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.
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.
- 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 $ax + by + c = 0$.
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 $c = 3$.
Right: $5 = 3(2) + c$ → $5 = 6 + c$ → $c = -1$. Always multiply $m$ by the x-coordinate first.
Wrong: Forgetting to convert to general form with integer coefficients. $y = \dfrac{1}{2}x + 3$ becomes $x + 2y + 6 = 0$ (signs wrong).
Right: $y = \dfrac{1}{2}x + 3$ → $2y = x + 6$ → $x - 2y + 6 = 0$.
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.
$m = -\dfrac{3}{4}$. $c = 5$. $y = -\dfrac{3}{4}x + 5$.
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 $y - y_1 = m(x - x_1)$.
$y = 2x + c$. $7 = 2(3) + c$. $c = 1$.
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.
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.
$y = \dfrac{1}{2}x + 3$ → $x - 2y + 6 = 0$.
Substitute directly: $y = -\dfrac{3}{4}x + 5$
Multiply by 4: $4y = -3x + 20$
Rearrange: $3x + 4y - 20 = 0$
Write: $y = 2x + c$
Substitute $(3, 7)$: $7 = 2(3) + c$
$7 = 6 + c$ → $c = 1$
Answer: $y = 2x + 1$
Gradient: $m = \dfrac{-1 - 5}{4 - (-2)} = \dfrac{-6}{6} = -1$
Find c: $y = -x + c$. Using $(-2, 5)$: $5 = -(-2) + c$ → $5 = 2 + c$ → $c = 3$
Check: Second point: $-1 = -(4) + 3 = -1$ ✓
Brain Trainer
4 quick-fire drills. Beat the clock.
5 MCQs and 3 short-answer questions. Target: 80% accuracy.
Your answer:
$y = \dfrac{3}{2}x + c$
$-1 = \dfrac{3}{2}(4) + c = 6 + c$
$c = -7$
$y = \dfrac{3}{2}x - 7$
Your answer:
(a) $m = \dfrac{0 - 4}{5 - (-3)} = \dfrac{-4}{8} =$ $-\dfrac{1}{2}$
(b) $y = -\frac{1}{2}x + c$. Using $(5, 0)$: $0 = -\frac{5}{2} + c$ → $c = \dfrac{5}{2}$
$y = -\dfrac{1}{2}x + \dfrac{5}{2}$
(c) $2y = -x + 5$ → $x + 2y - 5 = 0$
Your answer:
(a) $m = \dfrac{360 - 180}{5 - 2} = \dfrac{180}{3} =$ $60$. The gradient is $60 per hour (the hourly rate).
(b) $C = 60t + c$. Using $(2, 180)$: $180 = 60(2) + c$ → $c = 60$
$C = 60t + 60$
(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.
Consolidate and reflect before moving on.
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 $y = mx + c$.)
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.
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.
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.
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?
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