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Lesson 15 ~45 min Linear Relationships C · Path +95 XP

Rotations About the Origin

A rotation needs three things: a centre, an angle and a direction. For quarter turns about the origin the coordinate rules are short, and each one can be rebuilt from the picture rather than memorised in isolation.

Today's hook: Rotate $(4, 1)$ a quarter turn anticlockwise about the origin. The answer is $(-1, 4)$. The coordinates have swapped and one sign has changed, and knowing which one is the whole of this lesson.
0/5QUESTS
Think First
warm-up

Plot $(4, 0)$ and turn it a quarter turn anticlockwise about the origin. Where does it land? Now do the same for $(0, 3)$. What is happening to the pair of numbers?

Record your answer in your workbook.
1
The Big Idea
+5 XP to read

A quarter turn anticlockwise sends $(x, y)$ to $(-y, x)$: swap, then negate the new x. A half turn sends it to $(-x, -y)$. A quarter turn clockwise sends it to $(y, -x)$: swap, then negate the new y.

$$(x, y) \to (-y, x) \to (-x, -y) \to (y, -x)$$

Every one of these rules swaps the coordinates and changes a sign, except the half turn, which changes both signs and swaps nothing. If you can remember where $(1, 0)$ goes, you can rebuild any of them in seconds.

P(x, y) P′(−y, x) 90° O same distance, quarter turn round
$(-y, x), \ (-x, -y), \ (y, -x)$
Anticlockwise is the default
Unless a question says clockwise, a positive rotation turns anticlockwise, the same convention as the unit circle.
Test with a simple point
Track $(1, 0)$ through the turn. If your rule sends it to the right place, the rule is right.
The distance must survive
A rotation about the origin cannot change how far a point is from the origin. Check it when unsure.
2
What You'll Master
objectives

Know

  • that a rotation of $90°$ anticlockwise about the origin sends $P(x, y)$ to $P'(-y, x)$
  • that a rotation of $180°$ about the origin sends $P(x, y)$ to $P'(-x, -y)$, in either direction
  • that a rotation of $90°$ clockwise, equivalently $270°$ anticlockwise, sends $P(x, y)$ to $P'(y, -x)$

Understand

  • why a rotation about the origin leaves the distance from the origin unchanged
  • why the origin is the only invariant point of a rotation about the origin, other than for the full turn
  • why a rotation preserves orientation while a reflection reverses it

Can Do

  • find the image of a point or shape under a rotation of a multiple of $90°$ about the origin
  • convert between clockwise and anticlockwise descriptions of the same rotation
  • identify which rotation maps a given point onto a given image, and check it using distances
3
Words You Need
vocabulary
RotationA transformation that turns every point about a fixed centre through a fixed angle.
Centre of rotationThe fixed point everything turns about. Here it is always the origin.
AnticlockwiseThe positive direction of rotation, opposite to the movement of clock hands.
Quarter turnA rotation of $90°$. Four of them make a full revolution.
OrientationThe sense in which vertices are labelled. A rotation preserves it; a reflection reverses it.
4
What a Rotation Needs
+5 XP to read

A rotation is not fully described until three things are given: the centre, the angle and the direction.

In this lesson the centre is always the origin and the angle is always a multiple of $90°$, which keeps the arithmetic to sign changes. The direction still has to be stated, except for a half turn, where clockwise and anticlockwise give the same result.

The convention matches the unit circle: anticlockwise is positive. A question saying "rotate $90°$ about the origin" without further qualification means anticlockwise.

A rotation preserves lengths, angles and orientation. The image is congruent and faces the same rotational way round, which is what separates a rotation from a reflection.

5
Building the Quarter-Turn Rule
+5 XP to read

Rather than memorising the rule, watch two easy points.

The point $(1, 0)$ sits on the positive x-axis. A quarter turn anticlockwise takes it to the positive y-axis, at $(0, 1)$.

The point $(0, 1)$ sits on the positive y-axis. The same quarter turn takes it to the negative x-axis, at $(-1, 0)$.

So the x-direction becomes the y-direction, and the y-direction becomes the negative x-direction. A general point $(x, y)$ is $x$ steps in the first direction and $y$ steps in the second, so its image is $x$ steps up and $y$ steps left:

$P(x, y) \to P'(-y, x)$

Two points you can picture generate the rule, which is more reliable than a memorised string of symbols with a sign that may have drifted.

6
The Other Two Rules
+5 XP to read

Apply the quarter turn twice to get the half turn:

$(x, y) \to (-y, x) \to (-x, -y)$

So a rotation of $180°$ negates both coordinates. Turning the other way gives the same answer, which is why direction need not be stated for a half turn.

Apply it three times for the three-quarter turn:

$(-x, -y) \to (y, -x)$

A rotation of $270°$ anticlockwise is the same as $90°$ clockwise, so both are described by $(x, y) \to (y, -x)$.

Applying it a fourth time returns $(x, y)$, which confirms the chain closes correctly. If your four rules do not return you to the start, one of them has a sign wrong.

7
Clockwise and Anticlockwise
+5 XP to read

Any clockwise rotation can be renamed as an anticlockwise one by subtracting from $360°$.

$90°$ clockwise is $270°$ anticlockwise. $270°$ clockwise is $90°$ anticlockwise. $180°$ in either direction is the same transformation.

This matters because a question may state one and an answer key the other, and they are not disagreeing. Converting to a single convention before comparing avoids a false alarm.

The two quarter turns are easy to confuse, and swapping them is the usual error. The check takes seconds: track $(1, 0)$. Anticlockwise sends it up to $(0, 1)$; clockwise sends it down to $(0, -1)$.

8
Checking a Rotation
+5 XP to read

Two checks catch nearly every mistake.

The distance from the origin cannot change. Rotating $(4, 1)$ gives $(-1, 4)$, and both are $\sqrt{17}$ from the origin. If your image is a different distance away, the rule was applied wrongly.

For a quarter turn, the two radii are perpendicular. The gradient of $OP$ is $\tfrac{1}{4}$ and the gradient of $OP'$ is $\tfrac{4}{-1} = -4$, and their product is $-1$. This is the same negative-reciprocal fact from Lesson 8, appearing here as a consequence of turning through a right angle.

Notice also that the origin is the only point that does not move, since every other point is at a non-zero distance and gets carried round a circle. A rotation has exactly one invariant point, a reflection has a whole line of them, and a translation has none.

Watch Me Solve It · A quarter turn anticlockwise
+15 XP per step
Q1
PROBLEM
Find the image of $A(4, 1)$ after a rotation of $90°$ anticlockwise about the origin, and verify the answer.
  1. 1
    Apply the rule
    $(x, y) \to (-y, x)$
    $A(4, 1) \to A'(-1, 4)$
    Swap the coordinates to get $(1, 4)$, then negate the new first coordinate.
  2. 2
    Check the distance from the origin
    $OA = \sqrt{16 + 1} = \sqrt{17}$
    $OA' = \sqrt{1 + 16} = \sqrt{17}$
    Equal, as a rotation about the origin requires.
  3. 3
    Check the right angle
    $m_{OA} = \tfrac{1}{4}, \qquad m_{OA'} = \tfrac{4}{-1} = -4$
    $\tfrac{1}{4} \times (-4) = -1$
    The two radii are perpendicular, which confirms the turn was a quarter and not something else.
  4. 4
    Check the direction
    $A$ is in the first quadrant and $A'$ is in the second, which is the anticlockwise direction. Clockwise would have sent it to the fourth quadrant.
Answer$A'(-1, 4)$
Watch Me Solve It · A half turn on a triangle
+15 XP per step
Q2
PROBLEM
The triangle $PQR$ has $P(2, 3)$, $Q(-1, 5)$ and $R(4, -2)$. Find its image after a rotation of $180°$ about the origin.
  1. 1
    State the rule
    $(x, y) \to (-x, -y)$
    Both coordinates change sign. The direction need not be specified for a half turn.
  2. 2
    Apply it to each vertex
    $P(2, 3) \to P'(-2, -3)$
    $Q(-1, 5) \to Q'(1, -5)$
    $R(4, -2) \to R'(-4, 2)$
    Take care with the vertices that already have a negative coordinate.
  3. 3
    Check one distance
    $OQ = \sqrt{1 + 25} = \sqrt{26}$
    $OQ' = \sqrt{1 + 25} = \sqrt{26}$
    Unchanged, as required.
  4. 4
    Note what is preserved
    The image is congruent to the original and has the same orientation, since a rotation does not reverse it. Only the origin is invariant, and no vertex is at the origin here.
Answer$P'(-2, -3)$, $Q'(1, -5)$, $R'(-4, 2)$
Watch Me Solve It · A clockwise quarter turn
+15 XP per step
Q3
PROBLEM
Find the image of $B(-3, 5)$ after a rotation of $90°$ clockwise about the origin, and state the equivalent anticlockwise rotation.
  1. 1
    Recall the clockwise rule
    $(x, y) \to (y, -x)$
    Swap the coordinates, then negate the new second one. Checking on $(1, 0)$: it maps to $(0, -1)$, which is downwards, so the rule is indeed clockwise.
  2. 2
    Apply it
    $B(-3, 5) \to B'(5, 3)$
    Swapping gives $(5, -3)$, and negating the new second coordinate gives $(5, 3)$.
  3. 3
    Check the distance
    $OB = \sqrt{9 + 25} = \sqrt{34}$
    $OB' = \sqrt{25 + 9} = \sqrt{34}$
    Equal, as required.
  4. 4
    Give the equivalent
    $360° - 90° = 270°$
    A rotation of $90°$ clockwise is the same transformation as $270°$ anticlockwise.
Answer$B'(5, 3)$; equivalent to $270°$ anticlockwise
D
Brain Trainer · Swap, then fix the sign
5 problems

Five items. All rotations are about the origin, and anticlockwise unless stated otherwise.

  1. 1 Rotate $(2, 7)$ by $90°$ anticlockwise.

    $(x, y) \to (-y, x)$.$(-7, 2)$
  2. 2 Rotate $(2, 7)$ by $180°$.

    Negate both coordinates.$(-2, -7)$
  3. 3 Rotate $(2, 7)$ by $90°$ clockwise.

    $(x, y) \to (y, -x)$.$(7, -2)$
  4. 4 Rotate $(-5, -1)$ by $90°$ anticlockwise.

    Swap to $(-1, -5)$, then negate the first.$(1, -5)$
  5. 5 Which point is invariant under a rotation of $90°$ about the origin?

    Every other point is carried round a circle of non-zero radius.Only the origin
Complete in your workbook.
MC1
The anticlockwise quarter turn
+10 XP

Rotated $90°$ anticlockwise about the origin, the point $(6, 2)$ maps to:

MC2
Naming the rotation
+10 XP

A rotation about the origin maps $(3, -8)$ to $(-3, 8)$. It is:

MC3
Converting the direction
+10 XP

A rotation of $270°$ clockwise about the origin is the same as:

MC4
The check that catches errors
+10 XP

A student rotates $(5, 2)$ about the origin and obtains $(2, 5)$. This must be wrong as a quarter turn because:

MC5
Invariant points
+10 XP

A rotation of $90°$ about the origin has:

Q6
Rotate and verify
+15 XP
Q6
SHORT ANSWER
The point $A(-6, 2)$ is rotated $90°$ anticlockwise about the origin to give $A'$. Find $A'$, and verify your answer with two independent checks.
Write your working in your book.
Q7
Rotate a shape
+15 XP
Q7
SHORT ANSWER
The triangle $LMN$ has $L(1, 4)$, $M(5, 2)$ and $N(3, -1)$. Find the image after a rotation of $90°$ clockwise about the origin, and state which properties of the triangle are unchanged.
Write your working in your book.
Q8
Tell the three transformations apart
+15 XP
Q8
SHORT ANSWER
A single transformation maps $(4, 3)$ to $(-3, 4)$. Determine which transformation it is, and explain how the number of invariant points distinguishes rotations, reflections and translations from one another.
Write your working in your book.
S
Stretch Challenge · Rotating a line, and a rotation that is not about the origin
+25 XP
S
CHALLENGE
Find the equation of the image of the line $y = 2x + 3$ after a rotation of $90°$ anticlockwise about the origin, and explain how the gradient of the image relates to the original. Then work out the rule for a rotation of $180°$ about a general point $(p, q)$ rather than the origin.
R
Quick Review
recap

Three rules, all swaps and signs

$90°$ anticlockwise gives $(-y, x)$; $180°$ gives $(-x, -y)$; $90°$ clockwise, or $270°$ anticlockwise, gives $(y, -x)$.

Rebuild rather than memorise

Track $(1, 0)$ and $(0, 1)$ through the turn. Anticlockwise sends $(1, 0)$ up to $(0, 1)$; clockwise sends it down to $(0, -1)$.

Two checks catch most errors

The distance from the origin must be unchanged, and for a quarter turn the two radii must have gradients multiplying to $-1$.

Invariant points identify the type

A translation has none, a rotation has exactly one, and a reflection has a whole line of them. Counting them names the transformation.

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