Work backwards from geometric information to algebraic equations. Learn three ways to find the rule for any straight line.
45 min 5 MCQ + 3 SAQ MAS-LIN-C-01
Think First
Before we begin: A line passes through $(0, 3)$ and goes up 2 units for every 1 unit to the right. Can you write an equation for this line? What information did you use?
Come back to this after you have worked through the lesson.
1
The Big Idea
Finding the equation of a line means working backwards from geometric facts to an algebraic rule. If you know the gradient and y-intercept, you can write the equation instantly. If you know the gradient and any point, substitute to find the missing intercept. If you only know two points, calculate the gradient first, then use either point to find the equation.
Know
The three methods for finding the equation of a line. The general form $ax + by + c = 0$.
Understand
How $y = mx + c$ connects gradient and y-intercept. How to use $y - y_1 = m(x - x_1)$ when a point and gradient are known.
Can Do
Find the equation of a line given gradient and y-intercept, gradient and one point, or two points.
2
Lesson Objectives
Write the equation of a line when given its gradient and y-intercept.
Write the equation of a line when given its gradient and any point on the line.
Write the equation of a line when given two points on the line.
Convert between $y = mx + c$ and general form $ax + by + c = 0$.
3
Key Vocabulary
Point-gradient form
$y - y_1 = m(x - x_1)$, used when you know a point and the gradient.
General form
$ax + by + c = 0$, where $a$, $b$, and $c$ are integers and $a \geq 0$.
Gradient-intercept form
$y = mx + c$, the most common form for graphing.
4
Spot the Trap
Wrong: Substituting a point into $y = mx + c$ but solving for $m$ instead of $c$. For example, using $(2, 5)$ with $m = 3$ to get $5 = 2 + c$ and concluding $c = 3$.
The simplest case: when you already know $m$ and $c$, just write $y = mx + c$. This is the fastest method when both pieces of information are given directly. No calculation needed beyond substitution.
If you know both $m$ and $c$, the equation is $y = mx + c$. No solving needed.
Watch negative signs
$m = -2$ and $c = -3$ gives $y = -2x - 3$. Both negatives stay.
General form
Multiply to clear fractions, then rearrange to $ax + by + c = 0$ with integers.
6
From Gradient and One Point
+5 XP
When you know the gradient and any point on the line, write $y = mx + c$, substitute the known gradient for $m$, then substitute the coordinates of the known point for $x$ and $y$. Solve for $c$, then write the final equation. Alternatively, use the point-gradient form $y - y_1 = m(x - x_1)$.
$y = 2x + c$. $7 = 2(3) + c$. $c = 1$.
substitute, then solve
Multiply m by x first
$7 = 2(3) + c$ means $7 = 6 + c$. Do not add 2 and 3 first.
Point-gradient form
$y - y_1 = m(x - x_1)$ avoids solving for $c$. Rearrange to $y = mx + c$ at the end.
Check your answer
Substitute the original point back into your final equation. It must satisfy it.
7
From Two Points
+5 XP
When given two points, first calculate the gradient using $m = \dfrac{y_2 - y_1}{x_2 - x_1}$. Then substitute this gradient into $y = mx + c$ and use either point to solve for $c$. Write the final equation and verify using the second point.
Find $m$. Find $c$. Verify.
gradient first, then c
Calculate m first
You cannot find the equation without the gradient. Always start with $m = \dfrac{\Delta y}{\Delta x}$.
Either point works
After finding $m$, use either point to find $c$. Both give the same answer.
Verify with the other point
Substitute the second point into your final equation. If it does not work, find your error.
8
General Form
+5 XP
The general form of a straight line is $ax + by + c = 0$ where $a$, $b$, and $c$ are integers and $a \geq 0$. To convert from $y = mx + c$, multiply through by the denominator to eliminate fractions, then bring all terms to one side.
$y = \dfrac{1}{2}x + 3$ → $x - 2y + 6 = 0$.
clear fractions first
Eliminate fractions
Multiply every term by the denominator to get integer coefficients.
Move everything left
$ax + by + c = 0$ means all terms on the left, equal to zero on the right.
Make a positive
If $a$ is negative, multiply the entire equation by $-1$.
Worked Example 1 — Gradient and Y-Intercept
Q.Find the equation of the line with gradient $-\dfrac{3}{4}$ and y-intercept 5. Give your answer in general form.
Solution
1
Substitute directly: $y = -\dfrac{3}{4}x + 5$
2
Multiply by 4: $4y = -3x + 20$
3
Rearrange: $3x + 4y - 20 = 0$
Answer: $3x + 4y - 20 = 0$
Worked Example 2 — Gradient and Point
Q.Find the equation of the line with gradient 2 passing through $(3, 7)$.
Solution
1
Write: $y = 2x + c$
2
Substitute $(3, 7)$: $7 = 2(3) + c$
3
$7 = 6 + c$ → $c = 1$
4
Answer: $y = 2x + 1$
Answer: $y = 2x + 1$
Worked Example 3 — Two Points
Q.Find the equation of the line through $(-2, 5)$ and $(4, -1)$.
Multiply every term by the denominator (3) to clear the fraction. Then move all terms to the left side, keeping $x$ positive.
5.The y-intercept of $3x + 2y = 6$ is:
Correct! Set $x = 0$: $2y = 6$ → $y = 3$. The y-intercept is $(0, 3)$.
The y-intercept occurs where $x = 0$. Substitute $x = 0$ into the equation and solve for $y$.
6.Find the equation of the line with gradient $\dfrac{3}{2}$ passing through $(4, -1)$. Give your answer in the form $y = mx + c$.
Your answer:
Equation set up with correct gradient
Point substituted correctly
Final equation correct
Model Answer
$y = \dfrac{3}{2}x + c$
$-1 = \dfrac{3}{2}(4) + c = 6 + c$
$c = -7$
$y = \dfrac{3}{2}x - 7$
7.A line passes through $A(-3, 4)$ and $B(5, 0)$. (a) Find the gradient. (b) Find the equation in the form $y = mx + c$. (c) Convert to general form $ax + by + c = 0$.
8.A plumber charges a call-out fee plus an hourly rate. For a 2-hour job the cost is $180. For a 5-hour job the cost is $360. (a) Find the gradient and interpret its meaning. (b) Find the equation relating cost $C$ and time $t$. (c) Determine the call-out fee and explain your reasoning.
Your answer:
Gradient calculated and interpreted (1 mark)
Equation found correctly (2 marks)
Call-out fee identified with reasoning (2 marks)
Model Answer
(a) $m = \dfrac{360 - 180}{5 - 2} = \dfrac{180}{3} =$ $60$. The gradient is $60 per hour (the hourly rate).
(c) The call-out fee is $60, which is the y-intercept ($C$ when $t = 0$). This is the fixed cost before any hours are worked.
Review
Consolidate and reflect before moving on.
Stretch Challenge
A line passes through $(1, -2)$ and has the same gradient as the line $2x + 4y = 8$. Find the equation of this line in general form. (Hint: first find the gradient of $2x + 4y = 8$ by rearranging to $y = mx + c$.)
Key Idea
Three ways to find a line equation: gradient + y-intercept (instant), gradient + point (solve for c), two points (find m first). Always verify your answer.
Common Trap
Substituting a point but forgetting to multiply $m$ by $x$ first. Also, sign errors when converting to general form. Multiply every term by the denominator.
Connection
This lesson brings together gradient, intercepts, and equations. The next lessons explore parallel and perpendicular lines, and how to use equations to model real-world situations.
Interactive: Equation Builder — drag points on a coordinate plane and watch the equation update in real time.
Equation Finder
General Form Converter
Real-World Modeller
Daily Challenge
Find the equation of the line through $(-3, 4)$ and $(5, 0)$, then convert it to general form. Time yourself — can you do it in under 90 seconds?
Printable Worksheets
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