If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

High school geometry

Course: High school geometry > Unit 1

Lesson 4: Rotations
Math
High school geometry
Performing transformations
Rotations
© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

Determining rotations

Google Classroom
Learn how to determine which rotation brings one given shape to another given shape.
There are two properties of every rotation—the center and the angle.

Determining the center of rotation

Rotations preserve distance, so the center of rotation must be equidistant from point P and its image P′. That means the center of rotation must be on the perpendicular bisector of PP′―.
Point P and point P prime are equidistant from the line of reflection. There are 4 points marked on the line of reflection. The segment from P to each marked point is congruent to the corresponding segment from point P prime to the same point.
If we took the segments that connected each point of the image to the corresponding point in the pre-image, the center of rotation is at the intersection of the perpendicular bisectors of all of those segments.

Example

Let's find the center of rotation that maps △ABC to △A′B′C′.
A pre image triangle A B C and its image A prime B prime C prime. For the pre image triangle, Vertex A is at seven o clock, Vertex B is at about twelve o click, and Vertex C is at about three o clock. For the image triangle, vertex A prime is at two o clock, vertex B prime is at seven o clock, and vertex c prime is located at ten o clock.
The center of rotation must be on the perpendicular bisector of AA′―
A pre image triangle A B C and its image A prime B prime C prime. For the pre image triangle, Vertex A is at seven o clock, Vertex B is at about twelve o click, and Vertex C is at about three o clock. For the image triangle, vertex A prime is at two o clock, vertex B prime is at seven o clock, and vertex c prime is located at ten o clock. A solid line segment has endpoints from vertex A to vertex A prime. A dashed line bisects the solid line making a ninety degree angle.
The center of rotation also must be on the perpendicular bisector of BB′―.
A pre image triangle A B C and its image A prime B prime C prime. For the pre image triangle, Vertex A is at seven o clock, Vertex B is at about twelve o click, and Vertex C is at about three o clock. For the image triangle, vertex A prime is at two o clock, vertex B prime is at seven o clock, and vertex c prime is located at ten o clock. A solid line segment has endpoints from vertex A to vertex A prime. A dashed line bisects the solid line making a ninety degree angle. Another solid line segment has endpoints at vertex B and vertex B prime. A dashed line bisects this solid line segment at a ninety degree angle. A point is placed where the two dashed lines intersect.
We could also check the perpendicular bisector of CC′―, but we don't need to. Since all of the bisectors intersect at the same point, checking two is enough.
A pre image triangle A B C and its image A prime B prime C prime. For the pre image triangle, Vertex A is at seven o clock, Vertex B is at about twelve o click, and Vertex C is at about three o clock. For the image triangle, vertex A prime is at two o clock, vertex B prime is at seven o clock, and vertex c prime is located at ten o clock. A solid line segment has endpoints from vertex A to vertex A prime. A dashed line bisects the solid line making a ninety degree angle. Another solid line segment has endpoints at vertex B and vertex B prime. A dashed line bisects this solid line segment at a ninety degree angle. A third solid line segment has endpoints at vertex C and vertex C prime. A third dashed line bisects this line segment at a ninety degree angle. A point is placed where the three dashed lines intersect.

Let's try it!

Problem 1.1
△A′B′C′ is the image of △ABC after a rotation.
Triangle ABC is rotated to form triangle A prime, B prime, C prime. Point N is closer to point B than point B prime. Point P is closer to point A prime than point A. Point M is closer to point A prime than point A, but not as close as point P. Point Q is equidistant from point B and point B prime.
Which point is the center of rotation?
Choose 1 answer:

Determining angle of rotation

Once we have found the center of rotation, we have several options for determining the angle of the rotation.
Finally, we need to determine whether the rotation is counterclockwise, with a positive angle of rotation, or clockwise, with a negative angle of rotation.

Example

Let's estimate the angle of rotation that maps △ABC to △A′B′C′ about point P.
Triangle ABC is rotated about point P to form triangle A prime, B prime, C prime. Point B is at about twelve o'clock relative to point P. Point B prime is at about seven o'clock relative to point P.
We can compare m∠APA′ to benchmark angles.
Triangle ABC is rotated about point P to form triangle A prime, B prime, C prime. Both of the triangles are drawn on a circular compass. Angle A P A prime is labeled and the angle is closer to one eighty degrees in measure than ninety degrees.
The angle measure is a little closer to 180° than to 90°. We could split the circle into more equal parts to get a closer estimate.
Triangle ABC is rotated about point P to form triangle A prime, B prime, C prime. Both of the triangles are drawn on a circular compass. Angle A P A prime is labeled and the angle is about halfway between one hundred thirty-five degrees and one hundred eighty degrees.
We might estimate that the angle is around 150° to 160°, but we'd have to measure to be sure.
We also could have measured clockwise, but then we would need to use a negative angle measure. We go a little more than a half turn clockwise, so we could estimate the angle measure to be around −200°.

Let's try it!

Problem 2.1
Triangle △A′B′C′ is the image of △ABC under a rotation about point P.
The pre image of a triangle A B C and the image of a triangle A prime B prime C prime. The pre image triangle is above the image triangle. The vertex A is at seven o clock. The vertex B is at ten o clock. The vertex C is at five o clock. Vertex A prime is at four o clock. B prime is at seven o clock. C prime is at two o clock.
What is the best estimate of the angle of rotation?
Choose 1 answer:

Want to join the conversation?

Log in
  • aqualine ultimate style avatar for user amxw
    Posted 8 years ago. Direct link to amxw's post “why are positive rotation...”
    why are positive rotations counter clockwise?
    (86 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user eiloncohen
      Posted 8 years ago. Direct link to eiloncohen's post “The reason for this actua...”
      Incredible Answer
      Great Answer
      Good Answer
      The reason for this actually comes from trigonometry. The representation of angles on the coordinate plane is called Standard Position. As regular angles are made up of two rays, so are angles in standard position. The rays in standard position angles, however, have specific names; the initial side, which always stays on the positive x-axis (to the right of the origin), and the terminal side, which rotates about the origin to create the angle. If the angle is positive, the terminal side rotates counter clockwise, and if the angle is negative, the terminal side rotates clockwise.

      For example, if the terminal side was on the the positive y-axis (above the origin), then the angle made would be 90 degrees, because the terminal side rotated 90 degrees counter clockwise. Hope this helps!
      (135 votes)
  • blobby green style avatar for user jami5693
    Posted 5 years ago. Direct link to jami5693's post “How will this help me out...”
    How will this help me out with daily life
    (44 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user 1033458
    Posted 4 years ago. Direct link to 1033458's post “None of this makes any se...”
    None of this makes any sense
    (27 votes)
    Default Khan Academy avatar avatar for user
  • piceratops ultimate style avatar for user greenneyeddragonn
    Posted 3 years ago. Direct link to greenneyeddragonn's post “Alright, I am writing thi...”
    Alright, I am writing this for guys from the future who will come for help here in the comments...So it is said that the center of rotation is always some point on the perpendicular bisector of the two points that we want to rotate.

    But what exactly is the perpendicular bisector? It is a line that intersects a segment, splits it in two halves and creates a right angle. That is cool but why is it useful? It is useful because EVERY POINT ON THIS LINE (the perpendicular bisector)is at the same distance from the two end points of the segment. So the CENTER, CENTER, CENTER of rotation must be somewhere on this perpendicular bisector because IT'S THE CENTER of rotation meaning that it is equidistant from the starting point and the ending point. Cheers!
    (25 votes)
    Default Khan Academy avatar avatar for user
  • stelly yellow style avatar for user Lizzy.Pryor
    Posted 7 months ago. Direct link to Lizzy.Pryor's post “I just want to go home”
    I just want to go home
    (15 votes)
    Default Khan Academy avatar avatar for user
  • aqualine tree style avatar for user max hawley
    Posted 8 years ago. Direct link to max hawley's post “how would you go about do...”
    how would you go about doing this on graph paper
    (14 votes)
    Default Khan Academy avatar avatar for user
  • ohnoes default style avatar for user incandescentVirtuoso
    Posted 8 years ago. Direct link to incandescentVirtuoso's post “I'm still a bit unsure as...”
    I'm still a bit unsure as to how I am supposed to find the degree of rotation? The article is helpful but I don't feel like I should rely on the Khan Academy tool to figure out a rotation? Are there any videos/articles that explain how to do this without relying on the rotation tool? Sorry if this question is confusing!
    (14 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user vijayhm
      Posted 7 years ago. Direct link to vijayhm's post “If you wanted to find the...”
      If you wanted to find the angle of rotation on graph paper you would:

      1) Draw a line segment from one point on the original shape (let's call it Point A) to the center of rotation.

      2) Draw a line segment from the image of Point A (Point A') to the center of rotation.

      3) Measure the angle that they form using a protractor.

      There's no formula to figure it out as far as I know of. You would either use a tool on a graphing calculator similar to the Khan Academy tool or the method I described above for solving on graph paper.
      (8 votes)
  • starky seedling style avatar for user marlene brianna gutierrez
    Posted 6 years ago. Direct link to marlene brianna gutierrez's post “Why are positive notation...”
    Why are positive notations countered clockwise and why are negative notations clockwise?
    (14 votes)
    Default Khan Academy avatar avatar for user
    • primosaur ultimate style avatar for user Max Bentley
      Posted 3 years ago. Direct link to Max Bentley's post “copying from another comm...”
      copying from another comment (I did not write this, @eiloncohen did):
      The reason for this actually comes from trigonometry. The representation of angles on the coordinate plane is called Standard Position. As regular angles are made up of two rays, so are angles in standard position. The rays in standard position angles, however, have specific names; the initial side, which always stays on the positive x-axis (to the right of the origin), and the terminal side, which rotates about the origin to create the angle. If the angle is positive, the terminal side rotates counter clockwise, and if the angle is negative, the terminal side rotates clockwise.

      For example, if the terminal side was on the the positive y-axis (above the origin), then the angle made would be 90 degrees, because the terminal side rotated 90 degrees counter clockwise. Hope this helps!
      (1 vote)
  • blobby green style avatar for user Kezyah Roberson
    Posted 7 years ago. Direct link to Kezyah Roberson's post “how do you know a close e...”
    how do you know a close estimate?
    (14 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user garciaa0860
    Posted 6 months ago. Direct link to garciaa0860's post “confusing very confusing ...”
    confusing very confusing i think i might explode
    (11 votes)
    Default Khan Academy avatar avatar for user