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Short Study Notes — Equation of a Line
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Equation of a Straight Line — Cambridge IGCSE 0580 Maths Extended (2026)
Two reliable forms — y=mx+c and y−y1=m(x−x1) — let you write the equation of any line from a point and a gradient, or from two points. Cambridge tests this in nearly every Paper 4 coordinate-geometry question.
What you’ll learn
Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).
E3.2 — Interpret and obtain the equation of a straight-line graph in the form y=mx+c.
Find the equation of a line through given points or with given gradient and a point.
Slope-intercept form
y=mx+c. Read off gradient and y-intercept directly.
y=mx+c.
m = gradient (slope).
c = y-intercept (where the line crosses the y-axis).
To plot: mark the y-intercept at (0,c), then use the gradient to step rise/run for a second point.
c is where the line cuts the y-axis; m is the rise over the run of any step along it.
Worked. Plot y=−2x+5.
y-intercept at (0,5).
Gradient −2 means: every 1 right, go 2 down. So (1,3), (2,1), etc.
To find c given a point and gradient. Substitute.
Worked. Line has gradient 3 and passes through (2,7). Find c.
7=3(2)+c⇒c=1.
Equation: y=3x+1.
m = gradient.
c = y-intercept.
Substitute one point to find c if gradient is known.
Point-gradient form
y−y1=m(x−x1). The fastest way when you have one point and a gradient.
Formula.y−y1=m(x−x1).
This works when you know:
A specific point (x1,y1) ON the line.
The gradient m.
Worked. Line with gradient 4 through (2,−3).
y−(−3)=4(x−2).
y+3=4x−8.
y=4x−11 (rearranged to slope-intercept).
Worked. Line through (−1,5) with gradient −21.
y−5=−21(x−(−1)).
y−5=−21(x+1).
y=−21x−21+5=−21x+29.
When the gradient is a fraction, you may want to multiply through to clear fractions.
y−y1=m(x−x1).
Best when you have a point + a gradient.
Rearrange to y=mx+c for the final form.
From two points
Compute the gradient, then plug into point-gradient form using either point.
Method.
Compute the gradient m=x2−x1y2−y1.
Use point-gradient form with EITHER given point.
Rearrange.
Worked. Line through A(1,4) and B(5,12).
m=5−112−4=48=2.
Using A: y−4=2(x−1)⇒y=2x+2.
Check with B: 2(5)+2=12 ✓.
Find m from the two points.
Use point-gradient with EITHER point — same final equation.
Always check the OTHER point satisfies your equation.
Parallel and perpendicular lines
Parallel: same gradient. Perpendicular: negative reciprocal.
Parallel. Two lines are parallel if and only if they have the SAME gradient.
Worked. Line parallel to y=3x+1 through (2,4).
Gradient: 3.
y−4=3(x−2)⇒y=3x−2.
Perpendicular. Two lines are perpendicular when their gradients multiply to −1. Each is the negative reciprocal of the other.
Parallel lines share a gradient; perpendicular lines have gradients that multiply to −1.
Worked. Line perpendicular to y=2x+5 through (0,3).
Original gradient: 2.
Perpendicular gradient: −21.
y−3=−21(x−0)⇒y=−21x+3.
Special case. If a line is vertical (x=a), the perpendicular is horizontal (y=b). Their gradients are undefined and zero respectively — the "m1⋅m2=−1" rule doesn't apply directly.
Parallel: same gradient.
Perpendicular: m2=−m11.
Vertical and horizontal are perpendicular (special case).
Use point-gradient form with the given point and the new gradient.
Horizontal and vertical lines
y=c is horizontal. x=a is vertical. Don't try to put them in y=mx+c form.
Horizontal line y=c. Constant y. Gradient 0. Cannot use y=mx+c in the usual way (it works, with m=0, but the equation is just y=c).
Vertical line x=a. Constant x. Gradient undefined. CANNOT be written as y=mx+c — the slope-intercept form simply doesn't apply.
Recognising them.
"Find the line through (3,4) parallel to the y-axis" → vertical → x=3.
"Find the line through (3,4) parallel to the x-axis" → horizontal → y=4.
Horizontal: y=c (gradient 0).
Vertical: x=a (gradient undefined).
Vertical can't be written as y=mx+c.
How it’s examined
Equation-of-a-line questions appear on every Paper 4 — typically 4-6 marks combining gradient, point, and parallel/perpendicular logic. Paper 2 has 2-3 mark single-line items. Examiner reports flag forgetting to substitute the point's coordinates correctly and computing −1/m wrongly.
Worked examples, formulae, definitions and the mistakes examiners flag — everything you need to push from a pass to an A*.
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Step-by-step worked examples — Equation of a Line
Step-by-step solutions to past-paper-style questions on equation of a line, written exactly the way a tutor would explain them at the board.
1Equation from gradient and a point
Core• point-gradient form
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Question
Find the equation of the line with gradient 3 passing through (2,5).
Step-by-step solution
Step 1
Use y−y1=m(x−x1).
y−5=3(x−2)
Step 2
Expand and simplify.
y=3x−1
Answer
y=3x−1
2Equation from two points
Extended• Adapted from 0580/42 May/Jun 2024 Q11• two points
▼
Question
Find the equation of the line passing through A(−1,4) and B(3,−8).
Step-by-step solution
Step 1
Find the gradient.
m=3−(−1)−8−4=4−12=−3
Step 2
Use point-gradient form with A.
y−4=−3(x+1)
Step 3
Expand.
y=−3x+1
Answer
y=−3x+1
3Find a line parallel to a given one
Extended• parallel
▼
Question
Find the equation of the line through (4,7) that is parallel to y=2x−5.
Step-by-step solution
Step 1
Parallel → same gradient m=2.
Step 2
Use point-gradient form.
y−7=2(x−4)⟹y=2x−1
Answer
y=2x−1
4Find a line perpendicular to a given one
Extended• Adapted from 0580/42 Oct/Nov 2024 Q9• perpendicular
▼
Question
Find the equation of the line through (2,6) that is perpendicular to y=−31x+4.
Step-by-step solution
Step 1
Original gradient is −31. Perpendicular = negative reciprocal.
m⊥=3
Step 2
Apply point-gradient form.
y−6=3(x−2)⟹y=3x
Answer
y=3x
5Rearrange a general form into y=mx+c
Core• rearrange
▼
Question
State the gradient and y-intercept of 4x+2y=10.
Step-by-step solution
Step 1
Solve for y.
2y=−4x+10⟹y=−2x+5
Step 2
Read off.
m=−2,c=5
Answer
Gradient −2; y-intercept 5
6Find the intersection of two lines
Extended• Adapted from 0580/22 Oct/Nov 2023 Q14• intersection, simultaneous
▼
Question
Find the point of intersection of y=3x−4 and 2x+y=11.
Step-by-step solution
Step 1
Substitute the first equation into the second.
2x+(3x−4)=11
Step 2
Simplify.
5x−4=11⟹x=3
Step 3
Substitute back.
y=3(3)−4=5
Answer
(3,5)
7Equation of a horizontal or vertical line
Extended• horizontal, vertical
▼
Question
Write the equation of (a) the horizontal line through (2,−5) and (b) the vertical line through (7,3).
Step-by-step solution
Step 1
(a) Horizontal lines have the form y=c, where c is the y-coordinate.
y=−5
Step 2
(b) Vertical lines have the form x=k, where k is the x-coordinate.
x=7
Answer
(a) y=−5 (b) x=7
Examiner tip
The examiner report flags candidates often swap x and y for these two cases. Memorise: HORIZONTAL → y=…, VERTICAL → x=….
8Line through the midpoint of two points
Extended• midpoint, two points
▼
Question
Find the equation of the line passing through the midpoint of A(2,4) and B(8,−2) with gradient −3.
Step-by-step solution
Step 1
Midpoint of AB.
M=(22+8,24+(−2))=(5,1)
Step 2
Apply point-gradient form.
y−1=−3(x−5)
Step 3
Expand.
y=−3x+16
Answer
y=−3x+16
9Find the perpendicular bisector
Challenge• Adapted from 0580/42 May/Jun 2024 Q14• perpendicular bisector, midpoint
▼
Question
Find the equation of the perpendicular bisector of the segment from A(1,2) to B(7,6).
Step-by-step solution
Step 1
Find the midpoint.
M=(4,4)
Step 2
Gradient of AB.
mAB=7−16−2=32
Step 3
Perpendicular gradient.
m⊥=−23
Step 4
Apply point-gradient form through M.
y−4=−23(x−4)⟹y=−23x+10
Answer
y=−23x+10
Examiner tip
The mark scheme awards method marks for the midpoint, the original gradient and the perpendicular gradient separately. The examiner report flags candidates often forget the perpendicular bisector passes through M, not A or B.
10Find a line equidistant from two points
Challenge• perpendicular bisector, equidistant
▼
Question
A point P(x,y) is equidistant from A(0,0) and B(6,4). Show that P lies on the line 3x+2y=13.
Step-by-step solution
Step 1
Equidistant means P lies on the perpendicular bisector of AB. Use squared distances to avoid surds.
x2+y2=(x−6)2+(y−4)2
Step 2
Expand the right side.
x2+y2=x2−12x+36+y2−8y+16
Step 3
Cancel and simplify.
0=−12x−8y+52⟹12x+8y=52
Step 4
Divide by 4.
3x+2y=13
Answer
P lies on 3x+2y=13 (the perpendicular bisector of AB).
Examiner tip
The examiner report flags candidates often forget to square both distances to avoid square roots. The mark scheme awards method marks for the equidistance equation in distance-squared form.
Key Formulae — Equation of a Line
The formulae you need to memorise for equation of a line on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.
Slope-intercept form
y=mx+c
m
gradient
c
y-intercept
When to use
Standard form for a straight line — easiest to graph and read off.
Point-gradient form
y−y1=m(x−x1)
m
gradient
(x1,y1)
any known point on the line
When to use
Quickest way to write a line equation when you know a point and the slope.
General form
ax+by+c=0
When to use
Some answers are required in this form — rearrange to read off m and c.
Key Definitions and Keywords — Equation of a Line
Definitions to memorise and the exact keywords mark schemes credit for equation of a line answers — sharpened from recent examiner reports for the 2026 0580 sitting.
Equation of a line
Examiner keyword
An algebraic relationship between x and y that every point on the line satisfies.
y-intercept
Examiner keyword
The y-coordinate of the point where the line crosses the y-axis (x=0).
x-intercept
The x-coordinate where the line crosses the x-axis (y=0).
Collinear points
Three or more points lying on the same straight line.
Common Mistakes and Misconceptions — Equation of a Line
The traps other students keep falling into on equation of a line questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.
✕Forgetting to divide all terms when isolating y
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Why it happens
2y=−4x+10 gets simplified to y=−4x+5 instead of y=−2x+5.
How to avoid it
Divide EVERY term by the coefficient of y.
✕Using the reciprocal without changing sign
0580/42 — recurring
▼
Why it happens
Half-remembering the rule.
How to avoid it
Perpendicular = NEGATIVE reciprocal: flip AND change sign.
✕Saying parallel lines have the same equation
▼
Why it happens
Confusing "same gradient" with "same equation".
How to avoid it
Parallel = same m, generally different c. The lines are distinct unless they're identical.
✕Computing the gradient of a vertical line as ∞ instead of saying "undefined"
▼
Why it happens
Vertical lines have Δx=0, division by zero.
How to avoid it
Vertical line x=k has UNDEFINED gradient. Horizontal line y=k has gradient 0.
Practice questions
Exam-style questions with step-by-step worked solutions. Try one before checking the method.
Past paper style quiz
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Video lesson
Short walkthrough of the concepts students most often get stuck on.
Equation of a Line — frequently asked questions
The things students keep getting wrong in this sub-topic, answered.
Equation of a Line – Study Notes & Past Paper Style Questions | Cambridge IGCSE Maths 0580 Extended | Tutopiya