A9: Straight Line Graphs
Plot graphs of equations; y = mx + c; find equation of line through two points; parallel and perpendicular lines (Higher)
📋 Key Concepts
Linear equation: An equation of the form y = mx + c produces a straight line graph.
y = mx + c
m = gradient (steepness of the line)
c = y-intercept (where the line crosses the y-axis)
m = gradient (steepness of the line)
c = y-intercept (where the line crosses the y-axis)
📝 Plotting Straight Line Graphs
Method:
- Create a table of x values (usually -2, -1, 0, 1, 2)
- Substitute each x into the equation to find y
- Plot the points and draw a straight line through them
Example 1
Draw the graph of y = 2x + 1
Solution:
| x | -2 | -1 | 0 | 1 | 2 |
|---|---|---|---|---|---|
| y | -3 | -1 | 1 | 3 | 5 |
Plot points: (-2, -3), (-1, -1), (0, 1), (1, 3), (2, 5)
Draw a straight line through all points.
📝 Finding the Gradient
Gradient (m): Measures how steep a line is.
Gradient = change in y⁄change in x = vertical⁄horizontal
Example 2
Find the gradient of the line passing through (1, 2) and (4, 8)
Solution:
Gradient = 8 - 2⁄4 - 1 = 6⁄3 = 2
Types of gradient:
- Positive gradient: Line goes uphill (from left to right)
- Negative gradient: Line goes downhill
- Zero gradient: Horizontal line
- Undefined gradient: Vertical line
📝 Finding the Equation of a Line
Method:
- Find the gradient (m)
- Find the y-intercept (c)
- Write as y = mx + c
Example 3
Find the equation of the line with gradient 3 that passes through (0, -2)
Solution:
Gradient m = 3
y-intercept c = -2 (passes through (0, -2))
Equation: y = 3x - 2
Example 4
Find the equation of the line passing through (1, 3) and (3, 7)
Solution:
Step 1: Find gradient m = 7 - 3⁄3 - 1 = 4⁄2 = 2
Step 2: Use y = mx + c with point (1, 3):
3 = 2(1) + c
3 = 2 + c
c = 1
Equation: y = 2x + 1
📝 Parallel and Perpendicular Lines (Higher)
Parallel lines: Have the same gradient
Example 5
Find the equation of the line parallel to y = 3x - 1 that passes through (0, 4)
Solution:
Parallel lines have the same gradient: m = 3
Passes through (0, 4) so y-intercept c = 4
Equation: y = 3x + 4
Perpendicular lines: Gradients multiply to give -1
If one line has gradient m, a perpendicular line has gradient -1⁄m
Example 6
Find the gradient of a line perpendicular to y = 2x + 5
Solution:
Original gradient m = 2
Perpendicular gradient = -1⁄2
📝 Special Lines
Horizontal lines: y = c (gradient = 0)
Vertical lines: x = a (gradient undefined)
Vertical lines: x = a (gradient undefined)
Example 7
What is the equation of:
a) A horizontal line passing through (3, 5): y = 5
b) A vertical line passing through (3, 5): x = 3
❓ Practice Questions
Q1: State the gradient and y-intercept of y = 4x - 3.
Q2: Find the gradient of the line passing through (2, 1) and (5, 10).
Q3: Find the equation of the line with gradient -2 and y-intercept 5.
Q4: Find the equation of the line passing through (0, 3) and (2, 11).
Q5: Find the equation of the line parallel to y = 5x - 2 passing through (0, 1).
Q6: Find the gradient of a line perpendicular to y = 3x + 1.
✅ Answers
- Gradient = 4, y-intercept = -3
- Gradient = 3
- y = -2x + 5
- y = 4x + 3
- y = 5x + 1
- -1⁄3
🎯 Exam Tips
- y = mx + c: m is gradient, c is y-intercept
- Use y = mx + c or draw a table to plot lines
- Parallel lines have equal gradients
- Perpendicular gradients multiply to -1
- Horizontal lines have gradient 0
- Always check your line passes through the given points
🧠 Problem-Solving Strategies
Problem-Solving
To find a line's equation: calculate the gradient from two points, then substitute one point into y = mx + c to find c. For parallel lines, use the same gradient. For perpendicular lines, use the negative reciprocal of the gradient.
Multi-Step Problem
Line L passes through (1, 3) and (4, 12). (a) Find the equation of L. (b) Find the equation of the line perpendicular to L through (1, 3). (c) Find where the perpendicular line crosses the x-axis.
Solution:
(a) m = 12-3⁄4-1 = 3. y = 3x + c → 3 = 3(1) + c → c = 0. So y = 3x
(b) Perpendicular gradient = -1⁄3. y = -1⁄3x + c → 3 = -1⁄3(1) + c → c = 10⁄3. So y = -x⁄3 + 10⁄3
(c) When y = 0: x⁄3 = 10⁄3 → x = 10. Crosses at (10, 0)
⚠️ Common Errors
Watch Out!
1. Wrong: Calculating gradient as x₂-x₁⁄y₂-y₁ Correct: Gradient = y₂-y₁⁄x₂-x₁ (change in y over change in x)
2. Wrong: For y = 3x - 2, the y-intercept is (2, 0) Correct: The y-intercept is (0, -2) — it's where x = 0, not where the number appears
3. Wrong: Parallel to y = 5x + 2 has gradient -5 Correct: Parallel lines have the SAME gradient: m = 5
✍️ 6-Mark Exam Question
Extended Answer
6 marks: Line A passes through (0, 4) and (6, 0). (a) Find the equation of Line A. (b) Line B is perpendicular to Line A and passes through (3, 7). Find the equation of Line B. (c) Find the coordinates where Line A and Line B intersect.
(a) m = 0-4⁄6-0 = -2⁄3. y-intercept = 4. So y = -2⁄3x + 4
(b) Perpendicular gradient = 3⁄2. y = 3⁄2x + c → 7 = 3⁄2(3) + c → 7 = 9⁄2 + c → c = 5⁄2. So y = 3x⁄2 + 5⁄2
(c) Set equal: -2⁄3x + 4 = 3⁄2x + 5⁄2. Multiply by 6: -4x + 24 = 9x + 15 → 9 = 13x → x = 9⁄13. y = -2⁄3(9⁄13) + 4 = 50⁄13. Point (9⁄13, 50⁄13)
Mark scheme: (a) 2 marks. (b) 2 marks. (c) 2 marks for solving simultaneous equations.
📊 AO3: Reason & Interpret
Reasoning and Interpretation
Company A charges C = 25h + 50. Company B charges C = 40h + 20, where C = cost (£), h = hours.
(a) Which company is cheaper for 2 hours?
(b) After how many hours do both companies charge the same?
(c) A customer needs 5 hours of work. Which company should they choose and how much do they save?
Answers: (a) A: £100, B: £100 — same cost. (b) 25h + 50 = 40h + 20 → 30 = 15h → h = 2 hours. (c) A: £175, B: £220. Choose A, saving £45.
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