What is the gradient of a line and how is it calculated in IB MYP Mathematics?

The gradient of a line, also known as the slope, measures the steepness and direction of the line. In IB MYP Mathematics, it is calculated using the formula (y2 - y1) / (x2 - x1), where (x1, y1) and (x2, y2) are two distinct points on the line. A positive gradient indicates an upward slope, while a negative gradient indicates a downward slope.

How do you find the distance between two points in a coordinate plane?

To find the distance between two points in a coordinate plane, you can use the distance formula derived from the Pythagorean theorem: √((x2 - x1)² + (y2 - y1)²). This formula calculates the straight-line distance between the points (x1, y1) and (x2, y2). This concept is fundamental in IB MYP Geometry and Trigonometry.

What is the midpoint formula and how is it used in Geometry?

The midpoint formula is used to find the exact center point between two points on a coordinate plane. It is calculated as ((x1 + x2) / 2, (y1 + y2) / 2). This formula is essential in IB MYP Geometry for dividing a line segment into two equal parts and is often used in various geometric constructions and proofs.

Why is understanding the gradient important in Trigonometry?

Understanding the gradient is crucial in Trigonometry as it relates to the tangent of an angle in a right triangle. The gradient of a line can be interpreted as the tangent of the angle that the line makes with the positive x-axis. This relationship helps in solving problems involving angles and slopes, which are key components of the IB MYP Mathematics curriculum.

How can the concepts of distance and midpoint be applied in real-world scenarios?

In real-world scenarios, the concepts of distance and midpoint are widely used in fields such as navigation, architecture, and computer graphics. For example, calculating the distance is essential for determining the shortest path between two locations, while finding the midpoint is useful for designing symmetrical structures. These applications highlight the practical importance of these concepts taught in IB MYP Geometry and Trigonometry.

Why is "Forgetting the square root in the distance formula." a common mistake in Gradient, Distance and Midpoint?

Why it happens: Quick computation. How to avoid it: Distance = √(( x)^2 + ( y)^2). Without the root you just have (distance)^2.

Why is "Adding coordinates instead of averaging." a common mistake in Gradient, Distance and Midpoint?

Why it happens: Slight formula confusion. How to avoid it: Midpoint AVERAGES — divide each sum by 2.

Why is "Reversing $\Delta y$ and $\Delta x$." a common mistake in Gradient, Distance and Midpoint?

Why it happens: Direction confusion. How to avoid it: Always: m = change in ychange in x. Y on top.

Geometry and TrigonometryIB MYP Maths Extended Level

Gradient, Distance and Midpoint

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Gradient, Distance, and Midpoint — IB MYP Mathematics (Extended): the three core formulae of coordinate geometry

Three formulae let you analyse any line segment in the plane: gradient, distance and midpoint. This MYP Mathematics Extended note covers all three.

What you’ll learn

Mapped to the IB MYP Mathematics subject guide (2026 onwards).

MYP Mathematics A — Apply the three formulae.
MYP Mathematics A — Use these formulae together to classify shapes.
MYP Mathematics C — Show clear substitution into each formula.

The three formulae

Gradient (rise/run), distance (Pythagoras), midpoint (average).

Given two points $A(x_1, y_1)$ and $B(x_2, y_2)$ :

Gradient (slope) of $AB$ : $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{rise}}{\text{run}}.$

Distance $|AB|$ : $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$

This is the Pythagorean theorem applied to the right triangle whose horizontal leg is $|x_2 - x_1|$ and vertical leg is $|y_2 - y_1|$ .

Midpoint of $AB$ : $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right).$

Just average the x's and average the y's separately.

The distance between two points is the hypotenuse of the right triangle whose legs are the horizontal and vertical separations.

Worked example. $A(1, 2)$ and $B(7, 10)$ .

Gradient: $m = \dfrac{10 - 2}{7 - 1} = \dfrac{8}{6} = \dfrac{4}{3}$ .

Distance: $d = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ .

Midpoint: $M = \left(\dfrac{1 + 7}{2}, \dfrac{2 + 10}{2}\right) = (4, 6)$ .

$m = \dfrac{\Delta y}{\Delta x}$ .
$d = \sqrt{(\Delta x)^2 + (\Delta y)^2}$ .
$M = \left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$ .
Order doesn't matter.

Putting them together

Three formulae let you classify shapes and verify properties.

Worked example: is this a right-angled triangle?

Points $A(0, 0)$ , $B(4, 3)$ , $C(3, -4)$ . Show that $ABC$ is right-angled.

Strategy 1 — gradients. A right angle iff two adjacent sides have perpendicular gradients ( $m_1 m_2 = -1$ ).

$m_{AB} = \dfrac{3 - 0}{4 - 0} = \dfrac{3}{4}$ .
$m_{AC} = \dfrac{-4 - 0}{3 - 0} = -\dfrac{4}{3}$ .
$m_{AB} \times m_{AC} = \dfrac{3}{4} \times -\dfrac{4}{3} = -1$ . ✓
So $AB \perp AC$ , meaning the right angle is at $A$ .

Strategy 2 — Pythagoras. Compute all three side lengths and check $a^2 + b^2 = c^2$ for the longest.

Worked example: classify a quadrilateral by midpoints. $A(1, 0)$ , $B(5, 2)$ , $C(7, 6)$ , $D(3, 4)$ . Find the midpoints of the diagonals.

$AC$ midpoint: $((1+7)/2, (0+6)/2) = (4, 3)$ .
$BD$ midpoint: $((5+3)/2, (2+4)/2) = (4, 3)$ .

Same midpoint! In a quadrilateral whose DIAGONALS BISECT each other, the shape is a PARALLELOGRAM (could also be a rectangle, rhombus, etc.).

Worked example: which is closer? Point $P(2, 3)$ . Is $P$ closer to $A(5, 7)$ or $B(-1, 6)$ ?

$|PA| = \sqrt{(5-2)^2 + (7-3)^2} = \sqrt{9 + 16} = 5$ .
$|PB| = \sqrt{(-1-2)^2 + (6-3)^2} = \sqrt{9 + 9} = \sqrt{18} \approx 4.24$ .
$P$ is closer to $B$ .

Right angle: check $m_1 m_2 = -1$ OR use Pythagoras.
Parallelogram: diagonals bisect each other (same midpoint).
'Closer to': compute both distances and compare.

How it’s examined

Criterion A: apply all three formulae. Criterion C: clear substitution and reasoning when classifying shapes.

Sources: IB MYP Mathematics subject guide (IBO, official). Last reviewed 2026-05-25.

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

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

1Apply all three formulae
Getting started• formulae
Question
$A(2, 1)$ , $B(8, 9)$ . Find the gradient, distance and midpoint of $AB$ .
Step-by-step solution
1. Step 1
  Gradient: $m = \dfrac{9 - 1}{8 - 2} = \dfrac{8}{6} = \dfrac{4}{3}$ .
2. Step 2
  Distance: $d = \sqrt{6^2 + 8^2} = \sqrt{100} = 10$ .
3. Step 3
  Midpoint: $((2+8)/2, (1+9)/2) = (5, 5)$ .
Answer
$m = \tfrac{4}{3}$ , $d = 10$ , $M = (5, 5)$ .
2Show a triangle is right-angled
Building confidence• right angle
Question
$A(0, 0)$ , $B(2, 3)$ , $C(-3, 2)$ . Show that $\angle A = 90°$ .
Step-by-step solution
1. Step 1
  Gradient $AB = \dfrac{3-0}{2-0} = \dfrac{3}{2}$ .
2. Step 2
  Gradient $AC = \dfrac{2-0}{-3-0} = -\dfrac{2}{3}$ .
3. Step 3
  Product: $\tfrac{3}{2} \times -\tfrac{2}{3} = -1$ . So $AB \perp AC$ , and the angle at $A$ is $90°$ . ✓
Answer
Gradient product = $-1$ → right angle at $A$ .
3Perimeter of a triangle
Building confidence• distance
Question
Triangle with vertices $(0, 0)$ , $(6, 0)$ , $(3, 4)$ . Find its perimeter.
Step-by-step solution
1. Step 1
  Side 1 (origin to $(6,0)$ ): length 6.
2. Step 2
  Side 2 ( $(6,0)$ to $(3,4)$ ): $\sqrt{9 + 16} = 5$ .
3. Step 3
  Side 3 ( $(3,4)$ to origin): $\sqrt{9 + 16} = 5$ .
4. Step 4
  Perimeter = $6 + 5 + 5 = 16$ .
Answer
16 units.
4Find unknown coordinate of midpoint
Stretch• midpoint
Question
Midpoint of $PQ$ is $(4, 5)$ , and $P(1, 3)$ . Find $Q$ .
Step-by-step solution
1. Step 1
  Use midpoint formula: $(4, 5) = ((1 + x_Q)/2, (3 + y_Q)/2)$ .
2. Step 2
  $1 + x_Q = 8 \Rightarrow x_Q = 7$ .
3. Step 3
  $3 + y_Q = 10 \Rightarrow y_Q = 7$ .
Answer
$Q(7, 7)$ .

Key Definitions and Keywords — Gradient, Distance and Midpoint

Definitions to memorise and the exact keywords mark schemes credit for gradient, distance and midpoint answers — sharpened from recent examiner reports for the 2026 IB MYP Mathematics (Extended) sitting.

Gradient (coordinate geometry)
Examiner keyword
$m = \dfrac{y_2 - y_1}{x_2 - x_1}$ . The 'rise over run' between two points.
Distance between two points
Examiner keyword
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ — Pythagorean theorem applied to coordinate differences.
Midpoint
Examiner keyword
$M = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$ . The average of the two coordinates.

Common Mistakes and Misconceptions — Gradient, Distance and Midpoint

The traps other students keep falling into on gradient, distance and midpoint questions — taken from recent IB MYP Mathematics (Extended) examiner reports and mark schemes — and how to avoid them.

✕Forgetting the square root in the distance formula.
Why it happens
Quick computation.
How to avoid it
Distance = $\sqrt{(\Delta x)^2 + (\Delta y)^2}$ . Without the root you just have $(\text{distance})^2$ .
✕Adding coordinates instead of averaging.
Why it happens
Slight formula confusion.
How to avoid it
Midpoint AVERAGES — divide each sum by 2.
✕Reversing $\Delta y$ and $\Delta x$ .
Why it happens
Direction confusion.
How to avoid it
Always: $m = \dfrac{\text{change in } y}{\text{change in } x}$ . Y on top.

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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Gradient, Distance and Midpoint — frequently asked questions

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Gradient, Distance and Midpoint

Short Study Notes in the form of Slides

Short Study Notes — Gradient, Distance and Midpoint

Detailed Notes

What you’ll learn

How it’s examined

1Apply all three formulae

2Show a triangle is right-angled

3Perimeter of a triangle

4Find unknown coordinate of midpoint

Gradient (coordinate geometry)

Distance between two points

Midpoint

✕Forgetting the square root in the distance formula.

✕Adding coordinates instead of averaging.

✕Reversing Δy\Delta yΔy and Δx\Delta xΔx.

Practice questions

Past paper style quiz

Assess your understanding

Gradient, Distance and Midpoint — frequently asked questions

✕Reversing $\Delta y$ and $\Delta x$ .