Straight Line Models
SPEC 3.1c • COORDINATE GEOMETRY
Using linear equations to model real-world situations
Why Model with Lines?
Many real-world relationships can be approximated by straight lines, making them powerful tools for prediction and analysis.
Common Applications
Cost calculations (taxi fares, bills)
Distance-time relationships
Temperature conversions
Depreciation of assets
Simple growth models
When to Use
Constant rate of change
Fixed starting value + variable part
Two quantities with proportional relationship
Key Idea
In the model , every part has real-world meaning. Your job is to interpret what m and c represent in context.
Interpreting Gradient
The gradient in a linear model represents the rate of change - how much changes for each unit increase in .
Gradient Meaning
m = rate of change =
Example Context
Taxi fare: C = 2.50 + 1.40d (C = cost in £, d = distance in miles)
Gradient = 1.40 means "the cost increases by £1.40 for each additional mile"
Exam Tip
Always include units when interpreting gradient. Say "£ per mile" or "°C per minute" - this shows understanding!
Interpreting Y-Intercept
The y-intercept represents the value of when . In context, this is often a starting value or fixed amount.
Y-Intercept Meaning
c = value of y when x = 0Often: initial value, fixed charge, starting point
Example Context
Taxi fare: C = 2.50 + 1.40d
Y-intercept = 2.50 means "there is a fixed charge of £2.50 before any distance is travelled"
Watch Out
Sometimes x = 0 doesn't make sense in context (e.g., negative time). In these cases, acknowledge that the y-intercept may not have a practical interpretation.
01Worked Example
A plumber charges according to the formula C = 40 + 25t, where C is the total charge in pounds and t is the time in hours.
(a) Interpret the value 40 in context.
(b) Interpret the value 25 in context.
(c) Find the cost of a 3-hour job.
(a)
40 is the y-intercept (value when t = 0)
•
Interpretation: The plumber charges a fixed call-out fee of £40.
(b)
25 is the gradient (rate of change)
•
Interpretation: The plumber charges £25 per hour of work.
(c)
C = 40 + 25(3) = 40 + 75 = £115
Answer: (a) £40 call-out fee (b) £25 per hour (c) £115
Setting Up Models
To create a linear model from a description, identify the gradient and y-intercept from the context.
Method
Identify variables: What does x represent? What does y represent?
Find the gradient: Look for "per unit" or rate language
Find the y-intercept: Look for fixed/initial values
Write the equation: y = mx + c
Key Phrases
Gradient clues: "per hour", "for each", "rate of", "every additional"
Y-intercept clues: "fixed charge", "initial", "starting", "base cost", "standing charge"
Worked Example02
A mobile phone contract has a monthly fee of £12 plus 8p per minute for calls. Write an equation for the monthly cost C (in pounds) in terms of the number of minutes m used.
1
Identify variables: C = monthly cost (£), m = minutes used
2
Fixed monthly fee = £12 → this is the y-intercept (c = 12)
3
Cost per minute = 8p = £0.08 → this is the gradient (m = 0.08)
4
Write equation: C = 0.08m + 12
Answer: C = 0.08m + 12 (or C = 12 + 0.08m)
Worked Example03
Water freezes at 0°C (32°F) and boils at 100°C (212°F). Find a linear equation connecting F (Fahrenheit) to C (Celsius).
1
Two points: (0, 32) and (100, 212) where (C, F)
2
Find gradient: m =
3
Y-intercept: When C = 0, F = 32, so c = 32
4
Equation: F = C + 32
Answer: F = 1.8C + 32 (or F = ⁹⁄₅C + 32)
Making Predictions
Once you have a model, you can substitute values to make predictions.
Interpolation
Predicting within the range of known data.
Generally reliable ✓
Extrapolation
Predicting outside the range of known data.
Less reliable ✗
Important
Extrapolation assumes the linear relationship continues beyond the known data. This may not be true in reality!
Example
A model for plant height over weeks 1-8 may not predict week 20 accurately - growth often slows or stops!
04Worked Example
A savings model is given by S = 200 + 50t, where S is savings in pounds and t is time in months (for 0 ≤ t ≤ 24).
(a) Find the savings after 10 months.
(b) How long until the savings reach £800?
(c) Is it appropriate to use this model to predict savings after 5 years?
(a)
S = 200 + 50(10) = 200 + 500 = £700
(b)
800 = 200 + 50t → 600 = 50t → t = 12 months
(c)
5 years = 60 months, which is outside the given range (0 ≤ t ≤ 24)
•
This would be extrapolation. Savings patterns may change over longer periods (interest rates, spending habits), so the model may not be reliable.
Answer: (a) £700 (b) 12 months (c) No - extrapolation outside valid range
Limitations of Models
Linear models are approximations. You need to understand when they work and when they break down.
Common Limitations
Domain restrictions: Model only valid for certain x values
Negative values: May give impossible results (negative time, mass, etc.)
Extrapolation errors: Relationship may not continue outside data range
Oversimplification: Real relationships may be curved, not linear
Exam Skill
When asked to "comment on the validity" or "state a limitation", consider: Does the model work for all values? What happens at extreme values? Is the linear assumption reasonable?
05Worked Example
The height h (cm) of a plant t days after planting is modelled by h = 2 + 0.5t.
(a) What does the 2 represent?
(b) State two limitations of this model.
(a)
The 2 is the y-intercept, representing the initial height.
•
Interpretation: The plant was 2 cm tall when planted (at t = 0).
(b)
Limitation 1: The model predicts unlimited growth, but plants reach a maximum height.
•
Limitation 2: For negative t values, the model gives heights less than 2 cm, which doesn't make sense before planting.
•
Other valid answers: growth rate may not be constant; seasonal factors; plant may die
Answer: (a) Initial height of 2 cm (b) See above
Summary
Key Interpretations
In y = mx + cMeaning in Context
Gradient (m)Rate of change - how much y changes per unit of x (include units!)
Y-intercept (c)Starting/initial value - value of y when x = 0
Exam Checklist
Always interpret in context with units
Identify whether prediction is interpolation or extrapolation
State limitations when asked (domain, negative values, linearity assumption)
Check if x = 0 makes sense in the real-world context
Practice
Question 1: An electricity bill is calculated using B = 15 + 0.14u, where B is the bill in pounds and u is the number of units used.
(a) Interpret the value 15 in context.
(b) Interpret the value 0.14 in context.
(a)
15 is the y-intercept (value when u = 0)
•
Interpretation: There is a standing charge of £15 regardless of usage.
(b)
0.14 is the gradient (rate of change)
•
Interpretation: Each unit of electricity costs £0.14 (or 14p per unit).
Practice
Question 2: A car hire company charges a fixed fee of £30 plus £0.20 per mile driven.
(a) Write an equation for the total cost C in terms of miles driven m.
(b) Calculate the cost of driving 150 miles.
(a)
Fixed fee = £30 (y-intercept), Rate = £0.20 per mile (gradient)
•
Equation: C = 30 + 0.20m (or C = 0.2m + 30)
(b)
C = 30 + 0.20(150) = 30 + 30 = £60
Answer: (a) C = 30 + 0.2m (b) £60
Practice
Question 3: The temperature T (°C) of a cooling drink t minutes after being removed from the fridge is modelled by T = 5 + 2t for 0 ≤ t ≤ 15.
(a) Find the temperature after 8 minutes.
(b) Explain why this model should not be used to predict the temperature after 1 hour.
(a)
T = 5 + 2(8) = 5 + 16 = 21°C
(b)
1 hour = 60 minutes, which is outside the valid range (0 ≤ t ≤ 15).
•
The model would predict T = 5 + 2(60) = 125°C, which is impossible - the drink cannot exceed room temperature.
•
This is extrapolation; the linear model breaks down as the drink approaches room temperature.
Answer: (a) 21°C (b) Outside valid range; gives unrealistic temperature
Next: Why Model with Lines?