Transformations practice questions — Honors Math

The reflection sends $(x,y)$ to $(y,x)$. The clockwise quarter turn sends a point $(a,b)$ to $(b,-a)$, so it sends $(y,x)$ to $(x,-y)$.

The combination is therefore the single map $(x,y)\mapsto (x,-y)$, which is the reflection in the $x$-axis.

A point is invariant when $(x,-y)=(x,y)$, so $-y=y$ and $y=0$, with $x$ unrestricted: every point of the $x$-axis is invariant and no other point is.

$$y=0$$

Areas multiply by $k^2$, so

$$k^2={\displaystyle \frac{45}{80}}={\displaystyle \frac{9}{16}}\mathrm{.}$$

Taking square roots gives $\mathrm{\mid }k\mathrm{\mid }={\displaystyle \frac{3}{4}}$, so $k={\displaystyle \frac{3}{4}}$ or $k=-{\displaystyle \frac{3}{4}}$; the area ratio alone cannot separate these.

Writing $O$ for the centre, $\overrightarrow{O{P}^{\mathrm{\prime }}}=k\overrightarrow{OP}$. The image lying on the opposite side of $O$ means $\overrightarrow{O{P}^{\mathrm{\prime }}}$ points opposite to $\overrightarrow{OP}$, which forces $k<0$.

$$k=-{\displaystyle \frac{3}{4}}\mathrm{.}$$

Reflecting in $x=-1$ gives $x$-coordinate $2(-1)-5=-7$, so $P$ maps to $(-7,3)$.

Reflecting that image in $x=4$ gives $2(4)-(-7)=15$, so the final image is $(15,3)$.

The two mirrors are parallel and $5$ units apart, so the combination is a translation of $2\times 5=10$ units in the positive $x$-direction: $5+10=15$, with $y$ unchanged.

$$(15,3)$$

Midpoint of $A{A}^{\mathrm{\prime }}$: $M=({\displaystyle \frac{3+9}{2}},{\displaystyle \frac{8+2}{2}})=(6,5)$.

Gradient of $A{A}^{\mathrm{\prime }}$: ${\displaystyle \frac{2-8}{9-3}}=-1$. If the mirror has gradient $m$, then $(-1)m=-1$, so $m=1$.

Through $M(6,5)$ with gradient $1$: $y-5=1(x-6)$, giving $y=x-1$.

Check: reflection in a line $y=x+c$ sends $(x,y)$ to $(y-c,x+c)$; with $c=-1$ this sends $(3,8)$ to $(8+1,3-1)=(9,2)=A^{\mathrm{\prime }}$.

$$y=x-1$$

$\overrightarrow{AB}=(-1-7,3-9)=(-8,-6)$ and $\overrightarrow{A^{\mathrm{\prime }}{B}^{\mathrm{\prime }}}=(5-1,0-(-3))=(4,3)$.

Since $(4,3)=-{\displaystyle \frac{1}{2}}(-8,-6)$, the scale factor is $k=-{\displaystyle \frac{1}{2}}$ (negative, because image and object segments point in opposite directions).

Writing $C$ for the centre, $A^{\mathrm{\prime }}-C=-{\displaystyle \frac{1}{2}}(A-C)$. Multiplying by $2$: $2A^{\mathrm{\prime }}-2C=-A+C$, so $3C=2A^{\mathrm{\prime }}+A$.

$3C=(2\cdot 1+7,2\cdot (-3)+9)=(9,3)$.

$$C=(3,1)\mathrm{.}$$

Check on $B$: $\overrightarrow{CB}=(-4,2)$, and $-{\displaystyle \frac{1}{2}}(-4,2)=(2,-1)$, giving $C+(2,-1)=(5,0)=B^{\mathrm{\prime }}$.

The displacement from $C$ to $P$ has components $(5-1,3-2)=(4,1)$: from $C$, the point $P$ sits $4$ right and $1$ up.

A quarter turn clockwise sends a displacement of $p$ right and $q$ up to a displacement of $q$ right and $p$ down, i.e. $(p,q)\mapsto (q,-p)$. Test it on the unit step $(1,0)$, which a clockwise quarter turn sends to $(0,-1)$.

So the displacement from $C$ to the image is $(4,1)\mapsto (1,-4)$, and the image is $(1+1,2-4)=(2,-2)$.

Check: $\sqrt{4^2+1^2}=\sqrt{17}$ and $\sqrt{1^2+(-4{)}^2}=\sqrt{17}$, so the distance from the centre is unchanged, as a rotation demands.

$$(2,-2)$$

$\overrightarrow{O{P}^{\mathrm{\prime }}}=-{\displaystyle \frac{2}{3}}\overrightarrow{OP}$, so $\overrightarrow{O{P}^{\mathrm{\prime }}}$ points in the direction opposite to $\overrightarrow{OP}$. Hence $P$, $O$, $P^{\mathrm{\prime }}$ are collinear with $O$ lying between $P$ and $P^{\mathrm{\prime }}$, and

$$P{P}^{\mathrm{\prime }}=OP+O{P}^{\mathrm{\prime }}\mathrm{.}$$

The lengths satisfy $O{P}^{\mathrm{\prime }}={\displaystyle \frac{2}{3}}OP$, so

$35=OP+{\displaystyle \frac{2}{3}}OP={\displaystyle \frac{5}{3}}OP$, giving $OP=35\times {\displaystyle \frac{3}{5}}=21\text{ cm}$.

Therefore $O{P}^{\mathrm{\prime }}={\displaystyle \frac{2}{3}}\times 21$.

$$O{P}^{\mathrm{\prime }}=14\text{ cm}\mathrm{.}$$

$\overrightarrow{CA}=(4-1,3-2)=(3,1)$.

$\overrightarrow{C{A}^{\mathrm{\prime }}}=-2(3,1)=(-6,-2)$.

Adding this displacement to the centre: $A^{\mathrm{\prime }}=(1+(-6),2+(-2))$.

$$A^{\mathrm{\prime }}=(-5,0)\mathrm{.}$$

The negative factor puts $A^{\mathrm{\prime }}$ on the opposite side of $(1,2)$ from $A$, at twice the distance, which is consistent with $\overrightarrow{C{A}^{\mathrm{\prime }}}=(-6,-2)$ pointing opposite to $\overrightarrow{CA}=(3,1)$.

The mirror line $x=3$ is vertical, so the perpendicular through $P$ is the horizontal line $y=4$, and the image also has $y$-coordinate $4$.

$P$ lies $3-(-1)=4$ units to the left of the mirror line, so its image lies $4$ units to the right of it: the $x$-coordinate is $3+4=7$.

Check: the midpoint of $(-1,4)$ and $(7,4)$ is $({\displaystyle \frac{-1+7}{2}},4)=(3,4)$, which lies on $x=3$, and the segment joining the two points is horizontal, hence perpendicular to the vertical mirror line.

$$(7,4)$$

The top entry is the change in $x$ from object to image: $4-(-3)=7$.

The bottom entry is the change in $y$ from object to image: $-1-5=-6$.

Check: applying $\left(\begin{array}{c}7\\ -6\end{array}\right)$ to $A(-3,5)$ gives $(-3+7,5-6)=(4,-1)=A^{\mathrm{\prime }}$, as required.

$$\left(\begin{array}{c}7\\ -6\end{array}\right)$$