Triangle A B C A B C A BC Has Vertices A ( − 2 , − 3 ) , B ( − 5 , − 3 A(-2,-3), B(-5,-3 A ( − 2 , − 3 ) , B ( − 5 , − 3 ], And C ( − 4 , − 2 C(-4,-2 C ( − 4 , − 2 ]. The Triangle Is Rotated 90 ∘ 90^{\circ} 9 0 ∘ Counterclockwise Around The Origin.What Is The Correct Set Of Image Points For Triangle $A^{\prime}

Introduction

In geometry, rotation is a fundamental concept that involves turning a figure around a fixed point called the center of rotation. When a point is rotated counterclockwise around the origin in the coordinate plane, its coordinates change in a predictable way. In this article, we will explore how to rotate a triangle in the coordinate plane and find the image points of its vertices after a $90^{\circ}$ counterclockwise rotation around the origin.

Understanding Rotation in the Coordinate Plane

To rotate a point $(x, y)$ counterclockwise around the origin by an angle $\theta$, we can use the following rotation formulas:

$x' = x \cos \theta - y \sin \theta$ $y' = x \sin \theta + y \cos \theta$

where $(x', y')$ are the new coordinates of the point after rotation.

Rotating a Triangle: A Step-by-Step Example

Let's consider the triangle $ABC$ with vertices $A(-2,-3), B(-5,-3)$, and $C(-4,-2)$. We want to rotate this triangle $90^{\circ}$ counterclockwise around the origin.

Step 1: Rotate Vertex A

To rotate vertex $A(-2,-3)$ $90^{\circ}$ counterclockwise around the origin, we can use the rotation formulas:

$x' = x \cos \theta - y \sin \theta$ $y' = x \sin \theta + y \cos \theta$

Substituting $x = -2, y = -3, \theta = 90^{\circ}$, we get:

$x' = -2 \cos 90^{\circ} - (-3) \sin 90^{\circ} = 0 + 3 = 3$ $y' = -2 \sin 90^{\circ} + (-3) \cos 90^{\circ} = -2 + 0 = -2$

So, the new coordinates of vertex $A$ after rotation are $(3, -2)$.

Step 2: Rotate Vertex B

To rotate vertex $B(-5,-3)$ $90^{\circ}$ counterclockwise around the origin, we can use the rotation formulas:

$x' = x \cos \theta - y \sin \theta$ $y' = x \sin \theta + y \cos \theta$

Substituting $x = -5, y = -3, \theta = 90^{\circ}$, we get:

$x' = -5 \cos 90^{\circ} - (-3) \sin 90^{\circ} = 0 + 3 = 3$ $y' = -5 \sin 90^{\circ} + (-3) \cos 90^{\circ} = -5 + 0 = -5$

So, the new coordinates of vertex $B$ after rotation are $(3, -5)$.

Step 3: Rotate Vertex C

To rotate vertex $C(-4,-2)$ $90^{\circ}$ counterclockwise around the origin, we can use the rotation formulas:

$x' = x \cos \theta - y \sin \theta$ $y' = x \sin \theta + y \cos \theta$

Substituting $x = -4, y = -2, \theta = 90^{\circ}$, we get:

$x' = -4 \cos 90^{\circ} - (-2) \sin 90^{\circ} = 0 + 2 = 2$ $y' = -4 \sin 90^{\circ} + (-2) \cos 90^{\circ} = -4 + 0 = -4$

So, the new coordinates of vertex $C$ after rotation are $(2, -4)$.

Conclusion

In this article, we have explored how to rotate a triangle in the coordinate plane and find the image points of its vertices after a $90^{\circ}$ counterclockwise rotation around the origin. We have used the rotation formulas to find the new coordinates of each vertex after rotation. By following these steps, you can easily rotate any triangle in the coordinate plane and find its image points.

Example Problems

1. Rotate the triangle with vertices $A(2,3), B(5,3)$, and $C(4,2)$ $90^{\circ}$ counterclockwise around the origin.
2. Rotate the triangle with vertices $A(-3,-2), B(-2,-2)$, and $C(-3,-1)$ $90^{\circ}$ counterclockwise around the origin.

Solutions

1. To rotate the triangle with vertices $A(2,3), B(5,3)$, and $C(4,2)$ $90^{\circ}$ counterclockwise around the origin, we can use the rotation formulas:

$x' = x \cos \theta - y \sin \theta$ $y' = x \sin \theta + y \cos \theta$

Substituting $x = 2, y = 3, \theta = 90^{\circ}$, we get:

$x' = 2 \cos 90^{\circ} - 3 \sin 90^{\circ} = 0 - 3 = -3$ $y' = 2 \sin 90^{\circ} + 3 \cos 90^{\circ} = 2 + 0 = 2$

So, the new coordinates of vertex $A$ after rotation are $(-3, 2)$.

Substituting $x = 5, y = 3, \theta = 90^{\circ}$, we get:

$x' = 5 \cos 90^{\circ} - 3 \sin 90^{\circ} = 0 - 3 = -3$ $y' = 5 \sin 90^{\circ} + 3 \cos 90^{\circ} = 5 + 0 = 5$

So, the new coordinates of vertex $B$ after rotation are $(-3, 5)$.

Substituting $x = 4, y = 2, \theta = 90^{\circ}$, we get:

$x' = 4 \cos 90^{\circ} - 2 \sin 90^{\circ} = 0 - 2 = -2$ $y' = 4 \sin 90^{\circ} + 2 \cos 90^{\circ} = 4 + 0 = 4$

So, the new coordinates of vertex $C$ after rotation are $(-2, 4)$.

2. To rotate the triangle with vertices $A(-3,-2), B(-2,-2)$, and $C(-3,-1)$ $90^{\circ}$ counterclockwise around the origin, we can use the rotation formulas:

$x' = x \cos \theta - y \sin \theta$ $y' = x \sin \theta + y \cos \theta$

Substituting $x = -3, y = -2, \theta = 90^{\circ}$, we get:

$x' = -3 \cos 90^{\circ} - (-2) \sin 90^{\circ} = 0 + 2 = 2$ $y' = -3 \sin 90^{\circ} + (-2) \cos 90^{\circ} = -3 + 0 = -3$

So, the new coordinates of vertex $A$ after rotation are $(2, -3)$.

Substituting $x = -2, y = -2, \theta = 90^{\circ}$, we get:

$x' = -2 \cos 90^{\circ} - (-2) \sin 90^{\circ} = 0 + 2 = 2$ $y' = -2 \sin 90^{\circ} + (-2) \cos 90^{\circ} = -2 + 0 = -2$

So, the new coordinates of vertex $B$ after rotation are $(2, -2)$.

Substituting $x = -3, y = -1, \theta = 90^{\circ}$, we get:

$x' = -3 \cos 90^{\circ} - (-1) \sin 90^{\circ} = 0 + 1 = 1$ $y' = -3 \sin 90^{\circ} + (-1) \cos 90^{\circ} = -3 + 0 = -3$

So, the new coordinates of vertex $C$ after rotation are $(1, -3)$.

Conclusion

In this article, we have explored how to rotate a triangle in the coordinate plane and find the image points of its vertices after a $90^{\circ}$ counterclockwise rotation around the origin. We have used the rotation formulas to find the new coordinates of each vertex after rotation. By following these steps, you can easily rotate any triangle in the coordinate plane and find its image points.

Introduction

In our previous article, we explored how to rotate a triangle in the coordinate plane and find the image points of its vertices after a $90^{\circ}$ counterclockwise rotation around the origin. In this article, we will answer some frequently asked questions related to rotating a triangle in the coordinate plane.

Q1: What is the center of rotation?

A1: The center of rotation is the point around which the triangle is rotated. In the case of a $90^{\circ}$ counterclockwise rotation around the origin, the center of rotation is the origin $(0, 0)$.

Q2: How do I rotate a triangle by a different angle?

A2: To rotate a triangle by a different angle, you can use the rotation formulas:

$x' = x \cos \theta - y \sin \theta$ $y' = x \sin \theta + y \cos \theta$

where $\theta$ is the angle of rotation. For example, to rotate a triangle by $180^{\circ}$, you can use $\theta = 180^{\circ}$.

Q3: Can I rotate a triangle by a negative angle?

A3: Yes, you can rotate a triangle by a negative angle. To do this, you can use the rotation formulas with a negative angle. For example, to rotate a triangle by $-90^{\circ}$, you can use $\theta = -90^{\circ}$.

Q4: How do I find the image points of a triangle after rotation?

A4: To find the image points of a triangle after rotation, you can use the rotation formulas:

$x' = x \cos \theta - y \sin \theta$ $y' = x \sin \theta + y \cos \theta$

where $(x, y)$ are the coordinates of the original point and $(x', y')$ are the coordinates of the image point.

Q5: Can I rotate a triangle by a non-integer angle?

A5: Yes, you can rotate a triangle by a non-integer angle. To do this, you can use the rotation formulas with a non-integer angle. For example, to rotate a triangle by $45.5^{\circ}$, you can use $\theta = 45.5^{\circ}$.

Q6: How do I rotate a triangle by a multiple of $90^{\circ}$?

A6: To rotate a triangle by a multiple of $90^{\circ}$, you can use the rotation formulas with a multiple of $90^{\circ}$. For example, to rotate a triangle by $270^{\circ}$, you can use $\theta = 270^{\circ}$.

Q7: Can I rotate a triangle by a negative multiple of $90^{\circ}$?

A7: Yes, you can rotate a triangle by a negative multiple of $90^{\circ}$. To do this, you can use the rotation formulas with a negative multiple of $90^{\circ}$. For example, to rotate a triangle by $-270^{\circ}$, you can use $\theta = -270^{\circ}$.

Q8: How do I find the image points of a triangle after rotation by a multiple of $90^{\circ}$?

A8: To find the image points of a triangle after rotation by a multiple of $90^{\circ}$, you can use the rotation formulas:

$x' = x \cos \theta - y \sin \theta$ $y' = x \sin \theta + y \cos \theta$

where $(x, y)$ are the coordinates of the original point and $(x', y')$ are the coordinates of the image point.

Conclusion

In this article, we have answered some frequently asked questions related to rotating a triangle in the coordinate plane. We have covered topics such as the center of rotation, rotating by a different angle, rotating by a negative angle, finding the image points, rotating by a non-integer angle, rotating by a multiple of $90^{\circ}$, and rotating by a negative multiple of $90^{\circ}$. By following these steps, you can easily rotate any triangle in the coordinate plane and find its image points.

Example Problems

1. Rotate the triangle with vertices $A(2,3), B(5,3)$, and $C(4,2)$ by $180^{\circ}$ around the origin.
2. Rotate the triangle with vertices $A(-3,-2), B(-2,-2)$, and $C(-3,-1)$ by $-90^{\circ}$ around the origin.

Solutions

1. To rotate the triangle with vertices $A(2,3), B(5,3)$, and $C(4,2)$ by $180^{\circ}$ around the origin, we can use the rotation formulas:

$x' = x \cos \theta - y \sin \theta$ $y' = x \sin \theta + y \cos \theta$

Substituting $x = 2, y = 3, \theta = 180^{\circ}$, we get:

$x' = 2 \cos 180^{\circ} - 3 \sin 180^{\circ} = -2 + 0 = -2$ $y' = 2 \sin 180^{\circ} + 3 \cos 180^{\circ} = 0 - 3 = -3$

So, the new coordinates of vertex $A$ after rotation are $(-2, -3)$.

Substituting $x = 5, y = 3, \theta = 180^{\circ}$, we get:

$x' = 5 \cos 180^{\circ} - 3 \sin 180^{\circ} = -5 + 0 = -5$ $y' = 5 \sin 180^{\circ} + 3 \cos 180^{\circ} = 0 - 3 = -3$

So, the new coordinates of vertex $B$ after rotation are $(-5, -3)$.

Substituting $x = 4, y = 2, \theta = 180^{\circ}$, we get:

$x' = 4 \cos 180^{\circ} - 2 \sin 180^{\circ} = -4 + 0 = -4$ $y' = 4 \sin 180^{\circ} + 2 \cos 180^{\circ} = 0 - 2 = -2$

So, the new coordinates of vertex $C$ after rotation are $(-4, -2)$.

2. To rotate the triangle with vertices $A(-3,-2), B(-2,-2)$, and $C(-3,-1)$ by $-90^{\circ}$ around the origin, we can use the rotation formulas:

$x' = x \cos \theta - y \sin \theta$ $y' = x \sin \theta + y \cos \theta$

Substituting $x = -3, y = -2, \theta = -90^{\circ}$, we get:

$x' = -3 \cos (-90^{\circ}) - (-2) \sin (-90^{\circ}) = 0 + 2 = 2$ $y' = -3 \sin (-90^{\circ}) + (-2) \cos (-90^{\circ}) = 3 + 0 = 3$

So, the new coordinates of vertex $A$ after rotation are $(2, 3)$.

Substituting $x = -2, y = -2, \theta = -90^{\circ}$, we get:

$x' = -2 \cos (-90^{\circ}) - (-2) \sin (-90^{\circ}) = 0 + 2 = 2$ $y' = -2 \sin (-90^{\circ}) + (-2) \cos (-90^{\circ}) = 2 + 0 = 2$

So, the new coordinates of vertex $B$ after rotation are $(2, 2)$.

Substituting $x = -3, y = -1, \theta = -90^{\circ}$, we get:

$x' = -3 \cos (-90^{\circ}) - (-1) \sin (-90^{\circ}) = 0 + 1 = 1$ $y' = -3 \sin (-90^{\circ}) + (-1) \cos (-90^{\circ}) = 3 + 0 = 3$

So, the new coordinates of vertex $C$ after rotation are $(1, 3)$.

Conclusion

In this article, we have answered some frequently asked questions related to rotating a triangle in the coordinate plane. We have covered topics such as the center of rotation, rotating by a different angle, rotating by a negative angle, finding the image points, rotating by a non-integer angle, rotating by a multiple of $90^{\circ}$, and rotating by a negative multiple of $90^{\circ}$. By following these steps, you can easily rotate any triangle in the coordinate plane and find its image points.