Equation of a straight line, and parallel lines
Finding y = mx + c from a gradient and a point, from two points, and recognising parallel lines
About three lessons of 45 to 60 minutes
Two points, or one point and a slope, is all a straight line needs
A straight line's equation, y = mx + c, has exactly two unknowns: the gradient m (how steep it is) and the y-intercept c (where it crosses the y-axis). Once you know m and c, you know EVERY point on the line, forever, without ever needing to plot it.
You never actually need to be TOLD both m and c directly. If you know the gradient and just one point the line passes through, you can find c by substitution. If you know two points, you can find the gradient first (the same 'rise over run' calculation from finding a gradient between two coordinates), then use either point to find c the same way.
- A phone plan costs £10 plus £2 per GB usedy = 2x + 10: gradient 2 (the price per GB), y-intercept 10 (the fixed monthly cost)
- Two ramps built at the same steepness (gradient)distinct lines with the SAME gradient but different y-intercepts never meet, so they are parallel
- Given just two points a line passes throughfind the gradient first, then substitute one point in to find the y-intercept
- A new road built parallel to an existing onethe new road's equation must share the existing road's gradient exactly
What students will be able to do
Students will find the equation of a straight line, in the form y = mx + c, given either a gradient and one point or two points, recognise distinct parallel lines by equal gradients and different intercepts, and construct the equation of a distinct parallel line through a new point.
- I can find c by substituting a known gradient m and a point (x, y) into y = mx + c.
- I can find the gradient between two points using (change in y) / (change in x), then use it to find the full equation.
- I know that two distinct lines are parallel when they have the same gradient (the same m) and different y-intercepts.
- I can find the equation of a line that is parallel to a given line and passes through a new point.
Standards this unit teaches
- GCSE Algebra #9UK GCSE Mathematics (DfE, England)The equation of a straight line, and parallel lines
Subject content statement (Department for Education, "GCSE mathematics: subject content and assessment objectives", published 1 November 2013, reference DFE-00233-2013, "Algebra" section, "Graphs", item 9, https://www.gov.uk/government/publications/gcse-mathematics-subject-content-and-assessment-objectives): students should "plot graphs of equations that correspond to straight-line graphs in the coordinate plane; use the form y = mx + c to identify parallel and perpendicular lines; find the equation of the line through two given points, or through one point with a given gradient". "Use the form y = mx + c to identify parallel" and both "find the equation..." clauses are underlined type (Foundation, assessed for every student); "and perpendicular lines" is bold type (Higher tier only, since it needs the negative-reciprocal-gradient rule this unit deliberately excludes).
Prior knowledge
This unit builds on skills students should already have met. Revisit any that are shaky first.
Words to teach and display
- Gradient (m)
- how steep a line is, calculated as (change in y) / (change in x); the coefficient of x in y = mx + c
- y-intercept (c)
- the point where a line crosses the y-axis, i.e. its value of y when x = 0
- Parallel lines
- two distinct straight lines with the same gradient and different intercepts, which therefore never meet however far they are extended
Teach it: concrete, pictorial, abstract
The lesson moves from things students can hold, to pictures and diagrams, to the written maths. The diagrams below are drawn from data, so they are accurate and print cleanly. Teach straight from them.
1. Equation from a gradient and one point
ConcreteIf you know the gradient m and a single point (x1, y1) the line passes through, substitute both into y = mx + c and solve for the one remaining unknown, c.
Find the equation of the line with gradient 4 through (2, 3). Substitute: 3 = 4(2) + c, so 3 = 8 + c, so c = 3 - 8 = -5. The equation is y = 4x - 5.
Find the equation of the line with gradient -2 that passes through (5, 1).
- Substitute into y = mx + c: 1 = -2(5) + c.
- 1 = -10 + c.
- c = 1 + 10 = 11.
Answer: y = -2x + 11
- Why does substituting the point into y = mx + c leave only ONE unknown to solve for?
- How could you check your answer without redoing the whole calculation?
2. Equation from two points, and parallel lines
PictorialGiven two points, first find the gradient the same way as finding a gradient between coordinates, then use either point to find c exactly as before. Distinct lines are parallel when their gradients match and their intercepts differ.
Find the equation through (1, 4) and (3, 10). Gradient = (10 - 4)/(3 - 1) = 6/2 = 3. Substitute (1, 4): 4 = 3(1) + c, so c = 1. The equation is y = 3x + 1.
A line is parallel to y = 3x + 1 and passes through (2, 0). Find its equation.
- Parallel lines share the same gradient, so this new line also has gradient 3.
- Substitute (2, 0) into y = 3x + c: 0 = 3(2) + c, so 0 = 6 + c.
- c = 0 - 6 = -6.
Answer: y = 3x - 6
- Why is finding a line's equation from two points really a two-step process (gradient, then c), not one step?
- If two lines have the same c but different gradients, are they parallel?
Common misconceptions and how to address them
MisconceptionTwo lines are parallel if they look similar, or if their y-intercepts (c values) are close together.
Why it happens: Students judge parallel-ness visually or by comparing the wrong number (c instead of m).
How to address it: Parallel is ONLY about the gradient (m), never the y-intercept (c). Two lines with the SAME m but DIFFERENT c are parallel (they run alongside each other, never meeting); two lines with the same c but different m cross exactly once, at the y-axis, so they are NOT parallel.
MisconceptionWhen substituting a point into y = mx + c, students substitute x and y into the wrong places, or forget that c is what is being solved for.
Why it happens: The x and y labels of the given point are easy to swap by accident, especially under time pressure.
How to address it: Always write out y = mx + c with the KNOWN gradient substituted for m first, then substitute the point's y-value on the left and x-value multiplied by m on the right, matching the point (x, y) exactly as given.
Guided practice (with answers)
1. Find the equation of the line with gradient 5 through (1, 2).
Answer: y = 5x - 3, because 2 = 5(1) + c gives c = -3.
2. Find the equation of the line with gradient -1 through (-2, 4).
Answer: y = -x + 2, because 4 = -1(-2) + c gives 4 = 2 + c, so c = 2.
3. Find the equation of the line through (0, 1) and (2, 9).
Answer: y = 4x + 1, because gradient = (9-1)/(2-0) = 4, and substituting (0,1) gives c = 1 directly.
4. Are y = 2x + 5 and y = 2x - 3 parallel?
Answer: Yes, because they have the same gradient, 2.
5. Find the equation of the line parallel to y = -3x + 4 that passes through (1, 0).
Answer: y = -3x + 3, because the gradient must be -3, and 0 = -3(1) + c gives c = 3. Its different intercept confirms it is a distinct parallel line.
Independent practice worksheets
Practise building a line's equation from a gradient and a point, from two points, and working with parallel lines, all with computed, never-wrong answer keys.
Differentiation
- Always write the general form y = mx + c first, then substitute known values into it one at a time, rather than trying to combine steps mentally.
- For the two-point case, calculate and label the gradient FIRST as its own separate step before attempting to find c.
- Use the coordinateLine figure's rise/run brackets as a physical template: count grid squares up, then across, before dividing.
- Given a line's equation, find TWO different points that also lie on a line parallel to it.
- Investigate why every vertical line (x = a number) has an undefined gradient, and cannot be written in the form y = mx + c.
- Preview: research what makes two lines PERPENDICULAR (rather than parallel) as an extension beyond this unit's Foundation-tier scope.
Assessment: exit ticket
A three-question exit ticket sampling gradient+point, two points, and parallel lines.
1. Find the equation of the line with gradient 2 through (3, 5).
Answer: y = 2x - 1, because 5 = 2(3) + c gives c = -1.
2. Find the equation of the line through (0, -2) and (4, 6).
Answer: y = 2x - 2, because gradient = (6 - -2)/(4-0) = 8/4 = 2, and (0,-2) gives c = -2 directly.
3. Is y = -4x + 1 parallel to y = 4x + 1? Explain.
Answer: No, because the gradients -4 and 4 are different, even though the y-intercepts match.
Teacher notes and timings
- Rough timing: Lesson 1 gradient and point (section 1), Lesson 2 two points and parallel lines (section 2), Lesson 3 mixed practice and the exit ticket.
- Curriculum note: DfE GCSE Algebra item 9's 'identify PARALLEL lines' and both 'find the equation' clauses are Foundation; its 'and perpendicular lines' clause is Higher-only (bold) and deliberately excluded, since the negative-reciprocal-gradient rule it needs is a separate, harder skill.
- Distinct from this codebase's UK GCSE real-life graphs unit (batch 3): that unit reads a gradient directly off a described real-world graph as a rate (speed, a conversion rate); this unit builds the full algebraic EQUATION of an abstract line from a gradient+point or two points, a more general skill the real-life-graphs unit does not attempt.
- Present and print both work: use the Print button for a clean handout, or project the coordinateLine figure and build the rise/run gradient calculation live with the class.