LESSON PLAN

Linear Relations and Equations

Apothem Team

Grade 9 · Algebra & Patterning

LESSON AT A GLANCE

Warm-up

5 min

Explore

15 min

Formalize

10 min

Practice

12 min

Exit ticket

3 min

Warm-up

The graph-story matching opener: four graphs on the board (steep up through origin; gentle up with high start; flat; downhill), four stories (a wage, a taxi, a flat fee, a draining pool). Match in pairs, defend in sentences — slope and intercept vocabulary resurfaces on its own.

Grade 9 linear relations add the FORMS: slope-intercept, standard, and the skill of moving between graph, equation, table, and story with any one as the starting point.

Explore

Form-fluency workshop: (1) from two points to everything: through $(2, 7)$ and $(5, 16)$ — slope 3, then $y - 7 = 3(x - 2)$ or plug into $y = mx + b$ → $b = 1$ : $y = 3x + 1$ ; (2) standard form's superpower: $3x + 2y = 12$ — intercepts fall out by zeroing ( $x = 0 \to y = 6$ ; $y = 0 \to x = 4$ ), graph in ten seconds; (3) parallel-and-perpendicular discovery on grids: parallel lines share $m$ ; perpendicular slopes multiply to $-1$ (draw, measure, conjecture — proof optional at this grade).

Close with the interpretation round: for each form, WHICH questions does it answer fastest? (Slope-intercept: rate and start. Standard: both intercepts. Point-slope: building from data.)

Formalize

Formalize the three forms as three tools:

$y = mx + b \qquad Ax + By = C \qquad y - y_1 = m(x - x_1)$

Parallel: equal slopes, different intercepts. Perpendicular: $m_1 m_2 = -1$ (negative reciprocals). Horizontal $y = k$ has $m = 0$ ; vertical $x = k$ has no slope-intercept form at all — the forms have BLIND SPOTS, and knowing them is part of owning the forms.

Practice

Practice: equation from two points (twice); convert between all three forms; graph from standard via intercepts; one parallel and one perpendicular construction through a given point; one story-to-equation-to-question pipeline.

Exit ticket: line through $(4, -1)$ parallel to $y = -2x + 7$ . ( $y = -2x + 7$ shares… $m = -2$ : $y = -2x + 7$ ? No — through $(4,-1)$ : $y = -2x + 7$ gives $-1 = -8 + 7$ ✓!! The point lies ON it — trick question; the "parallel line" is the line itself, and noticing that IS the answer.)

Exit ticket

TIP Assign the exit-ticket trick deliberately: students who CHECK the point against the original line catch it; students on autopilot produce a "parallel" duplicate. The catch teaches more than ten clean exercises.

WORKED EXAMPLES

Example 1 — From data to decision: the tutoring-business line

The data: a tutor charged one client \$85 for a 2-hour session and another \$130 for 3.5 hours (same pricing structure — a base fee plus hourly rate).

Step 1: Slope from the two points $(2, 85), (3.5, 130)$ : $m = \frac{130 - 85}{3.5 - 2} = \frac{45}{1.5} = 30$ dollars/hour.

Step 2: Intercept via point-slope: $y - 85 = 30(x - 2) \to y = 30x + 25$ . Base fee: $25.

Step 3: Interpret both parameters in the story: \$30/hour teaching rate, \$25 booking fee.

Step 4: Deploy: a 5-hour exam-prep package: $30(5) + 25 = \$175$ . And the client question — "why isn't 5 hours just 2.5× my 2-hour bill?" — has a structural answer: the base fee doesn't scale. Linear ≠ proportional, and the intercept is exactly the difference.

Example 2 — Standard form in its natural habitat: the budget line

The situation: \$60 to spend on snacks: chips at \$3 a bag, drinks at \$2 a bottle. All combos?

Step 1: The constraint: $3c + 2d = 60$ — standard form ARISES here; nobody built it from slope.

Step 2: Intercepts by zeroing: all chips: $c = 20$ ; all drinks: $d = 30$ . Graph the segment between $(20, 0)$ and $(0, 30)$ .

Step 3: Read combos off the line: $c = 10 \to 2d = 30 \to d = 15$ ✓ affordable exactly.

Step 4: Slope's meaning HERE: $-\frac{3}{2}$ — every extra 2 bags of chips costs 3 bottles of drinks. Trade-off rate, not motion.

Step 5: The form-choice moral: budget constraints, recipes, and mixtures are BORN in standard form; converting to $y = mx+b$ is possible but answers the wrong instincts. Meet each form where it lives.

Example 3 — Perpendicularity earns its keep: the shortest path to the road

The scene: a straight road follows $y = \frac{1}{2}x + 1$ ; a farmhouse sits at $(6, 9)$ . The driveway should meet the road at right angles (shortest path). Where?

Step 1: Driveway slope: perpendicular to $\frac{1}{2}$ → $m = -2$ .

Step 2: Driveway line through the house: $y - 9 = -2(x - 6) \to y = -2x + 21$ .

Step 3: Intersection with the road: $\frac{1}{2}x + 1 = -2x + 21 \to \frac{5}{2}x = 20 \to x = 8, y = 5$ . The driveway meets the road at $(8, 5)$ .

Step 4: Driveway length (Pythagoras, cameo): $\sqrt{(8-6)^2 + (5-9)^2} = \sqrt{4 + 16} = \sqrt{20} \approx 4.5$ units.

Step 5: The assembly noted: perpendicular slopes + point-slope form + solving a system + distance — four tools, one problem. Grade 9 is where the toolbox starts working as a SET.

MATERIALS

•Graph-story match cards

•Grids

•Two-point data cards

•Form-conversion relay sheets

•Practice set (PDF)

WATCH FOR

!Standard form graphed by slope-hunting instead of intercept-zeroing — legal but slow; each form has its fast lane.

!Perpendicular slope taken as the plain reciprocal (

\frac{1}{m}

, sign forgotten).

-1

product, both flips.

!Vertical lines given slope 0 (confusing with horizontal). Undefined vs zero — the wall vs the floor.