Forms of the Equation of a Line
The same straight line can be written as y = mx + c or as ax + by + c = 0. Neither form is more correct; each makes a different fact obvious, and being fluent in moving between them is what the rest of this focus area assumes.
Take $y = 2x + 6$ and move every term to the left-hand side. What do you get? Now start from $2x - y + 6 = 0$ and make $y$ the subject. Did you get back where you started?
Gradient–intercept form, $y = mx + c$, puts the two most useful numbers on display. General form, $ax + by + c = 0$, hides them but handles every line, including the vertical ones. Rearranging is ordinary equation solving, nothing more.
$$y = mx + c \qquad \longleftrightarrow \qquad ax + by + c = 0$$
There is only ever one line. The two equations are two descriptions of the same set of points, in the way that $\tfrac{1}{2}$ and $0.5$ are two descriptions of one number. Changing form changes nothing about the line at all.
Know
- gradient–intercept form $y = mx + c$, where $m$ is the gradient and $c$ is the y-intercept
- general form $ax + by + c = 0$, where $a$, $b$ and $c$ are constants and $a$ and $b$ are not both zero
- that from general form the gradient is $-\dfrac{a}{b}$ and the y-intercept is $-\dfrac{c}{b}$ when $b \neq 0$
Understand
- why the two forms describe the same line, and why rearranging changes nothing geometric
- why general form can represent a vertical line but gradient–intercept form cannot
- why the letter $c$ carries a different meaning in each form, and why that is a naming clash rather than a contradiction
Can Do
- rearrange from gradient–intercept form to general form, clearing fractions and using a positive leading coefficient
- rearrange from general form to gradient–intercept form and state the gradient and y-intercept
- find the gradient and y-intercept of a line given in general form without fully rearranging
Every non-vertical straight line can be written as $y = mx + c$. This is the form you have used since Year 9, and its advantage is that it announces its own gradient and y-intercept.
Every straight line at all, vertical ones included, can be written as $ax + by + c = 0$. This is called general form, and its advantage is exactly that generality.
These are not two different lines, or two different kinds of line. They are two ways of writing the same equation, related by ordinary algebra. Rearranging an equation never changes the set of points that satisfy it, and the line is that set of points.
A useful habit: when a question gives you general form, ask whether you actually need $m$ and $c$. Sometimes you do and should rearrange; sometimes, as when finding intercepts, general form is perfectly convenient as it stands.
Two steps, in this order.
Clear any fractions first. Multiply every term by the denominator. Starting from $y = \tfrac{2}{3}x - 5$, multiplying by $3$ gives $3y = 2x - 15$.
Then move everything to one side. Subtracting $3y$ gives $0 = 2x - 3y - 15$, which is written $2x - 3y - 15 = 0$.
Doing these in the other order works but leaves fractions to clear from a longer expression, so it costs more effort for the same result.
By convention the coefficient of $x$ is written positive. If your rearrangement produces $-2x + 3y + 15 = 0$, multiply the whole equation by $-1$. Both are correct equations for the same line; the convention just means two students' answers can be compared without a moment's doubt.
This direction is the more common one in practice, because it is how you extract the gradient.
Make $y$ the subject. From $4x + 2y - 7 = 0$, move the other terms across to get $2y = -4x + 7$, then divide every term by $2$:
$y = -2x + \dfrac{7}{2}$
So the gradient is $-2$ and the y-intercept is $\tfrac{7}{2}$.
Divide every term, not just the first. Dropping the division on the constant is the single commonest slip here, and it produces a line with the right gradient in the wrong place, which no sketch will obviously contradict.
Doing the rearrangement in general, rather than with numbers, gives a shortcut worth having.
From $ax + by + c = 0$, move and divide: $by = -ax - c$, so $y = -\dfrac{a}{b}x - \dfrac{c}{b}$.
Comparing with $y = mx + c$ shows that the gradient is $-\dfrac{a}{b}$ and the y-intercept is $-\dfrac{c}{b}$.
So for $5x - 2y + 8 = 0$, the gradient is $-\dfrac{5}{-2} = \dfrac{5}{2}$ and the y-intercept is $-\dfrac{8}{-2} = 4$, both obtained without writing a single line of rearrangement.
Use the shortcut for speed, but be able to derive it. A remembered formula with the sign forgotten is worse than no formula, and the derivation takes two lines.
There is a genuine naming clash between the two forms, and it catches people out.
In $y = mx + c$, the letter $c$ is the y-intercept. In $ax + by + c = 0$, the letter $c$ is just the constant term, and it is not the y-intercept.
For $4x + 2y - 7 = 0$ the constant is $-7$, while the y-intercept is $\tfrac{7}{2}$. They are not even the same sign.
This is a collision of conventions, not a piece of mathematics. Two different traditions both reached for the letter $c$, and both stuck. The defence is to notice which form you are looking at before you interpret any letter in it.
Where confusion is likely, say what you mean: "the constant term is $-7$" and "the y-intercept is $\tfrac{7}{2}$" cannot be misread, whereas "$c = -7$" can.
Watch Me Solve It · 3 examples
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1Clear the fraction$3y = 3 \times \frac{2}{3}x - 3 \times 5$$3y = 2x - 15$Multiply every term by $3$, including the constant.
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2Move everything to one side$0 = 2x - 3y - 15$$2x - 3y - 15 = 0$Subtracting $3y$ from both sides leaves the coefficient of $x$ positive, so no final sign change is needed.
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3Check with a point$x = 3: \quad y = \tfrac{2}{3}(3) - 5 = -3$$2(3) - 3(-3) - 15 = 6 + 9 - 15 = 0$The point $(3, -3)$ satisfies both equations, which confirms the rearrangement.
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1Isolate the y term$2y = -4x + 7$Move both other terms to the right-hand side, changing their signs.
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2Divide every term by the coefficient of y$y = -2x + \frac{7}{2}$Every term, including the constant. This is where marks are most often lost.
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3Read off the two valuesComparing with $y = mx + c$ gives $m = -2$ and $c = \tfrac{7}{2}$. Note that the constant in the original equation was $-7$, which is neither of these.
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1Identify a, b and c$a = 5, \quad b = -2, \quad c = 8$The signs belong to the coefficients. Here $b$ is negative, which is what the minus sign in front of $2y$ means.
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2Apply the shortcut$m = -\frac{a}{b} = -\frac{5}{-2} = \frac{5}{2}$$\text{y-intercept} = -\frac{c}{b} = -\frac{8}{-2} = 4$Two negatives in each expression, so both results come out positive.
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3Verify by rearranging$-2y = -5x - 8$$y = \frac{5}{2}x + 4$Dividing by $-2$ changes the sign of every term. The result agrees with the shortcut.
Brain Trainer · 5 problems
Five rearranging items. Where a form is requested, give it exactly, with the coefficient of $x$ positive in general form.
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1 Write $y = 3x - 8$ in general form.
Move both terms across.$3x - y - 8 = 0$ -
2 Write $y = -\tfrac{1}{2}x + 4$ in general form.
Multiply by $2$ first, then collect.$x + 2y - 8 = 0$ -
3 Find the gradient of $6x + 3y - 9 = 0$.
$m = -\dfrac{a}{b} = -\dfrac{6}{3}$$m = -2$ -
4 Find the y-intercept of $2x - 5y + 15 = 0$.
$-\dfrac{c}{b} = -\dfrac{15}{-5}$$3$ -
5 Write $3x - 4y + 12 = 0$ in gradient–intercept form.
$-4y = -3x - 12$, then divide every term by $-4$.$y = \tfrac{3}{4}x + 3$
Multiple Choice · 5 questions
Written in general form, the line $y = 5x - 2$ is:
The gradient of the line $3x + 6y - 12 = 0$ is:
For the line $2x + 4y - 9 = 0$, which statement is correct?
Written in general form, the line $y = \tfrac{3}{5}x + 2$ is:
A student rearranges $2x - y + 6 = 0$ into $y = 2x + 6$. The two equations:
Short Answer · 3 questions
Two forms, one line
$y = mx + c$ displays the gradient and y-intercept. $ax + by + c = 0$ covers every line including vertical ones. Rearranging between them changes nothing geometric.
Fractions first, then collect
Going to general form, multiply through by the denominator before moving terms, and finish with a positive coefficient of $x$ by convention.
Divide every term
Coming back, isolate the $y$ term and then divide the whole equation by its coefficient. Dividing only the $x$ term is the commonest error in the topic.
The two meanings of c
In $y = mx + c$ it is the y-intercept; in $ax + by + c = 0$ it is only the constant, and the y-intercept is $-\dfrac{c}{b}$. Check which form you are reading before interpreting any letter.
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