$
* $L = (4, 0)$
Apply Rule $(x, y) \rightarrow (y, -x)$:
* $J(1, 3) \rightarrow J'(3, -1)$ *(Swap to $3, 1$, then negate second: $3, -1$)*
* $K(3, 3) \rightarrow K'(3, -3)$ *(Swap to $3, 3$, then negate second: $3, -3$)*
* $L(4, 0) \rightarrow L'(0, -4)$ *(Swap to $0, 4$, then negate second: $0, -4$)*
New Coordinates: $J'(3, -1), K'(3, -3), L'(0, -4)$
---
Original Points:
The vertices of the triangle are:
* $U = (2, -1)$
* $V = (4, 0)$
* $W = (1, -3)$
Apply Rule $(x, y) \rightarrow (y, -x)$:
* $U(2, -1) \rightarrow U'(-1, -2)$ *(Swap to $-1, 2$, then negate second: $-1, -2$)*
* $V(4, 0) \rightarrow V'(0, -4)$ *(Swap to $0, 4$, then negate second: $0, -4$)*
* $W(1, -3) \rightarrow W'(-3, -1)$ *(Swap to $-3, 1$, then negate second: $-3, -1$)*
New Coordinates: $U'(-1, -2), V'(0, -4), W'(-3, -1)$
---
Original Points:
The vertices of the triangle are:
* $G = (0, -4)$
* $H = (-1, -1)$
* $I = (3, -2)$
Apply Rule $(x, y) \rightarrow (-x, -y)$:
* $G(0, -4) \rightarrow G'(0, 4)$
* $H(-1, -1) \rightarrow H'(1, 1)$
* $I(3, -2) \rightarrow I'(-3, 2)$
New Coordinates: $G'(0, 4), H'(1, 1), I'(-3, 2)$
Final Answer:
1) $P'(2, -4), Q'(5, -2), R'(2, 2), S'(-2, 0)$
2) $P'(-1, 1), Q'(-3, 2), R'(-4, 3)$
3) $K'(-1, -4), L'(1, -3), M'(4, -5)$
4) $J'(3, -1), K'(3, -3), L'(0, -4)$
5) $U'(-1, -2), V'(0, -4), W'(-3, -1)$
6) $G'(0, 4), H'(1, 1), I'(-3, 2)$
1. Rotation $180^\circ$: The rule is $(x, y) \rightarrow (-x, -y)$. This flips the sign of both coordinates.
2. Rotation $90^\circ$ Counterclockwise: The rule is $(x, y) \rightarrow (-y, x)$. Swap the numbers and make the new first number negative.
3. Rotation $90^\circ$ Clockwise: The rule is $(x, y) \rightarrow (y, -x)$. Swap the numbers and make the new second number negative.
Let's solve each problem step-by-step by identifying the original points and applying the correct rule.
Problem 1: Rotation $180^\circ$ about the origin
Original Points:
Looking at the graph, the vertices of the quadrilateral are:
* $P = (-2, 4)$
* $Q = (-5, 2)$
* $R = (-2, -2)$
* $S = (2, 0)$
Apply Rule $(x, y) \rightarrow (-x, -y)$:
* $P(-2, 4) \rightarrow P'(2, -4)$
* $Q(-5, 2) \rightarrow Q'(5, -2)$
* $R(-2, -2) \rightarrow R'(2, 2)$
* $S(2, 0) \rightarrow S'(-2, 0)$
New Coordinates: $P'(2, -4), Q'(5, -2), R'(2, 2), S'(-2, 0)$
---
Problem 2: Rotation $180^\circ$ about the origin
Original Points:
The vertices of the triangle are:
* $P = (1, -1)$
* $Q = (3, -2)$
* $R = (4, -3)$
Apply Rule $(x, y) \rightarrow (-x, -y)$:
* $P(1, -1) \rightarrow P'(-1, 1)$
* $Q(3, -2) \rightarrow Q'(-3, 2)$
* $R(4, -3) \rightarrow R'(-4, 3)$
New Coordinates: $P'(-1, 1), Q'(-3, 2), R'(-4, 3)$
---
Problem 3: Rotation $90^\circ$ counterclockwise about the origin
Original Points:
The vertices of the triangle are:
* $K = (-4, 1)$
* $L = (-3, -1)$
* $M = (-5, -4)$
Apply Rule $(x, y) \rightarrow (-y, x)$:
* $K(-4, 1) \rightarrow K'(-1, -4)$ *(Swap to $1, -4$, then negate first: $-1, -4$)*
* $L(-3, -1) \rightarrow L'(1, -3)$ *(Swap to $-1, -3$, then negate first: $1, -3$)*
* $M(-5, -4) \rightarrow M'(4, -5)$ *(Swap to $-4, -5$, then negate first: $4, -5$)*
New Coordinates: $K'(-1, -4), L'(1, -3), M'(4, -5)$
---
Problem 4: Rotation $90^\circ$ clockwise about the origin
Original Points:
The vertices of the triangle are:
* $J = (1, 3)$
* $K = (3, 3)$
* $L = (4, 0)$
Apply Rule $(x, y) \rightarrow (y, -x)$:
* $J(1, 3) \rightarrow J'(3, -1)$ *(Swap to $3, 1$, then negate second: $3, -1$)*
* $K(3, 3) \rightarrow K'(3, -3)$ *(Swap to $3, 3$, then negate second: $3, -3$)*
* $L(4, 0) \rightarrow L'(0, -4)$ *(Swap to $0, 4$, then negate second: $0, -4$)*
New Coordinates: $J'(3, -1), K'(3, -3), L'(0, -4)$
---
Problem 5: Rotation $90^\circ$ clockwise about the origin
Original Points:
The vertices of the triangle are:
* $U = (2, -1)$
* $V = (4, 0)$
* $W = (1, -3)$
Apply Rule $(x, y) \rightarrow (y, -x)$:
* $U(2, -1) \rightarrow U'(-1, -2)$ *(Swap to $-1, 2$, then negate second: $-1, -2$)*
* $V(4, 0) \rightarrow V'(0, -4)$ *(Swap to $0, 4$, then negate second: $0, -4$)*
* $W(1, -3) \rightarrow W'(-3, -1)$ *(Swap to $-3, 1$, then negate second: $-3, -1$)*
New Coordinates: $U'(-1, -2), V'(0, -4), W'(-3, -1)$
---
Problem 6: Rotation $180^\circ$ about the origin
Original Points:
The vertices of the triangle are:
* $G = (0, -4)$
* $H = (-1, -1)$
* $I = (3, -2)$
Apply Rule $(x, y) \rightarrow (-x, -y)$:
* $G(0, -4) \rightarrow G'(0, 4)$
* $H(-1, -1) \rightarrow H'(1, 1)$
* $I(3, -2) \rightarrow I'(-3, 2)$
New Coordinates: $G'(0, 4), H'(1, 1), I'(-3, 2)$
Final Answer:
1) $P'(2, -4), Q'(5, -2), R'(2, 2), S'(-2, 0)$
2) $P'(-1, 1), Q'(-3, 2), R'(-4, 3)$
3) $K'(-1, -4), L'(1, -3), M'(4, -5)$
4) $J'(3, -1), K'(3, -3), L'(0, -4)$
5) $U'(-1, -2), V'(0, -4), W'(-3, -1)$
6) $G'(0, 4), H'(1, 1), I'(-3, 2)$
Parent Tip: Review the logic above to help your child master the concept of rotations worksheet answers.