Inverse and composite transformations in 2D

Unit 12 Inverse and composite transformations in 2D

Key unit competence: Execute combinations of transformations in solving geometric problems involving inverse and composite transformations of shapes.

Key ideas

• A composite transformation $g\circ f$ applies $f$ first, then $g$. In general the order matters: $g\circ f\ne f\circ g$.
• Two translations by $\vec{u}$ then $\vec{v}$ give one translation by $\vec{u}+\vec{v}$ (here order does not matter).
• Two central symmetries with centres $C_1$ then $C_2$ give a translation by $2\overrightarrow{C_1{C}_2}$.
• Two reflections in parallel lines give a translation (twice the distance between the lines, perpendicular to them); in lines meeting at angle $\alpha$ they give a rotation of $2\alpha$ about the meeting point.
• Two rotations about the same centre: angles add (${\theta }_1+{\theta }_2$).
• The inverse $f^{-1}$ undoes $f$: translation by $\vec{v}$ → translation by $-\vec{v}$; rotation of $\theta$ → rotation of $-\theta$ about the same centre; enlargement $k$ → enlargement $\frac{1}{k}$ (same centre); reflections and central symmetries are their own inverses.

Formulas to know

Common misconceptions

• "$g\circ f$ means do $g$ first." No: $g\circ f(P)=g(f(P))$, so $f$ is applied first.
• "The inverse of a reflection is a reflection in another line." No: a reflection is its own inverse (same line).

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. Triangle $A(-4,3)$, $B(-1,1)$, $C(-2,5)$ is translated by $\left(\begin{array}{c}6\\ 2\end{array}\right)$, then by $\left(\begin{array}{c}-1\\ -4\end{array}\right)$. Find the final images and the single translation.

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   Single vector $\left(\begin{array}{c}5\\ -2\end{array}\right)$: $A^{\prime \prime }(1,1)$, $B^{\prime \prime }(4,-1)$, $C^{\prime \prime }(3,3)$.

   Common mistake: Subtracting the vectors.

2. The square $ABCD$ is rotated $90^{\circ }$ about its centre four times. What is the final position, and why?

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   The original position: $4\times 90^{\circ }=360^{\circ }$, the identity.

   Common mistake: Saying it moves to a new place.

3. Find the inverse of each: translation by $\left(\begin{array}{c}-2\\ 7\end{array}\right)$; rotation of $60^{\circ }$ about $(1,1)$; enlargement centre $O$, $k=-3$.

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   Translation by $\left(\begin{array}{c}2\\ -7\end{array}\right)$; rotation of $-60^{\circ }$ about $(1,1)$; enlargement centre $O$, $k=-\frac{1}{3}$.

   Common mistake: Changing the centre of the rotation.

4. Explain why the composite of two rotations about the same centre is a rotation.

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   Each keeps the centre fixed and turns every point; turning by ${\theta }_1$ then by ${\theta }_2$ about the same point is turning by ${\theta }_1+{\theta }_2$.

   Common mistake: Thinking the angles multiply.

5. Show with $P(1,2)$ that reflecting in the x-axis then rotating $90^{\circ }$ about $O$ differs from doing them in the other order.

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   First order: $(1,-2)\to (2,1)$. Other order: $(-2,1)\to (-2,-1)$. Different, so the order matters.

   Common mistake: Using the clockwise rule for anticlockwise rotation.

6. Find the point $P$ whose image under the enlargement with centre $O$, $k=2$, followed by the translation $\left(\begin{array}{c}3\\ 1\end{array}\right)$, is $(9,5)$.

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   Undo in reverse order: $(9,5)-(3,1)=(6,4)$, then divide by 2: $P(3,2)$.

   Common mistake: Undoing the transformations in the wrong order.

7. Central symmetries with centres $C_1(2,0)$ and $C_2(2,3)$ are applied to $P(0,0)$ in that order. Find the image and the equivalent translation.

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   $S_{C_1}(P)=(4,0)$, $S_{C_2}(4,0)=(0,6)$. Translation $2\overrightarrow{C_1{C}_2}=\left(\begin{array}{c}0\\ 6\end{array}\right)$. ✓

   Common mistake: Using $\overrightarrow{C_1{C}_2}$ without doubling.

8. A point is reflected in $y=2$ and then in $y=5$. Describe the single transformation and find the image of $(1,0)$.

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   Translation by $\left(\begin{array}{c}0\\ 6\end{array}\right)$: $(1,6)$. Check: $(1,0)\to (1,4)\to (1,6)$.

   Common mistake: Translating by 3 only.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. The inverse of the translation by $\left(\begin{array}{c}3\\ -5\end{array}\right)$ is the translation by:	a. the rotation through $90^{\circ }$ clockwise about $(1,2)$
   2. Two enlargements with the same centre and factors 2 and 3 give an enlargement with factor:	b. an enlargement with centre $O$ and $k=\frac{1}{4}$
   3. The inverse of an enlargement with centre $O$ and $k=4$ is:	c. 6
   4. Find the inverse of the rotation through $90^{\circ }$ anticlockwise about $(1,2)$.	d. $\left(\begin{array}{c}-3\\ 5\end{array}\right)$

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   1 → d ($\left(\begin{array}{c}-3\\ 5\end{array}\right)$) · 2 → c (6) · 3 → b (an enlargement with centre $O$ and $k=\frac{1}{4}$) · 4 → a (the rotation through $90^{\circ }$ clockwise about $(1,2)$)

   • Reverse the vector.
   • Factors multiply.
   • Multiply by $\frac{1}{4}$ to undo.
   • Step 1. The inverse undoes the turn, so it turns back by the same angle.
     Step 2. Same centre, opposite direction: $90^{\circ }$ clockwise about $(1,2)$.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $y=-x$  ·  $120^{\circ }$  ·  $3500^{\circ }$  ·  a rotation of $60^{\circ }$
   1. Reflections in two lines meeting at $30^{\circ }$ give .
   2. Two rotations about $O$ of $50^{\circ }$ and $70^{\circ }$ give a rotation of .
   3. The composite in the previous question is equivalent to a reflection in the line .
   Show answer

   1: a rotation of $60^{\circ }$ · 2: $120^{\circ }$ · 3: $y=-x$

   • Twice the angle between the lines.
   • Angles add.
   • $(x,y)\to (-y,x)\to (-y,-x)$.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. The inverse of a rotation of $90^{\circ }$ anticlockwise about $O$ is:	a. $(7,3)$
   2. Enlarge $(3,2)$ with centre $O$ and factor 2, then translate by $\left(\begin{array}{c}1\\ -1\end{array}\right)$.	b. a rotation of $90^{\circ }$ clockwise about $O$
   3. $f$ is the reflection in $y=x$ and $g$ the translation by $\left(\begin{array}{c}2\\ 0\end{array}\right)$. Find $(g\circ f)(1,4)$.	c. a rotation of $180^{\circ }$ about $O$
   4. Reflect in the x-axis then in the y-axis. The result is:	d. $(6,1)$

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   1 → b (a rotation of $90^{\circ }$ clockwise about $O$) · 2 → a ($(7,3)$) · 3 → d ($(6,1)$) · 4 → c (a rotation of $180^{\circ }$ about $O$)

   • $R_{-90^{\circ }}$.
   • Step 1. Enlargement: $(6,4)$.
     Step 2. Translation: $(6+1,4-1)=(7,3)$.
   • Step 1. $g\circ f$ means apply $f$ first: $f(1,4)=(4,1)$.
     Step 2. Then $g$: $(4+2,1)=(6,1)$.
   • $(x,y)\to (x,-y)\to (-x,-y)$.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   a translation of 6 cm  ·  $(6,0)$  ·  $(8,0)$  ·  $\left(\begin{array}{c}4\\ 2\end{array}\right)$
   1. Reflections in two parallel lines 3 cm apart give .
   2. Reflect in the line $x=1$ and then in $x=4$. The composite maps $(0,0)$ to .
   3. Two central symmetries with centres $C_1(0,0)$ then $C_2(2,1)$ give the translation .
   Show answer

   1: a translation of 6 cm · 2: $(6,0)$ · 3: $\left(\begin{array}{c}4\\ 2\end{array}\right)$

   • Twice the distance between the lines.
   • Step 1. Two reflections in parallel lines give a translation by twice the distance between them: $2\times 3=6$.
     Step 2. Direction from the first line to the second: $(0,0)\mapsto (6,0)$. Check: $(0,0)\to (2,0)\to (6,0)$.
   • $2\overrightarrow{C_1{C}_2}$.

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