Every non-vertical straight line can be written as \(y = mx + c\). Read \(m\) and \(c\) and you know the line; find \(m\) and \(c\) and you can write its equation.
In \(y = mx + c\), \(m\) is the gradient and \(c\) is the \(y\)-intercept (the line crosses the \(y\)-axis at \((0, c)\)). To find an equation: get the gradient, then substitute one point to find \(c\). A vertical line has equation \(x = k\).
Fig. 1: y = 2x - 3 crosses the y-axis at (0, -3) and rises 2 for every 1 across, so m = 2 and c = -3
Section 01
Gradient and intercept.
In the lab, set \(m\) and \(c\) with the sliders. Change only \(c\): the line slides up and down but keeps its slope. Change only \(m\): the line pivots about the point \((0, c)\). Then set \(m = 2\), \(c = -3\) and compare with Fig. 1.
\(m\), the number multiplying \(x\), is the gradient.
\(c\), the number on its own, is the \(y\)-intercept: the line passes through \((0, c)\).
Order does not matter: \(y = 5 - 2x\) is \(y = -2x + 5\), so \(m = -2\) and \(c = 5\).
Special lines: \(y = k\) is horizontal (\(m = 0\)); \(x = k\) is vertical and has no \(y = mx + c\) form.
Worked example Reading the gradient and intercept
Write down the gradient and \(y\)-intercept of (a) \(y = 6x + 3\) (b) \(y = 5 - 2x\).
(a) The coefficient of \(x\) is 6 and the constant is 3.
(b) Rewrite as \(y = -2x + 5\): the coefficient of \(x\) is \(-2\) (the minus sign belongs to it).
A straight line crosses the \(y\)-axis at \((0, -3)\) and passes through \((2, 1)\). Find its equation.
The \(y\)-intercept is \(-3\), so \(c = -3\).
From \((0, -3)\) to \((2, 1)\): rise 4, run 2, so \(m = 4 \div 2 = 2\).
Put both into \(y = mx + c\).
Answer: \(y = 2x - 3\) (Fig. 1)
Examiner trap
Taking the first number as the gradient. In \(y = 5 - 2x\) the gradient is \(-2\), not 5: the gradient is always the number in front of \(x\), with its sign.
Section 02
Rearrange to y =.
(Extended) A line may be given as \(ax + by = c\), such as \(5x + 4y = 8\). To read the gradient and intercept, make \(y\) the subject first.
Worked example From ax + by = c (Extended)
Find the gradient and \(y\)-intercept of the line \(5x + 4y = 8\).
Subtract \(5x\) from both sides: \(4y = -5x + 8\).
Divide EVERY term by 4: \(y = -\dfrac{5}{4}x + 2\).
(Extended) Find the gradient and \(y\)-intercept of \(3x - 2y = 6\).
\(-2y = -3x + 6\), and dividing every term by \(-2\) gives \(y = \dfrac{3}{2}x - 3\). Gradient \(\dfrac{3}{2}\), \(y\)-intercept \(-3\).
Examiner trap
Dividing only some of the terms. From \(4y = -5x + 8\), the equation \(y = -5x + 2\) divides only the 8 by 4. Every term on the right must be divided.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
What is the gradient of the line \(y = 7 - 3x\)?
Quick check+10 XP
Where does the line \(y = 4x - 9\) cross the \(y\)-axis?
Quick check+10 XP
Which is the equation of the line with gradient 5 and \(y\)-intercept \(-2\)?
Quick check+10 XP
Which is the equation of the vertical line through \((3, -1)\)?
Section 03
Finding the equation.
Whatever you are given, the plan is the same: find the gradient \(m\), substitute the coordinates of one point on the line into \(y = mx + c\) to find \(c\), then write the equation in its simplest form.
Worked example Gradient and one point
Find the equation of the line with gradient 3 that passes through \((2, 1)\).
Find the equation of the line through \((4, -1)\) and \((4, 6)\).
Both points have \(x = 4\), so the line is vertical.
Every point on it has \(x\)-coordinate 4.
Answer: \(x = 4\)
Examiner trap
Putting the coordinates in the wrong places: for the point \((2, 1)\), \(x = 2\) and \(y = 1\). Writing \(2 = 3 \times 1 + c\) gives the wrong \(c\). Always check your final equation with the point.
Stage 2 · Exam
Exam-style questions
Unlocks once the four checks above are done. Worth more XP, written to IGCSE Paper 2 and Paper 4 standard.
Finish the four checks above to unlock the exam questions
Exam style+20 XP
(Extended) Find the gradient and \(y\)-intercept of the line \(5x + 4y = 8\).
Exam style+20 XP
Find the equation of the line with gradient 3 that passes through \((2, 1)\).
Exam style+20 XP
(Extended) Find the equation of the line through \((-2, -1)\) and \((4, 8)\).
Exam style+20 XP
A straight-line graph crosses the \(y\)-axis at \((0, 4)\) and the \(x\)-axis at \((6, 0)\). What is its equation?
Skill unlocked
Straight-line equations, cracked.
This skill is now recorded on your profile. Keep the chain going.