Worked examples — Transformations — translation, reflection, rotation, enlargement (basic)

This page is a workout. The parent note built the four moves and their rules; here we hit every awkward corner: negative shifts, all four quadrants, a rotation the parent skipped, a negative scale factor, a degenerate zero-shift, and a real-world problem. Before each answer you'll forecast — guess the result yourself, then check.

Everything is on the coordinate grid. A point is written $(x,y)$: $x$ = how far right (negative = left), $y$ = how far up (negative = down). We use tiny arrows and triangles throughout, so keep the figures beside you.

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The scenario matrix

Every transformation problem is really one of these case classes. The examples below are labelled with the cell they cover.

Cell	Transformation	The tricky ingredient
T1	Translation	vector with a negative component (down/left)
T2	Translation	zero vector $\left(\genfrac{}{}{0ex}{}{0}{0}\right)$ — degenerate, nothing moves
R1	Reflection	mirror is a diagonal line $y=x$ (coordinates swap)
R2	Reflection	point starts in quadrant III (both coords negative)
Ro1	Rotation	90° clockwise (parent only fully drilled anticlockwise)
Ro2	Rotation	a point on an axis (a degenerate, easy-to-check case)
E1	Enlargement	scale factor $0<k<1$ (shape shrinks)
E2	Enlargement	negative scale factor $k=-1$ (flips through the centre)
W	Word / real-world	a map/robot problem needing a translation
X	Exam twist	a composition, then "describe the single transformation"

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Example 1 — Translation with a negative component (Cell T1)

Forecast: $+4$ pushes right, $-5$ pushes down. Starting left-and-up, you should land somewhere right-and-lower. Guess the quadrant before reading on.

1. Write the rule. A translation by $\left(\genfrac{}{}{0ex}{}{a}{b}\right)$ sends $(x,y)\to (x+a,y+b)$. Here $a=4$, $b=-5$.
   • Why this step? A translation is literally "add the same displacement to every point" — no reference point needed, so we just add.
2. Add the horizontal part. $x^{\prime }=-2+4=2$.
   • Why this step? $a$ is positive so we move right; starting at $x=-2$, four steps right crosses zero into $x=2$. See the yellow arrow in the figure.
3. Add the vertical part. $y^{\prime }=3+(-5)=-2$.
   • Why this step? $b$ is negative, meaning down by 5; from $y=3$ that overshoots the $x$-axis to $y=-2$.
4. State the image. $A^{\prime }(2,-2)$ — quadrant IV, as forecast.

Verify: The displacement between $A$ and $A^{\prime }$ must equal the vector. $A^{\prime }-A=(2-(-2),-2-3)=(4,-5)$. ✓ Matches $\left(\genfrac{}{}{0ex}{}{4}{-5}\right)$.

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Example 2 — The zero vector (Cell T2, degenerate)

Forecast: A vector with no length means no push. Guess: the point doesn't move.

1. Apply the rule. $(x+0,y+0)=(7,-1)$.
   • Why this step? Adding zero changes nothing — this is the identity translation.
2. Interpret. The zero vector is the "do nothing" transformation; every point is its own image.

Verify: $B^{\prime }-B=(0,0)$. ✓ A degenerate case worth knowing: it confirms translations form a family that includes "stay still".

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Example 3 — Reflection in the diagonal $y=x$ (Cell R1)

Forecast: The parent note said $y=x$ swaps coordinates. So guess $(1,3)$ — and picture it as a mirror flip across the diagonal.

1. Recall the rule. Reflection in $y=x$ sends $(x,y)\to (y,x)$.
   • Why this step? The line $y=x$ is where "right-amount equals up-amount". Flipping across it trades your horizontal reach for your vertical reach — hence the swap.
2. Swap. $C(3,1)\to C^{\prime }(1,3)$.
   • Why this step? Mechanically exchange the two numbers.
3. Geometry check (figure). The green line $y=x$ is the mirror. $C$ sits below it, $C^{\prime }$ sits equally far above it. The segment $C{C}^{\prime }$ (red dashed) is perpendicular to the mirror and cut in half by it.
   • Why this step? That's the definition of a reflection: mirror is the perpendicular bisector of point-and-image.

Verify: Midpoint of $C{C}^{\prime }$ is $\left(\frac{3+1}{2},\frac{1+3}{2}\right)=(2,2)$, which lies on $y=x$. ✓ And the slope of $C{C}^{\prime }$ is $\frac{3-1}{1-3}=-1$, perpendicular to the mirror's slope $+1$. ✓

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Example 4 — Reflection with a point in quadrant III (Cell R2)

Forecast: $x$-axis reflection keeps $x$, flips the sign of $y$. From "left and down", guess "left and up".

1. Rule. $x$-axis reflection: $(x,y)\to (x,-y)$.
   • Why this step? The $x$-axis is $y=0$; reflecting flips how far above/below you are, so only $y$ changes sign.
2. Apply. $D(-4,-2)\to D^{\prime }(-4,-(-2))=D^{\prime }(-4,2)$.
   • Why this step? Watch the double negative: $-(-2)=+2$. This is exactly where beginners slip, so write it out.
3. Quadrant check. Started quadrant III (both negative); ended quadrant II (left, up). Correct — an $x$-axis flip moves you across the horizontal axis only.

Verify: Distance of $D$ below axis $=|-2|=2$; distance of $D^{\prime }$ above axis $=|2|=2$. Equal, on opposite sides. ✓

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Example 5 — Rotation 90° clockwise (Cell Ro1)

Forecast: The parent gave 90° anticlockwise as $(x,y)\to (-y,x)$. Clockwise is the opposite spin. Guess whether $x^{\prime }$ or $y^{\prime }$ picks up the minus sign.

1. Get the clockwise rule. Clockwise by 90° is anticlockwise by $-90°$. Plug $\theta =-90°$ into the parent's matrix ($\cos (-90°)=0$, $\sin (-90°)=-1$): $x^{\prime }=x\cos \theta -y\sin \theta =x(0)-y(-1)=y$ $y^{\prime }=x\sin \theta +y\cos \theta =x(-1)+y(0)=-x$ So $(x,y)\to (y,-x)$.
   • Why this step? Rather than memorise a second rule, we derive it by feeding a negative angle into the one formula we already trust.
2. Apply. $E(2,3)\to E^{\prime }(3,-2)$.
   • Why this step? Swap then negate the new second coordinate: $x^{\prime }=y=3$, $y^{\prime }=-x=-2$.
3. Direction check (figure). In the figure $E$ is in quadrant I (up-right). A clockwise quarter turn sweeps it down into quadrant IV (right-down) — exactly where $E^{\prime }(3,-2)$ sits. The blue arc shows the sweep direction.

Verify: Rotations preserve distance from the origin. $|OE|=\sqrt{2^2+3^2}=\sqrt{13}$; $|O{E}^{\prime }|=\sqrt{3^2+(-2{)}^2}=\sqrt{13}$. ✓ Also $OE\cdot O{E}^{\prime }=(2)(3)+(3)(-2)=0$, so they're perpendicular — a true 90° turn. ✓

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Example 6 — Rotating a point that sits on an axis (Cell Ro2, degenerate)

Forecast: 180° flips a point straight through the origin to the opposite side. $F$ is straight up the $y$-axis, so guess straight down.

1. Rule. 180° gives $(x,y)\to (-x,-y)$.
   • Why this step? At 180°, $\cos 180°=-1$, $\sin 180°=0$, so both coordinates just negate.
2. Apply. $F(0,5)\to F^{\prime }(-0,-5)=F^{\prime }(0,-5)$.
   • Why this step? $-0$ is just $0$ — the point stays on the $y$-axis, which is why this is a clean, degenerate check case.

Verify: $F$ and $F^{\prime }$ are equidistant from origin ($5$ each) and the origin is the midpoint of $F{F}^{\prime }$: $\left(\frac{0+0}{2},\frac{5+(-5)}{2}\right)=(0,0)$. ✓

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Example 7 — Enlargement that shrinks, $0<k<1$ (Cell E1)

Forecast: $k=\frac{1}{2}$ pulls each point halfway toward the centre. Guess: the image lands between $G$ and the centre.

1. Displacement from centre. $(x-a,y-b)=(6-2,4-0)=(4,4)$.
   • Why this step? Enlargement scales how far you are from the centre, so we first measure that reach.
2. Scale it. $k(4,4)=\frac{1}{2}(4,4)=(2,2)$.
   • Why this step? Multiplying by $\frac{1}{2}$ halves the reach — the shrink.
3. Add the centre back. $(2,0)+(2,2)=(4,2)$. So $G^{\prime }(4,2)$.
   • Why this step? The scaled displacement is measured from the centre, so we anchor it there.

Verify: Centre-to-$G$ distance $=\sqrt{4^2+4^2}=\sqrt{32}$; centre-to-$G^{\prime }$ distance $=\sqrt{2^2+2^2}=\sqrt{8}$. Ratio $\sqrt{8}/\sqrt{32}=\sqrt{1/4}=\frac{1}{2}=k$. ✓ And $G^{\prime }$ lies on the straight line from centre through $G$ (both have displacement direction $(1,1)$). ✓

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Example 8 — Negative scale factor $k=-1$ (Cell E2)

Forecast: A negative $k$ throws the point to the opposite side of the centre. With $|k|=1$ it lands the same distance away, but flipped. This is actually a 180° rotation about the centre — predict where.

1. Displacement from centre. $(5-1,1-1)=(4,0)$.
   • Why this step? Same first move: measure reach from the centre.
2. Scale by $-1$. $-1\cdot (4,0)=(-4,0)$.
   • Why this step? The minus sign reverses direction — instead of 4 units right of centre, we now go 4 units left. That's the flip-through.
3. Add centre. $(1,1)+(-4,0)=(-3,1)$. So $H^{\prime }(-3,1)$.

Verify: Centre is the midpoint of $H{H}^{\prime }$: $\left(\frac{5+(-3)}{2},\frac{1+1}{2}\right)=(1,1)$. ✓ Distance centre-to-$H$ = centre-to-$H^{\prime }$ = $4$ (since $|k|=1$). ✓

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Example 9 — Real-world word problem (Cell W)

Forecast: West is $-x$, north is $+y$. Guess the landing coordinate before computing.

1. Turn the words into a vector. "5 m west" $=-5$ in $x$; "2 m north" $=+2$ in $y$. Vector $=\left(\genfrac{}{}{0ex}{}{-5}{2}\right)$.
   • Why this step? Word directions must be converted to signed components using the stated axis convention, or every sign risks being wrong.
2. Translate. $(3+(-5),7+2)=(-2,9)$.
   • Why this step? The robot's move is a pure translation — add the displacement to the start.
3. Compare with the dock. Robot at $(-2,9)$; dock at $(-1,10)$. Not equal.

Verify: Gap $=$ dock $-$ robot $=(-1-(-2),10-9)=(1,1)$, a distance of $\sqrt{1^2+1^2}=\sqrt{2}\approx 1.41$ m. ✓ The robot misses the dock by about $1.41$ m.

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Example 10 — Exam twist: composition, then describe (Cell X)

Forecast: Two moves back to back. Predict the final quadrant, then guess what one clean transformation replaces the pair.

1. Step 1 — reflect in $y$-axis. $(x,y)\to (-x,y)$: $J(4,2)\to J_1(-4,2)$.
   • Why this step? $y$-axis is $x=0$; flip the sign of $x$ only.
2. Step 2 — rotate 90° anticlockwise. $(x,y)\to (-y,x)$: $J_1(-4,2)\to J^{\prime }(-2,-4)$.
   • Why this step? Apply the parent's anticlockwise rule to the output of step 1, in order.
3. Answer (a): $J^{\prime }(-2,-4)$.
4. Describe the single transformation (b). Track the net map: $(x,y)\to (-x,y)\to (-y,-x)$. So overall $(x,y)\to (-y,-x)$.
   • Why this step? Compose the two rules symbolically: feed $(-x,y)$ into $(x,y)\to (-y,x)$, giving $(-(y),(-x))=(-y,-x)$.
5. Recognise it. $(x,y)\to (-y,-x)$ is the parent's rule for reflection in the line $y=-x$.

Verify: Test on $J$: reflection in $y=-x$ gives $(4,2)\to (-2,-4)$, matching $J^{\prime }$. ✓ Test on another point $(1,0)$: two-step route $\to (-1,0)\to (0,-1)$; single rule $(-0,-1)=(0,-1)$. ✓ Same map.

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Recall Self-test (answer before revealing)

Translate $(-2,3)$ by $\left(\genfrac{}{}{0ex}{}{4}{-5}\right)$ ::: $(2,-2)$ Reflect $(3,1)$ in $y=x$ ::: $(1,3)$ Rotate $(2,3)$ 90° clockwise about O ::: $(3,-2)$ Enlarge $(6,4)$ from centre $(2,0)$ by $k=\frac{1}{2}$ ::: $(4,2)$ Reflect-in-$y$-axis then rotate-90°-anticlockwise equals which single reflection? ::: reflection in $y=-x$

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See also

• Vectors — translations are vectors; addition is the whole game.
• Coordinate Geometry & Distance Formula — every "Verify" line used distance to sanity-check.
• Trigonometry — the rotation rule came from $\cos ,\sin$ of angle sums.
• Matrices — rotations/enlargements as single matrix multiplications.
• Inverse Functions — reflection in $y=x$ is exactly graphing an inverse.
• Similar Triangles, Congruence, Symmetry — what each move preserves.