Try This First
Try this short activity before reading the lesson.
> Draw axes on graph paper and plot $(3,1)$. Now physically turn the page a quarter turn counterclockwise and read off where that point sits against the axes you drew. Turn it back. Fold the paper along the $y$-axis and find where $(3,1)$ lands. Write both images down before you read a single rule.
The Big Idea
A rigid motion is any move of the plane that keeps every distance: slide, flip, turn. Two figures are congruent exactly when one rigid motion carries the first onto the second, so the whole of congruence sits on top of these three moves. That is why they come first.
Coordinates turn each move into a rule you can apply without a picture. A translation adds a fixed pair. A reflection in an axis flips one sign. A rotation about the origin by a quarter turn swaps the coordinates and flips one sign. Every rule can be checked on a single easy point such as $(1,0)$, and you should always check.
Doing one motion after another is a composition. The second motion acts on the image of the first, never on the original. Slides and turns keep the figure's orientation, its clockwise or counterclockwise sense. A flip reverses it. Count the flips and you know the orientation of the final image before you compute a thing.
A rigid motion changes no distance, so two figures are congruent exactly when one rigid motion carries one onto the other. You will write translations, reflections, and rotations as coordinate rules, compose two of them into one rule, and say which motions keep orientation.
Words to Know
- rigid motion: A move of the plane that keeps every length and every angle. Translations, reflections, and rotations, and anything built by chaining them.
- image: Where a point or figure lands after a motion. The image of $P$ is written $P'$.
- composition: One motion followed by another. The second motion acts on the image of the first.
- orientation: The direction, clockwise or counterclockwise, in which the labeled vertices go around. Reflections reverse it; slides and turns keep it.
Formulas and Theorems
- Rotate 90° CCW: $(x,y)\mapsto(-y,\,x)$
- Rotate 180°: $(x,y)\mapsto(-x,\,-y)$
- Reflect in y = −x: $(x,y)\mapsto(-y,\,-x)$
- Translate by (a, b): $(x,y)\mapsto(x+a,\,y+b)$
Worked Examples and Your Turn
Study each example, then try the paired problem.
Example 1: A quarter turn as a rule
Problem: Write the coordinate rule for a rotation of $90^\circ$ counterclockwise about the origin, and apply it to $(2,5)$.
Solution:
- Watch one easy point. $(1,0)$ on the positive $x$-axis turns a quarter turn counterclockwise to $(0,1)$ on the positive $y$-axis. Why: a rule you can picture on one point is a rule you can trust.
- Watch a second point. $(0,1)$ turns to $(-1,0)$. So the old $x$ lands in the $y$ slot, and the old $y$ lands in the $x$ slot with its sign changed: the rule is $(x,y)\mapsto(-y,x)$. Why: two points on the axes pin down a rotation completely.
- Apply it: $(2,5)\mapsto(-5,2)$. Why: the rule does the turning; you just substitute.
Answer: $(x,y)\mapsto(-y,x)$, and $(2,5)$ goes to $(-5,2)$.
> Verification & Check: Both points are $\sqrt{29}$ from the origin, and $(2,5)$ was in the first quadrant, so a quarter turn counterclockwise should land in the second. $(-5,2)$ does.
Your Turn 1: Rotate $(-3,4)$ by $90^\circ$ counterclockwise about the origin.
Example 2: Two motions, one rule
Problem: Reflect $(2,5)$ in the $y$-axis, then rotate the image $90^\circ$ counterclockwise about the origin. Give the final point and the single rule for the whole composition.
Solution:
- First motion: reflection in the $y$-axis flips the sign of $x$: $(2,5)\mapsto(-2,5)$. Why: a point and its mirror image sit the same distance from the mirror, on opposite sides.
- Second motion acts on $(-2,5)$, not on $(2,5)$: $(-2,5)\mapsto(-5,-2)$. Why: composition is a chain; each link takes the previous output.
- For the single rule, chain the rules symbolically: $(x,y)\mapsto(-x,y)\mapsto(-y,-x)$. Why: the rotation rule $(a,b)\mapsto(-b,a)$ applied to $(a,b)=(-x,y)$ gives $(-y,-x)$.
Answer: $(-5,-2)$, and the composition is $(x,y)\mapsto(-y,-x)$, which is the reflection in the line $y=-x$.
> Notice: One reflection and one rotation: one flip. The result must reverse orientation, and it is a reflection, exactly as the flip count predicts.
Your Turn 2: Apply the same composition to $(3,1)$ and check that the single rule agrees.
Example 3: Which motion did it
Problem: Triangle $ABC$ has $A(1,2)$, $B(4,2)$, $C(1,6)$. Its image has $A'(2,1)$, $B'(2,4)$, $C'(6,1)$. Name the single rigid motion.
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Solution:
- Compare coordinates: every image has the coordinates of the original swapped. Why: a pattern in all three pairs is the rule.
- $(x,y)\mapsto(y,x)$ is the reflection in the line $y=x$. Why: points on that line are fixed by the swap, and a point and its swap sit symmetrically across it.
- Confirm with orientation. $A\to B\to C$ goes right then up, counterclockwise. $A'\to B'\to C'$ goes up then right, clockwise. Reversed, so a reflection fits. Why: a rotation could never reverse the sense.
Answer: Reflection in the line $y=x$.
Your Turn 3: The image of $P(3,-1)$ is $P'(1,3)$, and $O$ stays fixed. Which quarter turn about the origin was it?
Example 4: Clockwise or counterclockwise
Problem: Rotate $(4,1)$ by $90^\circ$ clockwise about the origin.
Solution:
- The wrong road: reach for the familiar rule $(x,y)\mapsto(-y,x)$ and write $(-1,4)$. Why it fails: that rule is the counterclockwise turn. Test it on $(1,0)$: it goes to $(0,1)$, which is up, and a clockwise turn of $(1,0)$ must go down.
- Build the clockwise rule from $(1,0)\mapsto(0,-1)$ and $(0,1)\mapsto(1,0)$: $(x,y)\mapsto(y,-x)$. Why: two axis points fix the rotation.
- Apply: $(4,1)\mapsto(1,-4)$. Why: $(4,1)$ is in the first quadrant; turning clockwise takes it to the fourth, and $(1,-4)$ is there.
Answer: $(1,-4)$. The counterclockwise rule would have given $(-1,4)$, on the wrong side of the plane.
> Notice: Never trust a remembered rule; trust $(1,0)$. Ten seconds on one axis point settles the direction every time.
Your Turn 4: Rotate $(-2,-3)$ by $270^\circ$ counterclockwise about the origin.
Example 5: Turning about a point that is not the origin
Problem: Rotate $(5,2)$ by $90^\circ$ counterclockwise about the point $(1,1)$.
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The chapter continues in the book.