Worked Examples: Translation

Translation adds the same displacement vector $(T_x, T_y)$ to every vertex of the object. Shape, size, and orientation remain unchanged.

Example 1: Translation by Vector (3, 2)

Given triangle vertices:

• $A(1, 2)$
• $B(4, 3)$
• $C(6, 1)$

Translation vector: $(T_x, T_y) = (3, 2)$

Translation formula:

$$x' = x + T_x,\quad y' = y + T_y$$

Apply translation:

• $A'(1+3,\;2+2) = (4, 4)$
• $B'(4+3,\;3+2) = (7, 5)$
• $C'(6+3,\;1+2) = (9, 3)$

New coordinates: $A'(4,4),\; B'(7,5),\; C'(9,3)$

Example 2: Translation 100 Units Right and 10 Units Up

Given triangle vertices:

• $A(20, 0)$
• $B(60, 0)$
• $C(40, 100)$

Translation parameters:

• $T_x = +100$ (right)
• $T_y = +10$ (up)

Apply translation:

• $A'(20+100,\;0+10) = (120, 10)$
• $B'(60+100,\;0+10) = (160, 10)$
• $C'(40+100,\;100+10) = (140, 110)$

New coordinates: $A'(120,10),\; B'(160,10),\; C'(140,110)$

Example 3: Translation with Negative and Positive Offsets

Given polygon vertices:

• $P_1(10, 10)$
• $P_2(15, 15)$
• $P_3(20, 10)$

Translation parameters:

• $T_x = -5$ (left)
• $T_y = +5$ (up)

Apply translation:

• $P_1'(10-5,\;10+5) = (5, 15)$
• $P_2'(15-5,\;15+5) = (10, 20)$
• $P_3'(20-5,\;10+5) = (15, 15)$

New coordinates: $P_1'(5,15),\; P_2'(10,20),\; P_3'(15,15)$

Worked Examples: Scaling

Scaling changes the size of an object by multiplying each coordinate with scaling factors. Unless stated otherwise, scaling is performed about the origin.

Example 1: Differential Scaling of a Rectangle

Given rectangle vertices:

• $P(2, 2)$
• $Q(4, 2)$
• $R(4, 5)$
• $S(2, 5)$

Scaling factors: $S_x = 2,\; S_y = 3$

Scaling formula (about origin):

$$x' = x \cdot S_x,\quad y' = y \cdot S_y$$

Apply scaling:

• $P'(2\cdot2,\;2\cdot3) = (4, 6)$
• $Q'(4\cdot2,\;2\cdot3) = (8, 6)$
• $R'(4\cdot2,\;5\cdot3) = (8, 15)$
• $S'(2\cdot2,\;5\cdot3) = (4, 15)$

New coordinates: $P'(4,6),\; Q'(8,6),\; R'(8,15),\; S'(4,15)$

Example 2: Uniform Scaling of a Triangle

Given triangle vertices:

• $A(0, 0)$
• $B(1, 1)$
• $C(5, 2)$

Uniform scaling factor: $S = 2$

Scaling formula:

$$x' = x \cdot S,\quad y' = y \cdot S$$

Apply scaling:

• $A'(0\cdot2,\;0\cdot2) = (0, 0)$
• $B'(1\cdot2,\;1\cdot2) = (2, 2)$
• $C'(5\cdot2,\;2\cdot2) = (10, 4)$

New coordinates: $A'(0,0),\; B'(2,2),\; C'(10,4)$

Example 3: Fixed Point (Composite) Scaling

Given square vertices:

• $A(0,0)$
• $B(2,0)$
• $C(2,2)$
• $D(0,2)$

Scaling factors: $S_x = 2,\; S_y = 2$

Fixed point: $(x_f, y_f) = (2, 2)$

Concept: To keep a point fixed during scaling, apply a composite transformation:

• Translate fixed point to origin
• Scale
• Translate back

Fixed-point scaling formula:

$$x' = x_f + (x - x_f)\cdot S_x$$

$$y' = y_f + (y - y_f)\cdot S_y$$

Apply scaling:

• $A'(2 + (0-2)\cdot2,\;2 + (0-2)\cdot2) = (-2, -2)$
• $B'(2 + (2-2)\cdot2,\;2 + (0-2)\cdot2) = (2, -2)$
• $C'(2 + (2-2)\cdot2,\;2 + (2-2)\cdot2) = (2, 2)$
• $D'(2 + (0-2)\cdot2,\;2 + (2-2)\cdot2) = (-2, 2)$

New coordinates: $A'(-2,-2),\; B'(2,-2),\; C'(2,2),\; D'(-2,2)$

Worked Examples: Rotation

Rotation turns points around a reference point (usually the origin) by an angle $\theta$ without changing shape or size. Counter-clockwise rotation is taken as positive.

Example 1: $90^\circ$ Counter-Clockwise Rotation About Origin

Given point: $P(5, 2)$

Rotation angle: $\theta = 90^\circ$ (counter-clockwise)

Rotation rule for $90^\circ$ CCW:

$$ (x, y) \rightarrow (-y, x) $$

Apply rotation:

• $P'( -2,\; 5 )$

New coordinates: $( -2, 5 )$

Example 2: $45^\circ$ Rotation of a Triangle About Origin

Given triangle vertices:

• $A(1,1)$
• $B(2,3)$
• $C(3,1)$

Rotation angle: $\theta = 45^\circ$

Rotation formulas:

$$ \begin{aligned} x' &= x\cos\theta - y\sin\theta \\ y' &= x\sin\theta + y\cos\theta \end{aligned} $$

For $45^\circ$:

$$ \cos45^\circ = \sin45^\circ = \frac{\sqrt{2}}{2} $$

Apply rotation:

• $A'(1,1)$ $$ x' = \frac{\sqrt{2}}{2}(1-1)=0,\quad y' = \frac{\sqrt{2}}{2}(1+1)=\sqrt{2} $$ ⇒ $A'(0,\sqrt{2})$
• $B'(2,3)$ $$ x' = \frac{\sqrt{2}}{2}(2-3)= -\frac{\sqrt{2}}{2},\quad y' = \frac{\sqrt{2}}{2}(2+3)= \frac{5\sqrt{2}}{2} $$ ⇒ $B'(-\tfrac{\sqrt{2}}{2},\tfrac{5\sqrt{2}}{2})$
• $C'(3,1)$ $$ x' = \frac{\sqrt{2}}{2}(3-1)= \sqrt{2},\quad y' = \frac{\sqrt{2}}{2}(3+1)= 2\sqrt{2} $$ ⇒ $C'(\sqrt{2},2\sqrt{2})$

New coordinates:

$A'(0,\sqrt{2}),\; B'(-\tfrac{\sqrt{2}}{2},\tfrac{5\sqrt{2}}{2}),\; C'(\sqrt{2},2\sqrt{2})$

Example 3: Rotation About a Pivot Point

Given point: $A(0,0)$

Pivot point: $P(-1,-1)$

Rotation angle: $45^\circ$ (counter-clockwise)

Concept: Rotation about a pivot uses a composite transformation:

• Translate pivot to origin
• Rotate
• Translate back

Step 1: Translate point relative to pivot

$$ (x_t, y_t) = (0+1,\;0+1) = (1,1) $$

Step 2: Rotate by $45^\circ$

$$ x_r = \frac{\sqrt{2}}{2}(1-1)=0,\quad y_r = \frac{\sqrt{2}}{2}(1+1)=\sqrt{2} $$

Step 3: Translate back

$$ x' = 0-1 = -1,\quad y' = \sqrt{2}-1 $$

New coordinates: $(-1,\; \sqrt{2}-1)$

Worked Examples: Reflection

Reflection produces a mirror image of an object about a specified axis or line. Distances are preserved, but orientation is reversed.

Example 1: Reflection About the Y-axis

Given point: $(4, 5)$

Reflection rule about Y-axis:

$$ (x, y) \rightarrow (-x, y) $$

Apply reflection:

• $(4, 5) \rightarrow (-4, 5)$

Reflected point: $(-4, 5)$

Example 2: Reflection of a Polygon About the Line $y = 2$

Given polygon vertices:

• $A(-1, 0)$
• $B(0, -2)$
• $C(1, 0)$
• $D(0, 2)$

Concept: Reflection about a horizontal line $y = c$ keeps $x$ same and mirrors $y$ as:

$$ y' = 2c - y $$

Here: $c = 2 \Rightarrow y' = 4 - y$

Apply reflection:

• $A'(-1,\;4-0) = (-1, 4)$
• $B'(0,\;4-(-2)) = (0, 6)$
• $C'(1,\;4-0) = (1, 4)$
• $D'(0,\;4-2) = (0, 2)$

Reflected polygon vertices: $A'(-1,4),\; B'(0,6),\; C'(1,4),\; D'(0,2)$

Example 3: Reflection About the Line $y = x$

Given triangle vertices:

• $A(2, 4)$
• $B(4, 6)$
• $C(2, 6)$

Reflection rule about $y = x$:

$$ (x, y) \rightarrow (y, x) $$

Apply reflection:

• $A(2,4) \rightarrow A'(4,2)$
• $B(4,6) \rightarrow B'(6,4)$
• $C(2,6) \rightarrow C'(6,2)$

Reflected triangle vertices: $A'(4,2),\; B'(6,4),\; C'(6,2)$

Worked Examples: Shearing

Shearing slants an object by shifting one coordinate in proportion to the other. Shape angles change, but parallel lines remain parallel.

Example 1: Shearing a Square with $Sh_x = 2$ and $Sh_y = 3$

Given square vertices:

• $A(0,0)$
• $B(1,0)$
• $C(1,1)$
• $D(0,1)$

Shear formulas (about origin):

$$ x' = x + Sh_x \cdot y,\quad y' = y + Sh_y \cdot x $$

Apply shearing:

• $A'(0+2\cdot0,\;0+3\cdot0) = (0,0)$
• $B'(1+2\cdot0,\;0+3\cdot1) = (1,3)$
• $C'(1+2\cdot1,\;1+3\cdot1) = (3,4)$
• $D'(0+2\cdot1,\;1+3\cdot0) = (2,1)$

New coordinates: $A'(0,0),\; B'(1,3),\; C'(3,4),\; D'(2,1)$

Example 2: Y-direction Shear About a Reference Line

Given point: $(1,3)$

Shear factor: $Sh_y = 2$

Reference line: $y_{ref} = -1$

Concept: Shearing relative to a reference line requires a composite operation:

• Translate reference line to origin
• Apply shear
• Translate back

Step 1: Translate point

$$ y_t = y - y_{ref} = 3 - (-1) = 4 $$

Step 2: Apply Y-shear

$$ y_s = y_t + Sh_y \cdot x = 4 + 2\cdot1 = 6 $$

Step 3: Translate back

$$ y' = y_s + y_{ref} = 6 - 1 = 5 $$

New coordinates: $(1,5)$

Example 3: X-direction Shear of a Triangle

Given triangle vertices:

• $A(5,5)$
• $B(10,5)$
• $C(7,10)$

Shear factor: $Sh_x = 2$

X-shear formula:

$$ x' = x + Sh_x \cdot y,\quad y' = y $$

Apply shearing:

• $A'(5+2\cdot5,\;5) = (15,5)$
• $B'(10+2\cdot5,\;5) = (20,5)$
• $C'(7+2\cdot10,\;10) = (27,10)$

New coordinates: $A'(15,5),\; B'(20,5),\; C'(27,10)$