How do you calculate the distance between two points in coordinate geometry?

In coordinate geometry, the distance between two points (x1, y1) and (x2, y2) is calculated using the distance formula: Distance = √((x2 - x1)² + (y2 - y1)²). This formula is derived from the Pythagorean theorem and is essential for solving problems in the Cambridge IGCSE Mathematics curriculum.

What is the formula for finding the midpoint of a line segment in coordinate geometry?

The midpoint of a line segment with endpoints (x1, y1) and (x2, y2) is found using the midpoint formula: Midpoint = ((x1 + x2)/2, (y1 + y2)/2). This formula gives the coordinates of the point that is exactly halfway between the two endpoints, a key concept in the Cambridge IGCSE Mathematics syllabus.

How is the gradient of a line determined in coordinate geometry?

The gradient (or slope) of a line in coordinate geometry is determined by the formula: Gradient = (y2 - y1) / (x2 - x1), where (x1, y1) and (x2, y2) are two distinct points on the line. The gradient indicates the steepness and direction of the line, which is a fundamental concept in the Cambridge IGCSE Mathematics curriculum.

Why is understanding the gradient important in coordinate geometry?

Understanding the gradient is crucial in coordinate geometry because it describes the rate of change between two variables and helps in identifying the direction and steepness of a line. This concept is widely used in various mathematical problems and real-world applications, making it an important part of the Cambridge IGCSE Mathematics curriculum.

Can you explain how distance, midpoint, and gradient are interconnected in coordinate geometry?

Distance, midpoint, and gradient are interconnected concepts in coordinate geometry. The distance formula helps determine the length of a line segment, the midpoint formula finds the center point of a segment, and the gradient indicates the slope of the line. Together, these concepts allow for a comprehensive analysis of geometric figures and their properties, which are essential skills in the Cambridge IGCSE Mathematics curriculum.

Why is "Forgetting to square the differences inside the distance formula" a common mistake in Distance, Midpoint and Gradient?

Why it happens: Students write d = √(x_2) - x_1 + √(y_2) - y_1. How to avoid it: Always: square each difference, ADD, then square root.

Why is "Subtracting instead of averaging in the midpoint formula" a common mistake in Distance, Midpoint and Gradient?

Why it happens: Confusion with the gradient formula. How to avoid it: Midpoint = ADD coordinates and divide by 2. Source: 0580/22 — recurring

Why is "Computing $\dfrac{x_{2} - x_{1}}{y_{2} - y_{1}}$ instead of $\dfrac{y_{2} - y_{1}}{x_{2} - x_{1}}$" a common mistake in Distance, Midpoint and Gradient?

Why it happens: Confusing rise/run with run/rise. How to avoid it: Gradient is RISE over RUN — y on top, x on bottom.

Why is "Using the reciprocal as the perpendicular gradient (without negating)" a common mistake in Distance, Midpoint and Gradient?

Why it happens: Half-remembering the rule. How to avoid it: Perpendicular gradient = NEGATIVE RECIPROCAL. Both flip and change sign.

Cambridge IGCSE Mathematics (0580)

Distance, Midpoint and Gradient

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Short Notes - Distance, Midpoint and Gradient

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Detailed Study Notes

Detailed notes on Coordinate Geometry for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Distance, Midpoint, Gradient — Cambridge IGCSE 0580 Maths Extended (2026)

Three formulas, three coordinate-geometry workhorses. Find the length of a line segment, the midpoint, and the gradient — building blocks for tangents, perpendiculars, and quadrilateral classification.

At a glance

Distance between $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ : $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ .
Midpoint: $\left(\dfrac{x_{1} + x_{2}}{2}, \dfrac{y_{1} + y_{2}}{2}\right)$ — the AVERAGE of coordinates.
Gradient: $\dfrac{y_{2} - y_{1}}{x_{2} - x_{1}}$ . Rise over run.
Parallel lines: SAME gradient.
Perpendicular lines: gradients multiply to $-1$ → $m_{2} = -\dfrac{1}{m_{1}}$ .
Horizontal line: gradient $0$ . Vertical line: gradient undefined.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E3.1 — Calculate the gradient and length of a straight line; find the midpoint of a line segment.
Recognise parallel and perpendicular lines from their gradients.

Distance between two points

Pythagoras applied to coordinates. Square the differences, add, square root.

Formula. Distance between $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$ : $d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}.$

This is just Pythagoras' theorem applied to the right triangle formed by the horizontal and vertical legs.

The distance is the hypotenuse — square the horizontal and vertical gaps, add, take the root.

Worked. Find the distance from $A(1, 2)$ to $B(4, 6)$ .

$d = \sqrt{(4 - 1)^{2} + (6 - 2)^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5$ .

Worked (negatives). Find the distance from $P(-2, 5)$ to $Q(3, -7)$ .

$d = \sqrt{(3 - (-2))^{2} + (-7 - 5)^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13$ .

Sign-doesn't-matter rule. Since each difference is squared, the order of subtraction doesn't matter: $(x_{2} - x_{1})^{2} = (x_{1} - x_{2})^{2}$ .

Square the differences.
Add.
Square root.
Order of subtraction doesn't matter (squared).

Midpoint of a line segment

Average each coordinate. Half-way along the line.

Formula. Midpoint of segment from $(x_{1}, y_{1})$ to $(x_{2}, y_{2})$ : $M = \left(\dfrac{x_{1} + x_{2}}{2}, \dfrac{y_{1} + y_{2}}{2}\right).$

Worked. Midpoint of $A(2, -1)$ and $B(8, 5)$ .

$M = \left(\dfrac{2 + 8}{2}, \dfrac{-1 + 5}{2}\right) = (5, 2)$ .

The midpoint sits exactly halfway — average the x-values and the y-values.

Reverse problem. "Given $M$ and one endpoint, find the other endpoint."

If $M = (5, 2)$ and one endpoint is $A(2, -1)$ , the other endpoint $B$ satisfies $\dfrac{2 + x}{2} = 5$ → $x = 8$ . Similarly $y = 5$ . So $B(8, 5)$ ✓.

Average each coordinate.
Same midpoint regardless of order of endpoints.
Reverse: solve for the missing endpoint.

Gradient of a line

Rise over run. Positive: rises left to right. Negative: falls.

Formula. Gradient of the line through $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ : $m = \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}}.$

The order matters HERE — but be consistent in numerator and denominator.

Sign of the gradient.

$m > 0$ : line rises left to right.
$m < 0$ : line falls left to right.
$m = 0$ : horizontal line.
Vertical line: $m$ is undefined (zero denominator).

Gradient tells you the steepness and direction — flat means zero, vertical means undefined.

Worked. Gradient through $A(1, 2)$ and $B(4, 11)$ .

$m = \dfrac{11 - 2}{4 - 1} = \dfrac{9}{3} = 3$ .

Parallel lines: SAME gradient.

Perpendicular lines. $m_{1} \times m_{2} = -1$ . Each is the negative reciprocal of the other.

If $m_{1} = 2$ , perpendicular gradient is $-\dfrac{1}{2}$ .
If $m_{1} = -\dfrac{3}{4}$ , perpendicular gradient is $\dfrac{4}{3}$ .

$m = \dfrac{\Delta y}{\Delta x}$ .
Positive: up. Negative: down. Zero: flat.
Parallel ⇔ same $m$ .
Perpendicular ⇔ $m_{1} \cdot m_{2} = -1$ .

Quick recap

Distance: Pythagoras on coordinate differences.
Midpoint: average each coordinate.
Gradient: rise over run.
Parallel: same gradient.
Perpendicular: gradients multiply to $-1$ .
Horizontal: $m = 0$ . Vertical: undefined.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Distance formula — $d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ .
Midpoint formula — $M = \left(\dfrac{x_{1} + x_{2}}{2}, \dfrac{y_{1} + y_{2}}{2}\right)$ .
Gradient formula — $m = \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .
Parallel — same gradient.
Perpendicular — gradients multiply to $-1$ .

How it’s examined

These three formulas appear inside almost every Paper 4 coordinate-geometry question (5-7 marks). Common combo: find midpoint, find gradient of perpendicular, write equation of perpendicular bisector. Examiner reports flag inverted gradient (run/rise instead of rise/run) and arithmetic slips with negatives.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E3.1); 0580/22 May/Jun 2024 — Q8 (gradient + midpoint); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Distance, Midpoint and Gradient

Step-by-step solutions to past-paper-style questions on distance, midpoint and gradient, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Distance, Midpoint and Gradient

Almost every distance, midpoint and gradient exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single instruction — find the distance / midpoint / gradient, or find a missing coordinate — using one of the three coordinate formulae.

How to approach it

Pick the right formula and substitute carefully: distance squares the differences and adds; midpoint averages the coordinates; gradient is rise over run. For a missing coordinate, set up the formula as an equation and solve.

Common trap

Subtracting instead of averaging in the midpoint formula, or flipping the gradient to $\frac{\Delta x}{\Delta y}$ . Examiner reports flag both — and forgetting the negative when finding a perpendicular gradient.

Show that / prove

Recognise it by

The words show that, prove or verify — typically asking you to demonstrate a triangle is isosceles, or that three points are collinear.

How to approach it

Compute the relevant distances or gradients, then finish with an explicit conclusion: state which two sides are equal, or that the gradients match and a point is shared.

Common trap

Computing the distances or gradients and stopping there. Examiner reports stress the final comparison line — and for collinearity, the shared-point clause — is where the last mark sits.

1Find the distance between two points
CoreDirect calculation• distance
Question
Find the distance between $A(1, 2)$ and $B(7, 10)$ .
Step-by-step solution
1. Step 1
  $d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ .
  $d = \sqrt{(7 - 1)^{2} + (10 - 2)^{2}}$
2. Step 2
  Compute.
  $= \sqrt{36 + 64} = \sqrt{100} = 10$
Answer
$d = 10$ units
2Find the midpoint of a segment
CoreDirect calculation• midpoint
Question
Find the midpoint of the segment joining $P(-3, 5)$ and $Q(7, -1)$ .
Step-by-step solution
1. Step 1
  Average each coordinate.
  $M = \left(\dfrac{-3 + 7}{2},\ \dfrac{5 + (-1)}{2}\right)$
2. Step 2
  Simplify.
  $= (2, 2)$
Answer
$M = (2, 2)$
3Find the gradient between two points
CoreDirect calculation• Adapted from 0580/22 May/Jun 2024 Q8• gradient
Question
Find the gradient of the line through $A(2, 3)$ and $B(8, 15)$ .
Step-by-step solution
1. Step 1
  $m = \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .
  $m = \dfrac{15 - 3}{8 - 2} = \dfrac{12}{6} = 2$
Answer
$m = 2$
4Find a missing coordinate using the gradient
ExtendedDirect calculation• gradient
Question
Points $P(1, k)$ and $Q(5, 11)$ lie on a line of gradient $2$ . Find $k$ .
Step-by-step solution
1. Step 1
  Set up the gradient equation.
  $\dfrac{11 - k}{5 - 1} = 2$
2. Step 2
  Solve.
  $11 - k = 8 \implies k = 3$
Answer
$k = 3$
5Find the gradient of a perpendicular line
ExtendedDirect calculation• gradient, perpendicular
Question
A line has gradient $\dfrac{3}{4}$ . Find the gradient of any line perpendicular to it.
Step-by-step solution
1. Step 1
  Perpendicular gradient is the negative reciprocal.
  $m_{\perp} = -\dfrac{4}{3}$
Answer
$-\dfrac{4}{3}$
6Find an endpoint given the midpoint
ExtendedDirect calculation• Adapted from 0580/22 May/Jun 2023 Q14• midpoint, endpoint
Question
The midpoint of segment $AB$ is $M(4, -1)$ . If $A = (-2, 5)$ , find the coordinates of $B$ .
Step-by-step solution
1. Step 1
  Let $B = (x, y)$ . Use the midpoint formula component-by-component.
  $\dfrac{-2 + x}{2} = 4 \implies x = 10$
2. Step 2
  And:
  $\dfrac{5 + y}{2} = -1 \implies y = -7$
Answer
$B = (10, -7)$
Examiner tip
The mark scheme awards method marks for setting up the two simultaneous component equations. The examiner report flags candidates often guess at $B$ by symmetry without writing the equations.
7Distance giving a surd answer
ExtendedDirect calculation• distance, surd
Question
Find the exact distance between $C(-3, 4)$ and $D(2, -1)$ , giving the answer in simplest surd form.
Step-by-step solution
1. Step 1
  Apply the distance formula.
  $d = \sqrt{(2 - (-3))^{2} + (-1 - 4)^{2}}$
2. Step 2
  Compute.
  $= \sqrt{25 + 25} = \sqrt{50}$
3. Step 3
  Simplify.
  $\sqrt{50} = 5\sqrt{2}$
Answer
$5\sqrt{2}$ units
8Gradient between two points on a curve
ExtendedDirect calculation• gradient, curve
Question
Points $P$ and $Q$ on the curve $y = x^{2} + 1$ have $x$ -coordinates $1$ and $3$ respectively. Find the gradient of line $PQ$ .
Step-by-step solution
1. Step 1
  Find the $y$ -coordinates.
  $P(1, 2),\ Q(3, 10)$
2. Step 2
  Apply the gradient formula.
  $m = \dfrac{10 - 2}{3 - 1} = \dfrac{8}{2} = 4$
Answer
$m = 4$
9Verify a triangle is isosceles
ChallengeShow that / prove• Adapted from 0580/42 Feb/Mar 2023 Q15• distance, isosceles
Question
Three points are $A(1, 3)$ , $B(5, 7)$ and $C(7, 1)$ . Show that triangle $ABC$ is isosceles.
Step-by-step solution
1. Step 1
  Compute $AB^{2}$ .
  $AB^{2} = (5 - 1)^{2} + (7 - 3)^{2} = 16 + 16 = 32$
2. Step 2
  Compute $BC^{2}$ .
  $BC^{2} = (7 - 5)^{2} + (1 - 7)^{2} = 4 + 36 = 40$
3. Step 3
  Compute $AC^{2}$ .
  $AC^{2} = (7 - 1)^{2} + (1 - 3)^{2} = 36 + 4 = 40$
4. Step 4
  $BC = AC = \sqrt{40}$ but $AB = \sqrt{32}$ — two equal sides, so triangle $ABC$ is isosceles.
Answer
$BC = AC = \sqrt{40}$ (or $2\sqrt{10}$ ); triangle is isosceles.
Examiner tip
The examiner report flags candidates often compute distances and stop without comparing. State which two sides are equal as the concluding line — that's where the final mark sits.
10Test for collinearity using gradients
ChallengeShow that / prove• gradient, collinear
Question
Show that the points $P(-1, -3)$ , $Q(2, 3)$ and $R(5, 9)$ are collinear.
Step-by-step solution
1. Step 1
  Gradient of $PQ$ .
  $m_{PQ} = \dfrac{3 - (-3)}{2 - (-1)} = \dfrac{6}{3} = 2$
2. Step 2
  Gradient of $QR$ .
  $m_{QR} = \dfrac{9 - 3}{5 - 2} = \dfrac{6}{3} = 2$
3. Step 3
  Both gradients are equal and they share the point $Q$ , so $P$ , $Q$ , $R$ lie on the same straight line.
Answer
$m_{PQ} = m_{QR} = 2$ , with $Q$ as a common point — the three points are collinear.
Examiner tip
The mark scheme awards the final mark for explicitly stating "same gradient AND shared point". The examiner report flags candidates often forget the shared-point clause.

Key Formulae — Distance, Midpoint and Gradient

The formulae you need to memorise for distance, midpoint and gradient on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Distance formula
$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$
$(x_{1}, y_{1})$
first point
$(x_{2}, y_{2})$
second point
When to use
Whenever you need the straight-line distance between two coordinate points.
Midpoint formula
$M = \left(\dfrac{x_{1} + x_{2}}{2},\ \dfrac{y_{1} + y_{2}}{2}\right)$
When to use
Average of the coordinates — finds the point exactly halfway.
Gradient formula
$m = \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}}$
When to use
Slope of a line through two given points — rise over run.
Perpendicular gradient
$m_{1} \cdot m_{2} = -1 \iff m_{2} = -\dfrac{1}{m_{1}}$
When to use
Two lines are perpendicular when the product of their gradients is $-1$ .

Key Definitions and Keywords — Distance, Midpoint and Gradient

Definitions to memorise and the exact keywords mark schemes credit for distance, midpoint and gradient answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Distance between two points
Examiner keyword
The straight-line length of the segment joining them, found via Pythagoras' theorem in coordinate form.
Midpoint
Examiner keyword
The point exactly halfway between two given points; coordinates are the average of the endpoints.
Gradient
Examiner keyword
Steepness of a line; rise over run.
Parallel lines
Lines with equal gradients ( $m_{1} = m_{2}$ ).
Perpendicular lines
Examiner keyword
Lines whose gradients multiply to $-1$ ( $m_{1} m_{2} = -1$ ).

Common Mistakes and Misconceptions — Distance, Midpoint and Gradient

The traps other students keep falling into on distance, midpoint and gradient questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Forgetting to square the differences inside the distance formula
Why it happens
Students write $d = \sqrt{x_{2} - x_{1}} + \sqrt{y_{2} - y_{1}}$ .
How to avoid it
Always: square each difference, ADD, then square root.
✕Subtracting instead of averaging in the midpoint formula
0580/22 — recurring
Why it happens
Confusion with the gradient formula.
How to avoid it
Midpoint = ADD coordinates and divide by $2$ .
✕Computing $\dfrac{x_{2} - x_{1}}{y_{2} - y_{1}}$ instead of $\dfrac{y_{2} - y_{1}}{x_{2} - x_{1}}$
Why it happens
Confusing rise/run with run/rise.
How to avoid it
Gradient is RISE over RUN — y on top, x on bottom.
✕Using the reciprocal as the perpendicular gradient (without negating)
Why it happens
Half-remembering the rule.
How to avoid it
Perpendicular gradient = NEGATIVE RECIPROCAL. Both flip and change sign.