Phase 1: Park Layout & Infrastructure (Lines, Angles, & Slopes)

1. "I Do" - Direct Instruction & Modeling

Every park design starts on a flat 2D plane (our coordinate system). Our main promenade is a straight line. If Promenade A runs through points $(1, 3)$ and $(5, 11)$, its slope ($m$) is:
m = (y₂ - y₁) / (x₂ - x₁) = (11 - 3) / (5 - 1) = 8 / 4 = 2

To keep traffic flowing seamlessly, a parallel monorail track must be built. Because parallel lines have equal slopes, the monorail line must also have a slope of $2$. If we want a service path to cross the promenade at a perfect 90-degree angle for safety crews, its slope must be the opposite reciprocal: $-1/2$.

2. "We Do" - Guided Practice

Let's design a security fence crossing two parallel paths. Suppose Path 1 is defined by the equation $y = \frac{3}{4}x + 2$ and Path 2 is $y = \frac{3}{4}x - 5$. A straight security fence (the transversal) cuts across both paths at an angle.

Interactive Question: If the angle between the security fence and Path 1 is $115^\circ$, what must the corresponding angle be where the fence cuts Path 2 to prove the paths are perfectly parallel?
Answer: It must also be $115^\circ$ because corresponding angles are congruent when parallel lines are cut by a transversal.

3. "You Do" - Blueprint Challenge Task 1

Phase 2: The Ferris Wheel & Circular Rides (Circle Geometry)

1. "I Do" - Direct Instruction & Modeling

The centerpiece of our park is a massive Ferris Wheel. We want to place its center at coordinate $(5, 8)$ with a radius of $4$ units. The standard equation of a circle is:
(x - h)² + (y - k)² = r²
Substituting our center $(5, 8)$ and radius $r = 4$, we get: (x - 5)² + (y - 8)² = 16.

To calculate the footprint of this attraction:

• Circumference (perimeter of track): $C = 2\pi r = 2 \cdot \pi \cdot 4 = 8\pi \approx 25.13$ units.
• Area (ground footprint space): $A = \pi r^2 = \pi \cdot 4^2 = 16\pi \approx 50.27$ square units.

2. "We Do" - Guided Practice

Suppose we need to relocate (translate) this Ferris Wheel to make room for a popcorn stand. We want to move it $3$ units to the left and $2$ units up.
Let's track the new center:
New $h = 5 - 3 = 2$
New $k = 8 + 2 = 10$
Our new circle equation becomes: (x - 2)² + (y - 10)² = 16.

Concept Check - Circle Similarity: Are the old Ferris Wheel and the newly translated Ferris Wheel similar? Yes! All circles are similar because you can map any circle onto another using only translation and dilation (scaling).

3. "You Do" - Blueprint Challenge Task 2

Phase 3: The Rollercoaster Scaffolding (Triangle Rigidity & Trigonometry)

1. "I Do" - Direct Instruction & Modeling

Rollercoasters rely heavily on triangular trusses because triangles are the most rigid geometric shape. Imagine a giant drop hill on our coaster. We need to build a structural support beam that creates a right-angled triangle with the ground.

The launch hill rises to a height of $30$ meters and extends horizontally along the ground for $40$ meters. Using the Pythagorean Theorem ($a^2 + b^2 = c^2$), we can find the length of the diagonal track ($c$):
30² + 40² = c² ➔ 900 + 1600 = c² ➔ 2500 = c² ➔ c = 50 meters

To ensure the coaster isn't too steep for safety regulations, we must calculate the angle of elevation ($\theta$) from the ground:
tan(θ) = Opposite / Adjacent = 30 / 40 = 0.75
Using a calculator to find the inverse tangent: θ = tan⁻¹(0.75) ≈ 36.87°. This is well within the safe zone of under $45^\circ$!

2. "We Do" - Guided Practice

Let's scale down this structural support design to build a miniature "Kids Coaster" that is mathematically similar to the big one. We want the scale factor to be $1:5$ (or $0.2$).

Let's calculate together:

• What is the new height? $30 \text{ m} \cdot 0.2 = 6\text{ m}$.
• What is the new base run? $40\text{ m} \cdot 0.2 = 8\text{ m}$.
• What is the new track length? $50\text{ m} \cdot 0.2 = 10\text{ m}$.
• Does the angle of elevation change? No! In similar triangles, corresponding angles are congruent. Let's verify: $\tan(\theta) = 6/8 = 0.75 \rightarrow \theta \approx 36.87^\circ$. The angles remain identical, ensuring a safe, consistent ride experience.

3. "You Do" - Blueprint Challenge Task 3

Phase 4: Advanced Attractions & FX (Transformations & Parabolas)

1. "I Do" - Direct Instruction & Modeling

Amusement parks love symmetry and repeating patterns. Let's design a series of identical game booths. If we design one triangular booth at coordinates $P(1, 1)$, $Q(3, 1)$, and $R(2, 3)$, we can replicate it using rigid transformations. Rigid transformations (translations, reflections, rotations) preserve the exact size and shape of our object.

Let's reflect this booth across the y-axis to create a mirroring stall on the opposite side of the walkway. The reflection rule across the y-axis is: (x, y) ➔ (-x, y). Our new coordinates are $P'(-1, 1)$, $Q'(-3, 1)$, and $R'(-2, 3)$. The stalls are congruent, maintaining visual balance!

Next, we need a spectacular parabolic water fountain display at the park's entrance. A parabola is mathematically defined as the set of all points equidistant from a fixed point (the focus) and a fixed straight line (the directrix). The middle point between them is the vertex.

2. "We Do" - Guided Practice

Let's design a water stream parabolic path where the vertex is at $(0,0)$ and the focus (where the nozzle is angled to point) is at $(0, 2)$.

Since the vertex $(0,0)$ is exactly halfway between the focus and the directrix, the directrix must be the horizontal line $y = -2$. The standard equation of a vertical parabola opening upward with its vertex at the origin is:
x² = 4py
Where $p$ is the distance from the vertex to the focus (here, $p = 2$). Substituting $p$:
x² = 8y ➔ y = (1/8)x²

3. "You Do" - Blueprint Challenge Task 4