Two facts fix a line: a gradient and a point, or simply two points. The point-gradient formula turns either into an equation in one step.
Today's hook, You will use the point-gradient formula in every calculus question that asks for a tangent. Learning it properly here saves you from re-deriving it under exam pressure later.
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Recall, your gut answer first
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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Recall, your gut answer first
+5 XP warm-up
A line has gradient 2 and passes through $(3, 5)$. Without any formula, work out its $y$-intercept by reasoning backwards from the point. Then write its equation.
Before you work it out, what is your instinct? Write it down, then check it against the lesson.
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One point and a gradient is enough
Work through the core explanation before applying it.
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One point and a gradient is enough
+5 XP to read
If a line has gradient $m$ and passes through $(x_1, y_1)$, then for any other point $(x, y)$ on it the gradient formula gives $\dfrac{y - y_1}{x - x_1} = m$. Multiplying out gives the point-gradient formula, which is the gradient formula rearranged, not a new fact.
$y - y_1 = m(x - x_1)$ two points: $m = \dfrac{y_2 - y_1}{x_2 - x_1}$ first, then substitute either point
It is the gradient formula rearranged
$\frac{y-y_1}{x-x_1} = m$ multiplied by $(x - x_1)$. If you forget the formula you can rebuild it in one line.
Either point works
With two points, substituting either one into the formula gives the same final equation. Pick the one with easier numbers.
Mind the double negative
Through $(-2, 5)$ the formula gives $y - 5 = m(x - (-2))$, which is $y - 5 = m(x + 2)$.
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What you'll master
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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What you'll master
Know
Key facts
The point-gradient formula is $y - y_1 = m(x - x_1)$.
It follows directly from the gradient formula, so it can be rebuilt if forgotten.
Given two points, find the gradient first, then substitute either point.
The result can be left in point-gradient form, or rearranged to gradient-intercept or general form as the question asks.
Understand
Concepts
Why one point and a gradient determine a line uniquely.
Why either of two given points may be substituted without changing the answer.
Why the formula is the gradient definition rearranged rather than something new to memorise.
Can do
Skills
Find the equation of a line from a point and a gradient.
Find the equation of a line through two points.
Present the answer in whichever form a question requests.
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Key terms
Point-gradient formulaA way to build a line equation from one point and the gradient. Like this: gradient 2 through $(3,5)$ gives $y - 5 = 2(x - 3)$.
SubstitutionReplacing $x_1$ and $y_1$ in the formula with the coordinates of the known point. Like this: for $(-2, 7)$ you substitute $x_1 = -2$ and $y_1 = 7$.
Two-point problemFinding a line when you are given two points and no gradient. Like this: through $(1,3)$ and $(5,11)$, first find $m = 2$, then substitute.
Collinear pointsPoints lying on the same straight line, which you can test by checking the gradients match. Like this: $(0,1)$, $(2,5)$ and $(4,9)$ all give gradient 2, so they are collinear.
Vertical line equationA line with undefined gradient, written $x = k$ and not expressible by the point-gradient formula. Like this: through $(4,1)$ and $(4,9)$ the equation is $x = 4$.
Required formThe arrangement the question asks for, which may be point-gradient, gradient-intercept or general. Like this: "in general form" means finish at $ax + by + c = 0$.
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From a point and a gradient
Work through the core explanation before applying it.
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From a point and a gradient
core concept
Let $(x, y)$ be any point on the line other than the known point $(x_1, y_1)$. The gradient between them must equal $m$, so $\dfrac{y - y_1}{x - x_1} = m$.
Multiplying both sides by $(x - x_1)$ gives $y - y_1 = m(x - x_1)$. That is the point-gradient formula, and it is just the gradient definition cleared of its fraction.
For gradient 2 through $(3, 5)$: $y - 5 = 2(x - 3)$. Expanding gives $y = 2x - 1$ if the question wants gradient-intercept form.
Watch negative coordinates. Through $(-2, 5)$ with gradient $-3$: $y - 5 = -3(x + 2)$, because $x - (-2) = x + 2$. Two sign slips here cancel and hide the error.
Quick check: which equation has gradient 4 and passes through $(1, 2)$?
Point-gradient formula $y - y_1 = m(x - x_1)$, obtained by clearing the fraction in the gradient definition. Substitute the point for $x_1$ and $y_1$, keeping the signs, then rearrange to whichever form is asked for.
Pause, copy the one-line derivation from the gradient formula, the worked gradient 2 through $(3,5)$, and the negative-coordinate warning, into your book.
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From two points
core concept
We just saw that a point and a gradient fix a line. That raises a question: what if you are given two points and no gradient at all? This card answers it → calculate the gradient from the two points first, then you are back to the previous case.
Step 1: find the gradient with $m = \dfrac{y_2 - y_1}{x_2 - x_1}$. Step 2: substitute that gradient and **either** point into the point-gradient formula.
Through $(1, 3)$ and $(5, 11)$: $m = \dfrac{11 - 3}{5 - 1} = 2$. Using $(1, 3)$: $y - 3 = 2(x - 1)$, so $y = 2x + 1$.
Using the other point gives $y - 11 = 2(x - 5)$, which expands to $y = 2x + 1$ as well. Either choice reaches the same line, so pick the friendlier numbers.
Two points with the same $x$. Then the run is zero, the gradient is undefined, and the point-gradient formula does not apply. The line is vertical: through $(4,1)$ and $(4,9)$ the equation is simply $x = 4$.
Fill the blank: the gradient of the line through $(2, 1)$ and $(6, 9)$ is .
With two points: find the gradient first, then substitute it and either point into $y - y_1 = m(x - x_1)$. Both points give the same final equation. If the two $x$-coordinates match, the line is vertical and its equation is $x = k$.
Pause, copy the two-step method, the worked $(1,3)$ and $(5,11)$ giving $y = 2x+1$ from either point, and the vertical-line exception, into your book.
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Presenting the answer, and checking it
core concept
We just saw both routes to an equation. That raises a question: when is the answer actually finished? This card answers it → when it is in the form the question asked for and both given points satisfy it.
Read the instruction. "In gradient-intercept form" means finish at $y = mx + b$. "In general form" means finish at $ax + by + c = 0$ with integer coefficients. Leaving it in point-gradient form when a form was specified loses a mark.
Always check by substituting a given point back in. For $y = 2x + 1$ and the point $(5, 11)$: $2(5) + 1 = 11$ ✓. If it fails, the gradient or the substitution is wrong.
Three points are collinear if the gradient between the first pair equals the gradient between the second pair. That is the standard way to prove collinearity.
Substituting back catches almost every slip. It costs one line and it tests the gradient and the arithmetic together. Do it on every equation-of-a-line question.
Which is NOT a way to check the equation of a line through two given points?
Finish in the form the question asks for. Check by substituting a given point back into your equation; if it does not satisfy it, the gradient or the arithmetic is wrong. Three points are collinear when the gradients between successive pairs are equal.
Pause, copy the three answer forms, the substitute-back check with a worked example, and the collinearity test, into your book.
Worked examples · 3 in a row, reveal as you go
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Work examples end to end
Follow the reasoning through complete worked solutions.
PROBLEM 1 · POINT AND GRADIENT
Find the equation of the line with gradient $-3$ passing through $(-2, 5)$. Give your answer in gradient-intercept form.
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$y - 5 = -3(x - (-2))$
Substitute $m = -3$, $x_1 = -2$, $y_1 = 5$.
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$y - 5 = -3(x + 2) = -3x - 6$
$x - (-2)$ becomes $x + 2$; expand carefully.
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$y = -3x - 1$
Add 5 to both sides. Check $(-2,5)$: $-3(-2) - 1 = 5$ ✓
PROBLEM 2 · TWO POINTS
Find the equation of the line through $(1, 3)$ and $(5, 11)$, in general form.
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$m = \dfrac{11 - 3}{5 - 1} = \dfrac{8}{4} = 2$
Gradient first.
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$y - 3 = 2(x - 1)$, so $y = 2x + 1$
Substitute the easier point.
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$2x - y + 1 = 0$
Move to general form. Check $(5,11)$: $10 - 11 + 1 = 0$ ✓
PROBLEM 3 · PROVING COLLINEARITY
Show that $A(0, 1)$, $B(2, 5)$ and $C(4, 9)$ are collinear.
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$m_{AB} = \dfrac{5 - 1}{2 - 0} = 2$
Gradient of the first pair.
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$m_{BC} = \dfrac{9 - 5}{4 - 2} = 2$
Gradient of the second pair.
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$m_{AB} = m_{BC}$ and they share point $B$, so $A$, $B$ and $C$ are collinear
Equal gradients through a common point means one straight line.
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Quick-fire practice
Work through the core explanation before applying it.
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Quick-fire practice
+10 XP
Find the equation of the line with gradient 5 through $(0, -2)$.
Find the equation through $(2, 1)$ and $(4, 7)$.
Find the equation with gradient $-\tfrac{1}{2}$ through $(6, 0)$.
What is the equation of the line through $(3, 2)$ and $(3, 9)$?
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Revisit your reasoning
Run the quick drill and copy the summary into your book.
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Revisit your reasoning
At the start you found the $y$-intercept of the line through $(3,5)$ with gradient 2 by reasoning backwards. Redo it with the point-gradient formula and confirm you get the same equation. Which method would you trust under time pressure, and why?
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Multiple choice
Answer the drill bank and rate your confidence.
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Multiple choice
+5 XP per correct · +25 XP all-correct
Pick your answer, then rate your confidence, that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
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Short answer
Write full responses, then check them against the model answers.
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Short answer
ApplyBand 43 marks
Q1. Find the equation of the line passing through $(-3, 4)$ and $(1, -8)$, giving your answer in general form. (3 marks)
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ApplyBand 43 marks
Q2. A line has gradient $\tfrac{2}{5}$ and passes through $(10, -1)$. Find its equation in gradient-intercept form and state its $y$-intercept. (3 marks)
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UnderstandBand 42 marks
Q3. Explain why substituting either of two given points into the point-gradient formula produces the same final equation. (2 marks)
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📖 Comprehensive answers (click to reveal)
Practice 1: $y = 5x - 2$. Practice 2: $m = 3$, $y = 3x - 5$. Practice 3: $y = -\frac{1}{2}x + 3$. Practice 4: $x = 3$, a vertical line with undefined gradient.
Q3 (2 marks): Both points lie on the same line, and that line has a single gradient $m$ which is used in either substitution [1]. The formula states that every point on the line satisfies $y - y_1 = m(x - x_1)$, so anchoring at either known point describes the same set of points and rearranges to the same equation [1].
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Review and finish
Take the module quiz if you are ready, then mark the lesson complete.
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Boss battle · Line Builder
earn bronze · silver · gold
Build line equations from points and gradients, finishing in whichever form is asked. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.