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

$$k^2 = \dfrac{45}{80} = \dfrac{9}{16}.$$

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

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

$$k = -\dfrac{3}{4}.$$

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 $AA'$: $M=\left(\dfrac{3+9}{2},\dfrac{8+2}{2}\right)=(6,5)$.

Gradient of $AA'$: $\dfrac{2-8}{9-3}=-1$. 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 $(8+1,\,3-1)=(9,2)=A'$.

$$y=x-1$$

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

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

Writing $C$ for the centre, $A' - C = -\dfrac{1}{2}\,(A - C)$. Multiplying by $2$: $2A' - 2C = -A + C$, so $3C = 2A' + A$.

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

$$C = (3,\,1).$$

Check on $B$: $\overrightarrow{CB} = (-4,\,2)$, and $-\dfrac{1}{2}(-4,\,2) = (2,\,-1)$, giving $C + (2,\,-1) = (5,\,0) = B'$.

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.

$$(2,-2)$$

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

$$PP' = OP + OP'.$$

The lengths satisfy $OP' = \dfrac{2}{3}\,OP$, so

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

Therefore $OP' = \dfrac{2}{3} \times 21$.

$$OP' = 14\text{ cm}.$$

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

$\overrightarrow{CA'} = -2\,(3,\,1) = (-6,\,-2)$.

Adding this displacement to the centre: $A' = (1 + (-6),\ 2 + (-2))$.

$$A' = (-5,\,0).$$

The negative factor puts $A'$ on the opposite side of $(1,\,2)$ from $A$, at twice the distance, which is consistent with $\overrightarrow{CA'} = (-6,\,-2)$ 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 $\left(\dfrac{-1+7}{2},4\right)=(3,4)$, 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 $\begin{pmatrix}7\\-6\end{pmatrix}$ to $A(-3,5)$ gives $(-3+7,\,5-6)=(4,-1)=A'$, as required.

$$\begin{pmatrix}7\\-6\end{pmatrix}$$