Free Printable Math Worksheets for Geometry - Free Printable

To solve these problems, we need to apply geometric transformations (rotations and reflections) to the given shapes on the coordinate grid. Let's go through each problem step by step.

Problem 1: Rotate $90^\circ$ clockwise about origin.
* Rule: When rotating a point $(x, y)$ $90^\circ$ clockwise around the origin $(0,0)$, the new coordinates become $(y, -x)$.
* Original Vertices: Looking at the triangle in Problem 1, the vertices are approximately:
* Point A: $(-2, 4)$
* Point B: $(-4, 1)$
* Point C: $(-1, 1)$
* Apply Rule $(x, y) \rightarrow (y, -x)$:
* $A(-2, 4) \rightarrow A'(4, -(-2)) = (4, 2)$
* $B(-4, 1) \rightarrow B'(1, -(-4)) = (1, 4)$
* $C(-1, 1) \rightarrow C'(1, -(-1)) = (1, 1)$
* Result: The new triangle has vertices at $(4, 2)$, $(1, 4)$, and $(1, 1)$. This matches the shape shown in the answer key area for #1 (a triangle in the first quadrant).

Problem 2: Reflect over the line $x = 2$.
* Rule: Reflection over a vertical line $x = k$ changes the x-coordinate. The distance from the point to the line is preserved on the other side. Formula: $x' = 2k - x$. Here $k=2$, so $x' = 4 - x$. The y-coordinate stays the same ($y' = y$).
* Original Vertices:
* Point D: $(1, 5)$
* Point E: $(3, 5)$
* Point F: $(3, 3)$
* Point G: $(1, 3)$ -- Wait, looking closely at image #2, it's a trapezoid or rectangle part. Let's look at the vertices visible.
* Top-left: $(1, 5)$
* Top-right: $(3, 5)$
* Bottom-right: $(3, 3)$
* Bottom-left vertex seems to be connected to another point? No, it looks like a right trapezoid with vertices $(1,5), (3,5), (3,3)$ and maybe $(2,3)$? Let's re-examine. Actually, it looks like a triangle with vertices $(1,5), (3,5), (3,3)$? No, there is a fourth point. Let's assume the main shape is defined by points $(1,5), (3,5), (3,3)$ and $(1,3)$ is not connected. Let's look at the reflected image provided in the solution spot.
* Let's just pick the key points.
* Point $(1, 5)$ reflects over $x=2$. Distance from $x=1$ to $x=2$ is 1 unit left. So new point is 1 unit right of $x=2$, which is $x=3$. New point: $(3, 5)$.
* Point $(3, 5)$ reflects over $x=2$. Distance from $x=3$ to $x=2$ is 1 unit right. So new point is 1 unit left of $x=2$, which is $x=1$. New point: $(1, 5)$.
* Point $(3, 3)$ reflects over $x=2$. Becomes $(1, 3)$.
* Point $(1, ?)$... let's look at the bottom left vertex. It looks like $(1,3)$? If so, it reflects to $(3,3)$.
* Essentially, the shape flips horizontally across the vertical line $x=2$. The part that was on the left moves to the right, and vice versa.

Problem 3: Rotate $180^\circ$ about origin.
* Rule: Rotating $180^\circ$ around the origin changes the sign of both coordinates: $(x, y) \rightarrow (-x, -y)$.
* Original Vertices:
* Point H: $(-2, 2)$
* Point I: $(-4, 2)$
* Point J: $(-3, 4)$
* Apply Rule $(x, y) \rightarrow (-x, -y)$:
* $H(-2, 2) \rightarrow H'(2, -2)$
* $I(-4, 2) \rightarrow I'(4, -2)$
* $J(-3, 4) \rightarrow J'(3, -4)$
* Result: The new triangle is in the fourth quadrant with vertices $(2, -2), (4, -2), (3, -4)$.

Problem 4: Reflect over the y-axis.
* Rule: Reflection over the y-axis changes the sign of the x-coordinate: $(x, y) \rightarrow (-x, y)$.
* Original Vertices:
* Point K: $(2, -2)$
* Point L: $(4, -2)$
* Point M: $(3, -4)$
* Apply Rule $(x, y) \rightarrow (-x, y)$:
* $K(2, -2) \rightarrow K'(-2, -2)$
* $L(4, -2) \rightarrow L'(-4, -2)$
* $M(3, -4) \rightarrow M'(-3, -4)$
* Result: The triangle flips from the fourth quadrant to the third quadrant.

Problem 5: Rotate $90^\circ$ counterclockwise about origin.
* Rule: Rotating $90^\circ$ counterclockwise follows the rule: $(x, y) \rightarrow (-y, x)$.
* Original Vertices:
* Point N: $(-2, -2)$
* Point O: $(-4, -2)$
* Point P: $(-3, -4)$
* Wait, let's look closer at #5. The shape is in the third quadrant.
* Vertex 1: $(-2, -2)$
* Vertex 2: $(-4, -2)$ ?? No, looking at the grid lines.
* Let's identify coordinates carefully.
* Top-right vertex of the shape: $(-2, -2)$? No, it looks like $(-1, -2)$? Let's assume standard grid spacing.
* Let's look at the "answer" drawn in blue ink for #5. It is in the second quadrant.
* Original shape vertices appear to be: $(-2, -2), (-4, -2), (-3, -4)$? Or maybe $(-2,-1)...$
* Let's use the visual transformation. A $90^\circ$ CCW rotation moves a shape from Quadrant III to Quadrant IV? No.
* Q1 $\rightarrow$ Q2 $\rightarrow$ Q3 $\rightarrow$ Q4 $\rightarrow$ Q1 is Clockwise.
* Counter-Clockwise: Q3 $\rightarrow$ Q2.
* So a shape in the bottom-left (Q3) moves to top-left (Q2).
* Let's check the coordinates of the blue drawing in #5.
* The blue drawing has vertices at roughly $(-2, 2), (-2, 4), (-4, 3)$.
* Let's reverse engineer the original. If result is $(-2, 2)$, original was $(2, -2)$? No.
* Rule: $(x,y) \rightarrow (-y, x)$.
* If original is $(-2, -2)$, new is $(2, -2)$. That's Q4.
* If original is $(-2, -4)$, new is $(4, -2)$.
* Let's re-read the graph for #5. The black shape is in Q3. Vertices: $(-2, -2), (-4, -2), (-3, -4)$?
* $(-2, -2) \rightarrow (2, -2)$
* $(-4, -2) \rightarrow (2, -4)$
* $(-3, -4) \rightarrow (4, -3)$
* This would put it in Q4. But the blue ink shows it in Q2?
* Ah, let's look at the text again. "Rotate $90^\circ$ counterclockwise".
* Maybe my coordinate reading is off. Let's look at the vertex closest to the origin in #5. It is at $(-2, -2)$? Or $(-1, -2)$?
* Let's look at the blue answer key. The vertex closest to origin is $(-2, 2)$? No, it's $(-2, 1)$?
* Let's try a different vertex. The "pointy" end. In black, it's at $(-3, -4)$. In blue, it's at $(-4, 3)$?
* If $(-3, -4)$ becomes $(-4, 3)$, then $x=-3, y=-4 \rightarrow -y=4, x=-3$. Result $(4, -3)$. That doesn't match $(-4,3)$.
* Wait, $(-y, x)$ for $(-3, -4)$ is $(4, -3)$.
* Is it possible the rotation is clockwise? No, text says CCW.
* Is it possible the original coordinates are different?
* Let's look at the blue shape in #5 again. Vertices: $(-2, 2), (-4, 2), (-3, 4)$?
* If the result is $(-3, 4)$, then $-y = -3 \Rightarrow y=3$ and $x=4$. Original $(4,3)$? No, original is in Q3.
* Let's restart the coordinate reading for #5 Black Shape.
* Vertex A: $(-2, -2)$
* Vertex B: $(-4, -2)$
* Vertex C: $(-3, -4)$ -- wait, is it $(-3, -4)$ or $(-3, -1)$? It looks like it goes down 2 units from the base. Base is at $y=-2$. Tip is at $y=-4$.
* Let's apply CCW rotation $(x,y) \rightarrow (-y, x)$.
* $A(-2, -2) \rightarrow (2, -2)$.
* $B(-4, -2) \rightarrow (2, -4)$.
* $C(-3, -4) \rightarrow (4, -3)$.
* This results in a shape in Quadrant 4.
* However, the student's handwritten answer (blue ink) is in Quadrant 2. Did the student make a mistake? Or did I misread the direction?
* "Rotate $90^\circ$ counterclockwise".
* Let's check if the black shape is actually in Quadrant 4? No, x is negative, y is negative. It is Q3.
* Rotation of Q3 by $90^\circ$ CCW goes to Q4?
* Start at 7 o'clock (Q3). Move CCW (towards 6, 5, 4...). Yes, it goes to Q4.
* Wait. CCW is opposite to clock hands.
* 12 -> 9 -> 6 -> 3.
* Q2 (top left) -> Q3 (bottom left) -> Q4 (bottom right) -> Q1 (top right).
* So Q3 rotated $90^\circ$ CCW lands in Q4.
* The blue ink drawing is in Q2. This implies the student might have rotated $90^\circ$ *Clockwise* (Q3 -> Q2) or $180^\circ$? No, $180$ would be Q1.
* Actually, let's look at the blue drawing in #5. It looks like the reflection of the black shape over the line $y=x$? Or just a rotation error?
* Regardless, the task is to solve it correctly.
* Correct Answer for #5: The shape should be in Quadrant 4 with vertices $(2, -2), (2, -4), (4, -3)$.

Problem 6: Rotate $180^\circ$ about origin.
* Rule: $(x, y) \rightarrow (-x, -y)$.
* Original Vertices:
* The shape is in Quadrant 4.
* Vertex R: $(2, -2)$
* Vertex S: $(4, -2)$
* Vertex T: $(3, -4)$ ? Or $(5, -2)$? Let's assume similar dimensions to previous triangles.
* Let's say vertices are $(2, -2), (4, -2), (3, -4)$.
* Apply Rule:
* $(2, -2) \rightarrow (-2, 2)$
* $(4, -2) \rightarrow (-4, 2)$
* $(3, -4) \rightarrow (-3, 4)$
* Result: The shape moves to Quadrant 2. This matches the position of the blue ink drawing in #6.

Summary of Steps for Final Output:
Since the user wants the solution, I will describe the final position and coordinates for each transformed shape.

1. Rotate $90^\circ$ CW: Shape moves from Q2 to Q1. Coordinates $(x,y) \rightarrow (y, -x)$.
2. Reflect over $x=2$: Shape flips horizontally across the vertical line $x=2$.
3. Rotate $180^\circ$: Shape moves from Q2 to Q4. Coordinates $(x,y) \rightarrow (-x, -y)$.
4. Reflect over y-axis: Shape flips horizontally from Q4 to Q3. Coordinates $(x,y) \rightarrow (-x, y)$.
5. Rotate $90^\circ$ CCW: Shape moves from Q3 to Q4. Coordinates $(x,y) \rightarrow (-y, x)$. *(Note: The handwritten answer in the image appears incorrect as it places the shape in Q2, which corresponds to a clockwise rotation).*
6. Rotate $180^\circ$: Shape moves from Q4 to Q2. Coordinates $(x,y) \rightarrow (-x, -y)$.

I will provide the specific coordinates for the vertices based on visual estimation from the grid, assuming each grid square is 1 unit.

Vertex Estimation:
* #1 (Triangle in Q2): Vertices $(-2, 4), (-4, 1), (-1, 1)$.
* Transform: $(y, -x)$.
* New Vertices: $(4, 2), (1, 4), (1, 1)$.
* #2 (Trapezoid/Triangle in Q1): Vertices $(1, 5), (3, 5), (3, 3), (1, 3)$? Let's assume it's the triangle with vertices $(1,5), (3,5), (3,3)$ and the segment to $(1,3)$ is implicit or it's a rectangle cut. Let's look at the shape. It's a right trapezoid with vertices $(1,5), (3,5), (3,3), (2,3)$? No, the bottom side is slanted? No, vertical line at $x=3$, horizontal at $y=5$. The left side connects $(1,5)$ to... $(1,3)$? And bottom connects $(1,3)$ to $(3,3)$? If it's a rectangle $(1,3)-(3,5)$, reflection over $x=2$ keeps it in the same place because it's symmetric around $x=2$?
* Center of rectangle $x=1$ to $3$ is $x=2$. Yes. If it is a rectangle from $x=1$ to $3$, reflecting over $x=2$ results in the exact same shape.
* However, the shape in #2 looks like a triangle on top of a rectangle? Or just a polygon.
* Let's look at the vertices: $(1,5), (3,5), (3,3)$ and $(1,4)$? There is a point at $(1,4)$. And a line from $(1,4)$ to $(3,3)$?
* Let's assume the vertices are $(1,5), (3,5), (3,3), (1,4)$.
* Reflect over $x=2$:
* $(1,5) \rightarrow (3,5)$
* $(3,5) \rightarrow (1,5)$
* $(3,3) \rightarrow (1,3)$
* $(1,4) \rightarrow (3,4)$
* New vertices: $(3,5), (1,5), (1,3), (3,4)$.
* #3 (Triangle in Q2): Vertices $(-2, 2), (-4, 2), (-3, 4)$.
* Transform: $(-x, -y)$.
* New Vertices: $(2, -2), (4, -2), (3, -4)$.
* #4 (Triangle in Q4): Vertices $(2, -2), (4, -2), (3, -4)$.
* Transform: $(-x, y)$.
* New Vertices: $(-2, -2), (-4, -2), (-3, -4)$.
* #5 (Triangle in Q3): Vertices $(-2, -2), (-4, -2), (-3, -4)$.
* Transform: $(-y, x)$.
* New Vertices: $(2, -2), (2, -4), (4, -3)$.
* #6 (Triangle in Q4): Vertices $(2, -2), (4, -2), (3, -4)$? Wait, looking at #6, the shape is slightly different.
* Vertices: $(1, -2), (3, -2), (2, -4)$?
* Let's check the grid. Origin is center.
* Right 1, Down 2: $(1, -2)$.
* Right 3, Down 2: $(3, -2)$.
* Right 2, Down 4: $(2, -4)$.
* Transform $180^\circ$: $(-x, -y)$.
* New Vertices: $(-1, 2), (-3, 2), (-2, 4)$.

Final Answer Formulation:
I will list the new coordinates for the vertices of each shape after the transformation.

Double Check Calculations:
1. $(-2,4) \rightarrow (4,2)$. Correct.
2. Reflection over $x=2$. Point $(1,5)$ is 1 unit left of 2. Image is 1 unit right of 2 $\rightarrow (3,5)$. Correct.
3. $(-3,4) \rightarrow (3,-4)$. Correct.
4. $(3,-4) \rightarrow (-3,-4)$. Correct.
5. $(-3,-4) \rightarrow (4,-3)$. Correct. (Student drew it wrong in the image, but I must provide the correct math answer).
6. $(2,-4) \rightarrow (-2,4)$. Correct.

The request asks to "Solve the problem accurately". I will provide the coordinates of the transformed vertices for each number.

Final Answer:
1. Rotate $90^\circ$ clockwise: The new vertices are $(4, 2)$, $(1, 4)$, and $(1, 1)$.
2. Reflect over $x = 2$: The shape flips across the vertical line $x=2$. For example, the vertex at $(1, 5)$ moves to $(3, 5)$, and $(3, 3)$ moves to $(1, 3)$.
3. Rotate $180^\circ$: The new vertices are $(2, -2)$, $(4, -2)$, and $(3, -4)$.
4. Reflect over y-axis: The new vertices are $(-2, -2)$, $(-4, -2)$, and $(-3, -4)$.
5. Rotate $90^\circ$ counterclockwise: The new vertices are $(2, -2)$, $(2, -4)$, and $(4, -3)$. *(Note: This places the shape in Quadrant IV).*
6. Rotate $180^\circ$: Assuming vertices at $(1, -2), (3, -2), (2, -4)$, the new vertices are $(-1, 2)$, $(-3, 2)$, and $(-2, 4)$.