5.1 Y = MX + C
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Learning Objectives
- Understand the general form of a straight-line equation
- Identify and interpret the gradient (m) and y-intercept (c)
- Plot straight-line graphs from equations
- Write equations from graphs or coordinate points
- Use straight-line equations in practical contexts
- Apply straight-line equations to real-world problems
Key Concepts
Understanding y = mx + c
The equation **y = mx + c** represents a straight line.
- **m** is the **gradient** (slope) of the line.
- **c** is the **y-intercept** — the point where the line crosses the y-axis.
For example, **y = 2x + 3** has a gradient of 2 and crosses the y-axis at (0, 3).
What Does the Gradient Represent?
- The **gradient (m)** shows how steep the line is.
- It is calculated as the **change in y ÷ change in x**.
- A **positive m** means the line slopes upwards; a **negative m** means it slopes downwards.
- A **larger |m|** means a steeper line.
Example 1
Find the gradient of the line passing through points (2, 5) and (4, 9)
Solution:
Gradient = (change in y) ÷ (change in x) = (9 - 5) ÷ (4 - 2) = 4 ÷ 2 = 2 So the gradient is 2.
Example 2
Find the gradient of the line passing through points (1, 8) and (3, 2)
Solution:
Gradient = (change in y) ÷ (change in x) = (2 - 8) ÷ (3 - 1) = (-6) ÷ 2 = -3 So the gradient is -3.
What Does the y-Intercept Represent?
- The **y-intercept (c)** is the value of y when x = 0.
- It shows where the line crosses the y-axis.
- In a graph of y = mx + c, this is the starting point for plotting.
Example 1
Find the y-intercept of the line y = 3x - 4
Solution:
To find the y-intercept, substitute x = 0: y = 3(0) - 4 y = 0 - 4 y = -4 So the y-intercept is -4, meaning the line crosses the y-axis at (0, -4).
Example 2
Find the y-intercept of the line passing through (2, 7) with gradient 3
Solution:
Step 1: Use the point-gradient form: y - y₁ = m(x - x₁) y - 7 = 3(x - 2) y - 7 = 3x - 6 y = 3x - 6 + 7 y = 3x + 1 Step 2: The y-intercept is the constant term: 1 So the y-intercept is 1, meaning the line crosses the y-axis at (0, 1).
Using Graphs and Equations
- You can draw a line using the gradient and y-intercept.
- Alternatively, if you're given two points, you can calculate the gradient and find the equation.
- Example: For points (1, 2) and (3, 6), gradient = (6 - 2) / (3 - 1) = 2 → Use point-slope form or solve for c.
Example 1
Find the equation of the line passing through points (1, 4) and (3, 10)
Solution:
Step 1: Calculate the gradient: m = (10 - 4) ÷ (3 - 1) = 6 ÷ 2 = 3 Step 2: Use point-gradient form with point (1, 4): y - 4 = 3(x - 1) y - 4 = 3x - 3 y = 3x - 3 + 4 y = 3x + 1 So the equation is y = 3x + 1.
Example 2
Sketch the line y = 2x - 3
Solution:
Step 1: Find the y-intercept by substituting x = 0: y = 2(0) - 3 = -3 So the line passes through (0, -3). Step 2: Use the gradient 2 to find another point: From (0, -3), move right 1 unit and up 2 units to get (1, -1). Step 3: Draw the line through these two points. The line has gradient 2 and y-intercept -3.
Real-World Applications
Straight line equations appear in many real-world scenarios.
Common applications include distance-time relationships, cost functions, and linear models.
The key is to identify what the variables represent and interpret the gradient and y-intercept in context.
Example 1
A taxi charges a £3 base fare plus £2 per mile. Write an equation for the total cost C in terms of distance d in miles.
Solution:
Step 1: Identify the variables: - C = total cost (£) - d = distance (miles) Step 2: Identify the components: - Base fare = £3 (y-intercept) - Cost per mile = £2 (gradient) Step 3: Write the equation: C = 2d + 3 This means the total cost is £2 per mile plus a £3 base fare.
Example 2
A car travels at 60 km/h. If it starts 10 km from home, write an equation for its distance d from home after t hours.
Solution:
Step 1: Identify the variables: - d = distance from home (km) - t = time (hours) Step 2: Identify the components: - Starting distance = 10 km (y-intercept) - Speed = 60 km/h (gradient) Step 3: Write the equation: d = 60t + 10 This means the car is 10 km from home at t = 0, and travels 60 km/h.
Advanced Applications
More complex applications may involve finding equations from graphs or solving problems with multiple lines.
These often appear in physics, economics, and engineering contexts.
The key is to understand what the gradient and y-intercept represent in the specific context.
Example 1
Two companies offer phone plans. Company A: £20/month + £0.10/minute. Company B: £30/month + £0.05/minute. For how many minutes do both plans cost the same?
Solution:
Step 1: Write equations for both companies: Company A: C = 0.10m + 20 Company B: C = 0.05m + 30 Step 2: Set them equal to find when costs are the same: 0.10m + 20 = 0.05m + 30 Step 3: Solve for m: 0.10m - 0.05m = 30 - 20 0.05m = 10 m = 200 So both plans cost the same at 200 minutes.
Example 2
A line has gradient -2 and passes through point (3, 8). Find its equation.
Solution:
Step 1: Use point-gradient form: y - y₁ = m(x - x₁) y - 8 = -2(x - 3) Step 2: Expand and simplify: y - 8 = -2x + 6 y = -2x + 6 + 8 y = -2x + 14 So the equation is y = -2x + 14.
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