Three things kept slipping in Lesson 17. This time we picture each one before touching a number.
This is a slow-down lesson — no new topic. We're going back to the three places slope tripped us up in Lesson 17: forgetting the minus sign when a line goes downhill, getting stuck when the run is the unknown, and not seeing that slope is just a rate. Each one is easy to see and easy to forget as a rule — so we draw the picture first, then do the maths. New scenery this time: gondolas, ski runs, and a plane coming in to land.
Part A · Uphill is +, downhill is −
Picture it: which way is the line heading?
Slope isn't only a size — it has a direction built in. Read the line left-to-right, like reading a sentence. If it climbs, the slope is positive. If it falls, the slope is negative. If it stays level, the slope is zero. Same three pictures every time:
The three signs of slope — read each line left → right
The run always goes across (→). The rise goes up for a climb and down for a fall. A downhill rise is a negative number, so the whole ratio comes out negative. If your line falls but your answer is positive, you dropped the minus sign.
Sign rule: $m = \dfrac{\text{rise}}{\text{run}}$. Going up as you move right → $+$. Going down as you move right → $-$. When a height drops, the rise is negative — e.g. a fall of $80$ m is a rise of $\textcolor{#d1495b}{-80}$.
Worked example — a plane coming in to land
A plane descends from $900$ m to the runway over a horizontal distance of $12{,}000$ m. Find the slope of its descent.
It goes down $900$ m, so the rise is $\textcolor{#d1495b}{-900}$, not $900$:
The minus sign is doing real work: it tells you the plane is descending. Drop it and you'd describe a plane climbing away from the runway — the opposite of landing.
Part B · When the run is the mystery
Picture it: you know the height and the steepness — how long is the base?
Sometimes the slope is a limit you're given ("no steeper than…"), you know how high you need to reach, and the question is how much floor space the ramp or slide needs. The triangle is the same triangle — one side is just now the unknown.
A slide to a 1.5 m platform, slope must be 0.5 — how long is the base?
You know the rise and the slope, and you want the run. So rearrange the formula to put the run on its own.
Rearranging: start from $m = \dfrac{\text{rise}}{\text{run}}$. Multiply both sides by run, then divide by $m$:
$$\text{run} = \frac{\text{rise}}{m}$$
Read it as: "how many slope-steps fit into the height." Slope $0.5$ means half a metre up per metre across, so to climb $1.5$ m you need $1.5 \div 0.5 = 3$ metres across.
Worked example — colour-coded rearrange
■ rise = the height to reach■ run = the base (unknown)
The slide reaches a platform $\textcolor{#2255cc}{1.5}$ m high and its slope must be exactly $0.5$. Find the run.
Check it forwards: a $3$ m base with a $1.5$ m rise gives $m = 1.5/3 = 0.5$ ✓. Whenever you solve for the run, plug it back into $\text{rise}/\text{run}$ — you should get the slope you started with.
Part C · Slope = steepness = rate of change
Picture it: bigger ratio = steeper line
Two gondola cables leave the same station. To decide which is steeper, you don't judge by how the drawing looks — you compare the numbers. The line with the bigger slope (bigger size, ignoring the sign) is the steeper climb.
Two cables from one station — compare the slopes
Both start at the station. The red cable gains more height for the same distance across — a bigger rise/run, so a bigger slope, so it's steeper. Compare the numbers, not the eye.
And here's the big idea that ties slope to the sequences you already know: slope is a rate — how much $y$ changes each time $x$ goes up by $1$. Plot an arithmetic sequence and the slope is the common difference.
The sequence 5, 8, 11, 14 plotted — each step is +3
Between any two neighbours the run is $1$ term and the rise is $3$ — so $m = 3/1 = 3$. That $3$ is exactly the common difference $d$. The slope of a plotted arithmetic sequence equals its common difference.
Slope as a rate: in $y = (\text{start}) + (\text{rate})\cdot x$, the rate is the slope and the start is where the line meets the $y$-axis. If the quantity grows, the rate is positive; if it shrinks (distance remaining, battery left, fuel), the rate is negative.
Worked example — a road trip counting down
A road trip is $1{,}200$ km. Mia's family drives $300$ km each day. Write the distance $y$ still remaining after $x$ days, and give the slope with its sign.
$$y = 1200 - 300x$$
The distance left shrinks, so the rate — the slope — is $\textcolor{#d1495b}{-300}$ km/day (negative). The $1200$ is the start: the distance before day $0$, where the line meets the $y$-axis.
Practice — sketch the triangle first, then the maths
Tip: before every answer, draw a tiny slope triangle — is the line going up or down? Which side do you know, and which are you solving for?
Problem 1 Sign
A mountain road climbs $45$ m over a horizontal distance of $500$ m. Find the slope. Is it positive or negative, and why?
A ski gondola descends $600$ m in altitude over a horizontal distance of $2{,}400$ m on the way back down. Find the slope — including its sign. What does the sign tell you?
Careful: the altitude drops, so the rise is negative.
write rise as a negative → divide → interpret the sign →
Problem 3 Find the run
A water slide must reach a platform $2$ m high with a slope of exactly $0.4$. What is the horizontal distance (the run) the slide needs? (Rearrange $m = \text{rise}/\text{run}$.)
rearrange → run = rise ÷ m → check forwards →
Problem 4 Compare
Two ski runs leave the same peak. Run A drops $90$ m over $600$ m across. Run B drops $80$ m over $400$ m across. (a) Find each slope (with sign). (b) Which run is steeper, and how do you know?
slope of each → compare the sizes → name the steeper →
Problem 5 Rate
The arithmetic sequence $6, 10, 14, 18, \ldots$ is plotted with term number as $x$ and value as $y$. (a) What is the slope of the line through the points? (b) How does that number connect to the common difference?
rise ÷ run between two points → state slope → link to $d$ →
Problem 6 Rate
A phone starts a road trip at $100\%$ battery and loses $8\%$ for every hour of navigation, $x$. Write an equation for the battery $y$ remaining after $x$ hours. Then state the slope (the rate) with its sign, and say what the $100$ represents.
turn words into $y = \ldots$ → slope + sign → what is 100? →
Problem 7 Find the runOpen
Design a gondola cable up a mountainside. It must: (1) have a slope between $0.3$ and $0.5$; (2) reach a top station at least $400$ m higher than the base. Choose a slope in range, then find the horizontal run your cable needs. Show the numbers and say why your slope is inside the allowed range.
pick a slope → run = rise ÷ m → justify →
Show answers
Problem 1
$m = 45/500 = 0.09$ · Positive, because the road climbs (altitude goes up as you move forward).
Problem 2
The altitude drops, so rise $= -600$: $m = -600/2400 = -0.25$ · The negative sign means the gondola is going downhill.
(a) Run A: $m = -90/600 = -0.15$ · Run B: $m = -80/400 = -0.20$. (b) Run B is steeper — its slope has the bigger size ($0.20 > 0.15$), even though both are negative.
Problem 5
(a) Rise $= 4$, run $= 1$ between consecutive terms, so $m = 4$. (b) It equals the common difference $d = 4$ — the slope of a plotted arithmetic sequence is always its common difference.
Problem 6
$y = 100 - 8x$ · slope (rate) $= -8$ %/hour — negative, because the battery drains · the $100$ is the starting battery (the $y$-intercept, before any navigation).
Problem 7
Open — any slope in $[0.3, 0.5]$ with rise $\geq 400$ m works. Example: slope $= 0.4$, run $= 400/0.4 = 1{,}000$ m horizontal. The slope $0.4$ sits between $0.3$ and $0.5$, so it's inside the allowed range.
Tutor notes (for Seba)
Why this lesson exists: in Lesson 17 Mia missed 6 of 11 — Problems 2, 4, 5, 6, 9, 10. They cluster into three patterns: the sign of slope on a descent (P2, and the rate in P10), rearranging the formula for the run (P4, P9), and reading slope as a rate / comparison (P5 compare, P6 = common difference). This lesson isolates exactly those three, visual-first, with fresh scenery (planes, slides, gondolas — not the horse-ramps that just failed).
Part A watch-point: the minus sign. If a line falls but her answer is positive, ask her to point at which way the line heads left-to-right. Have her say the sentence "it goes down, so the rise is negative" before dividing.
Part B watch-point: she can find slope forwards but froze when the run was unknown (P4 wrong, yet P7 — the same move — was right, so it's fragile, not absent). Drill the single rearrange $\text{run} = \text{rise}/m$ and always check forwards.
Part C watch-point: "steeper" = bigger size of slope, ignore the sign (both ski runs are negative in P4). And reinforce that the plotted sequence's slope is the common difference — this stitches slope to the arithmetic work from lessons 08–12a.
Ledger: tags slope-from-graph, slope-two-points, rate-of-change, equation-translation should be marked struggling from Lesson 17 and only cleared by the mastery criterion (≥90% across ≥5 attempts over ≥2 lessons) — a good day here is not enough. Keep them as review slices in lessons 18–19. (Local ledger mirror is stale; update D1 when auth is available.)
Coming up next → Lesson 18: Slope from Two Points
Once these three pictures feel automatic — the sign, the rearrange, the rate — Lesson 18 gives you the shortcut formula $m = \dfrac{y_2 - y_1}{x_2 - x_1}$ so you can find slope from two coordinates with no grid at all.