Transformations — Mathslive.ie

Section 1 of 3

Translations

A translation is a movement in a certain direction for a certain distance.

The shape never changes size and never turns — it just slides.

(i) Worked example — image under a translation

Find the image of $A(-3,1)$ under the translation that maps $P(1,4)$ to $Q(5,-1)$.

First get the rule — what the translation does to $x$ and to $y$.

$P(1,4) \to Q(5,-1)$

$x:\ 1 \to 5$ is $+4$. $y:\ 4 \to -1$ is $-5$.

Rule: $(x,\,y) \to (x+4,\ y-5)$

Now apply the same rule to $A$.

$A(-3,1)$

$x:\ -3+4 = 1$. $y:\ 1-5 = -4$.

Image $= (1,\,-4)$

(ii) Worked example — completing a parallelogram

$P(-1,3)$, $Q(2,-4)$, $R(-3,5)$. Find $S$ so that $PQRS$ is a parallelogram.

In a parallelogram, the rule that takes you along one side matches the opposite side.

The rule has to end with $S$. So $Q \to R$ is the same as $P \to S$.

$Q(2,-4) \to R(-3,5)$

$x:\ 2 \to -3$ is $-5$. $y:\ -4 \to 5$ is $+9$.

Rule: $(x,\,y) \to (x-5,\ y+9)$

Apply it to $P$:

$P(-1,3)$

$x:\ -1-5 = -6$. $y:\ 3+9 = 12$.

$S = (-6,\,12)$

YOU TRY · 1

Find the image of $B(2,-3)$ under the translation that maps $P(0,1)$ to $Q(4,4)$.

Get the rule from $P \to Q$ first, then apply it to $B$.

Rule: $(x,y) \to (x+4,\ y+3)$

$(6,\,0)$

YOU TRY · 2

$A(1,2)$, $B(5,3)$, $C(7,-1)$. Find $D$ so that $ABCD$ is a parallelogram.

Rule ends with $D$, so $B \to C$ is the same as $A \to D$.

$B \to C$: $(x,y) \to (x+2,\ y-4)$

Apply to $A(1,2)$.

$D = (3,\,-2)$

Section 2 of 3

Central Symmetry

Go through a point, and go the same distance again on the other side.

That point is the centre. The notation $S_P(Q)$ means the central symmetry of $Q$ through the point $P$.

(i) Worked example — symmetry in a point

Find the image of $(3,-1)$ under central symmetry in $(-2,4)$.

Step from the point to the centre, then take the same step again.

$(3,-1) \to (-2,4)$

$x:\ 3 \to -2$ is $-5$. $y:\ -1 \to 4$ is $+5$.

Apply $-5,\ +5$ again, starting at the centre $(-2,4)$:

$x:\ -2-5 = -7$. $y:\ 4+5 = 9$.

Image $= (-7,\,9)$

(ii) Worked example — the $S_P(Q)$ notation

$P(-1,3)$ and $Q(2,5)$ are two points. Find $S_P(Q)$.

$S_P(Q)$ — the symmetry of $Q$ through the point $P$.

$Q(2,5) \to P(-1,3)$

$x:\ 2 \to -1$ is $-3$. $y:\ 5 \to 3$ is $-2$.

Apply $-3,\ -2$ again, starting at $P$:

$x:\ -1-3 = -4$. $y:\ 3-2 = 1$.

$S_P(Q) = (-4,\,1)$

YOU TRY · 3

Find the image of $(4,1)$ under central symmetry in $(1,-2)$.

Step from the point to the centre, then the same step again.

$(4,1) \to (1,-2)$ is $-3,\ -3$.

Again from $(1,-2)$.

$(-2,\,-5)$

YOU TRY · 4

$P(2,-1)$ and $Q(-3,4)$. Find $S_P(Q)$.

$Q$ through $P$ — step $Q \to P$, then again.

$Q(-3,4) \to P(2,-1)$ is $+5,\ -5$.

Again from $P$.

$S_P(Q) = (7,\,-6)$

$(7,\,-6)$

Section 3 of 3

Axial Symmetry

Go through a line at $90^\circ$, and go the same distance again on the other side.

A reflection in a line. The image sits the same distance from the line, on the far side.

(i) The quick ones — axes and origin

Three you should know on sight:

Know on sight

1.$S_x$ — axial symmetry in the $x$-axis — change the sign of $y$.

2.$S_y$ — axial symmetry in the $y$-axis — change the sign of $x$.

3.$S_o$ — central symmetry in the origin — change the sign of both.

$S_x(1,3) = (1,\,-3)$

$S_y(1,3) = (-1,\,3)$

$S_o(1,3) = (-1,\,-3)$

(ii) Worked example — reflection in a vertical line

Find the image of $(-1,3)$ under axial symmetry in the line $x=1$.

The line $x=1$ is vertical, so only $x$ moves — the $y$ stays put.

$-1$ is a distance of $2$ from the line $x=1$. Go $2$ past the line: $1+2 = 3$.

Image $= (3,\,3)$

(iii) Worked example — reflection in a slanted line

Find the image of $(3,5)$ under axial symmetry in $x+y=4$.

Drop a perpendicular from the point to the line, find where they cross, then use central symmetry through that crossing point.

Perpendicular line: swap the sign — $x - y = k$.

Through $(3,5)$: $3-5 = k \Rightarrow k = -2$. So $x - y = -2$.

Solve with $x + y = 4$:

Add the two lines: $2x = 2 \Rightarrow x = 1$, then $y = 3$. They cross at $(1,3)$.

Now central symmetry of $(3,5)$ through $(1,3)$.

$(3,5) \to (1,3)$ is $-2,\ -2$. Again: $1-2 = -1$, $3-2 = 1$.

Image $= (-1,\,1)$

YOU TRY · 5

Write down $S_x(-2,5)$, $S_y(-2,5)$ and $S_o(-2,5)$.

$S_x$ flips $y$, $S_y$ flips $x$, $S_o$ flips both.

$S_x = (-2,-5)$, $S_y = (2,5)$, $S_o = (2,-5)$

$(-2,-5)$, $(2,5)$, $(2,-5)$

YOU TRY · 6

Find the image of $(4,1)$ under axial symmetry in the line $x=2$.

Vertical line — only $x$ moves. How far is $4$ from $x=2$?

$4$ is $2$ past the line; go $2$ back: $2-2 = 0$.

$(0,\,1)$

YOU TRY · 7

Find the image of $(2,3)$ under axial symmetry in $x+y=1$.

Perpendicular $x-y=k$, find $k$, solve with the line, then central symmetry through the crossing point.

$x-y=k$: $2-3=-1$, so $x-y=-1$.

With $x+y=1$: $2x=0 \Rightarrow x=0,\ y=1$. Cross at $(0,1)$.

$(2,3)\to(0,1)$ is $-2,-2$; again $\Rightarrow (-2,-1)$.

$(-2,\,-1)$

SUM

The lot in one box

Transformations toolkit

1.Translation: find the rule (change in $x$, change in $y$), then apply it to the point.

2.Parallelogram $PQRS$: the rule $Q \to R$ equals $P \to S$.

3.Central symmetry $S_P(Q)$: step point $\to$ centre, then take the same step again.

4.$S_x$ flips $y$; $S_y$ flips $x$; $S_o$ flips both.

5.Axial symmetry in a line: drop a perpendicular, find where it crosses, then central symmetry through that point.

End of lesson

Transformations — HL · Mathslive.ie