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Learning Objectives

  • Find the image of a point in a rotation in a coordinate plane.
  • Recognize that a rotation is an isometry.

Sample Rotations

In this lesson we will study rotations centered at the origin of a coordinate plane. We begin with some specific examples of rotations. Later we will see how these rotations fit into a general formula.

We define a rotation as follows: In a rotation centered at the origin with an angle of rotation of n^\circ, a point moves counterclockwise along an arc of a circle. The central angle of the circle measures n^\circ.

The original preimage point is one endpoint of the arc, and the image of the original point is the other endpoint of the arc:

  • Rotations centered at the origin move points ______________________ along an arc of a circle.
  • For a rotation of n^\circ, the central angle of the circle measures ____________.
  • The preimage point is one endpoint of the _________________ and the image is the other endpoint.

180^\circ Rotation

Our first example is rotation through an angle of 180^\circ:

In a 180^\circ rotation, the image of P (h, k) is the point P^\prime (-h, -k).

Notice:

  • P and P^\prime are the endpoints of a diameter of a circle.

\rightarrow This means that the distance from the point P to the origin (or the distance from the point P^\prime to the origin) is a radius of the circle.

The distance from P to the origin equals the distance from ________ to the origin.

  • The rotation is the same as a “reflection in the origin.”

\rightarrowThis means that if we use the origin as a mirror, the point P is directly across from the point P^\prime.

A 180^\circ _____________________________ is also a reflection in the origin.

In a rotation of 180^\circ, the x-coordinate and the y-coordinate of the __________________ become the negative versions of the values in the image.

A 180^\circ rotation is an isometry. The image of a segment is a congruent segment:

Use the Distance Formula:

PQ &= \sqrt{(k-t)^2  + (h-r)^2}\\P^\prime Q^\prime & = \sqrt{(-k-(-t))^2  + (-h-(-r))^2} = \sqrt{(-k+t)^2  + (-h+r)^2}\\& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad =  \sqrt{(t-k)^2  + (r-h)^2}\\& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad = \sqrt{(k-t)^2 + (h-r)^2}

So PQ = P^\prime Q^\prime

  • A 180^\circ rotation is an ___________________________, so distance is preserved.
  • When a segment is rotated 180^\circ (or reflected in the origin), its image is a _________________________________ segment.

90^\circ Rotation

The next example is a rotation through an angle of 90^\circ. The rotation is in the counterclockwise direction:

In a 90^\circ rotation, the image of P (h, k) is the point P^\prime (-k, h).

Notice:

  • \overline{PO} and \overline{P^\prime O} are both radii of the same circle, so PO = P^\prime O.

If PO and P^\prime O are both radii, then they are the same ______________________.

  • \angle POP^\prime is a right angle.
  • The acute angle formed by \overline{PO} and the x-axis and the acute angle formed by \overline{P^\prime O} and the x-axis are complementary angles.

Remember, complementary angles add up to ________________^\circ.

You can see by the coordinates of the preimage and image points, in a 90^\circ rotation:

  • the x- and y-coordinates are switched AND
  • the x-coordinate is negative.

In a 90^\circ rotation, switch the _________- and _________-coordinates and make the new x-coordinate _________________________. 

A 90^\circ rotation is an isometry. The image of a segment is a congruent segment.

Use the Distance Formula:

PQ &= \sqrt{(k-t)^2  + (h-r)^2}\\P^\prime Q^\prime &= \sqrt{(h-r)^2  + (-k-(-t))^2} = \sqrt{(h-r)^2  + (t-k)^2}\\&\qquad \qquad \qquad \qquad \qquad \qquad \quad = \sqrt{(k-t)^2 + (h-r)^2}

So PQ = P^\prime Q^\prime

Reading Check:

Which of the following are isometries? Circle all that apply:

& 30^\circ rotation && 45^\circ rotation && 60^\circ rotation\\& 90^\circ rotation && 150^\circ rotation	&& 180^\circ rotation\\& Reflection && Translation && Bisection

Example 1

What are the coordinates of the vertices of \Delta ABC in a rotation of 90^\circ?

Point A is (4, 6), B is (–4, 2), and C is (6, –2).

In a 90^\circ rotation, the x-coordinate and the y-coordinate are switched AND the new x-coordinate is made negative:

  • A becomes A^\prime : switch x and y to (6, 4) and make x negative (–6, 4)
  • B becomes B^\prime : switch x and y to (2, –4) and make x negative (–2, –4)
  • C becomes C^\prime : switch x and y to (–2, 6) and make x negative (- (-2), 6) = (2, 6)

So the vertices of \Delta A^\prime B^\prime C^\prime are (–6, 4), (–2, –4), and (2, 6).

Plot each of these points on the coordinate plane above and draw in each side of the new rotated triangle. Can you see how \Delta ABC is rotated 90^\circ to \Delta A^\prime B^\prime C^\prime?

Reading Check:

1. True or false: A rotation is always in the counterclockwise direction.

2. On the coordinate plane below, create a point anywhere you like, and label it P.

Then draw a second point W that is the image of point P rotated 180^\circ.

3. On the coordinate plane above, draw a third point R that is the image of your original point P rotated 90^\circ.

4. Is a 90^\circ rotation an isometry? Explain.

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5. Is a 180^\circ rotation an isometry? Explain.

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6. What type of rotation is the same as a reflection in the origin?

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