IB MYP Mathematics: Coordinate geometry and transformations (MYP Mathematics)

This study guide follows the IB Mathematics guide (for use from September 2020/January 2021), first assessment 2022, and covers the geometry rows of the spatial reasoning branch that deal with position and movement on a plane: coordinates and straight lines, the distance, midpoint and gradient formulae, symmetry and the transformations, congruence, similarity and tessellations. The guide’s framework has a column for MYP 1-3, then standard and extended columns for MYP 4-5, with enrichment ideas listed apart. Items here are labelled to match. MYP has no tiers.

Angles, circles, area, volume and trigonometry are covered elsewhere. The revision notes give a quick recap, and the practice set to test yourself. The course hub lists related pages, the printable checklist tracks progress, and the free diagnostics can point you to weak topics.

What this unit covers

Framework row	What you should be able to do	Column
Coordinates	Plot and read points	MYP 1-3
Symmetry and reflection	Find lines of symmetry; reflect shapes	MYP 1-3
y = mx + c, parallel gradients	Read and build line equations	Standard
Distance, midpoint, gradient formulae	Calculate and apply them	Standard
Rotation; isometric transformations; enlargements	Perform and describe them	Standard
Similarity, congruence, tessellations	Justify, prove and use	Standard
Perpendicular gradients; enlargement about a point, by a rational factor; identical representation	Use and show	Extended
Dilation, polar coordinates, vector notation, vector spaces	Meet the ideas	Enrichment

Coordinates, y = mx + c and parallel lines

A point is written (x, y): across first, then up. Gradient means vertical change over horizontal change. For a line y = mx + c, the gradient is m and the line meets the y-axis at (0, c), the y-intercept. To find the x-intercept, set y = 0 and solve.

Lines are parallel exactly when their gradients are equal.

Worked example 1. A line passes through A(−2, 5) and B(4, −7).

1. Gradient m = (−7 − 5)/(4 − (−2)) = −12/6 = −2.
2. Substitute A into y = −2x + c: 5 = −2(−2) + c, so 5 = 4 + c and c = 1.
3. The line is y = −2x + 1. Check with B: −2(4) + 1 = −7.
4. For the x-intercept, 0 = −2x + 1, so x = 1/2.
5. A line parallel to this one through (3, 4) also has m = −2. Then 4 = −6 + c, so c = 10 and the line is y = −2x + 10.

If a line is given as 6x + 2y = 10, rearrange it to y = −3x + 5 before reading the gradient.

Distance, midpoint and gradient formulae

For P(x₁, y₁) and Q(x₂, y₂):

• distance PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]
• midpoint = ((x₁ + x₂)/2, (y₁ + y₂)/2)
• gradient = (y₂ − y₁)/(x₂ − x₁)

Worked example 2. P(−3, 2) and Q(5, 8).

1. Horizontal change 8, vertical change 6.
2. PQ = √(8² + 6²) = √100 = 10.
3. Midpoint = ((−3 + 5)/2, (2 + 8)/2) = (1, 5).
4. Gradient = 6/8 = 3/4.

Worked example 3 (classifying a quadrilateral). A(1, 1), B(5, 2), C(6, 6), D(2, 5).

1. AB² = 4² + 1² = 17, BC² = 1² + 4² = 17, CD² = 4² + 1² = 17, DA² = 1² + 4² = 17. All four sides are √17 long.
2. Gradient of AB is 1/4 and of DC is (6 − 5)/(6 − 2) = 1/4. Gradient of BC is 4 and of AD is 4. Opposite sides are parallel.
3. ABCD is a rhombus. It is not a square: the gradients of AB and BC multiply to 1, not −1, so the angle at B is not a right angle.

Perpendicular gradients (extended)

Two lines at right angles have gradients whose product is −1. So the partner of m is −1/m. The diagonals of the rhombus above show it: AC has gradient 5/5 = 1 and BD has gradient (5 − 2)/(2 − 5) = −1, and 1 × (−1) = −1.

Worked example 4. Find the line through (4, 1) perpendicular to y = 2x − 3.

1. The new gradient is −1/2.
2. 1 = −(1/2)(4) + c, so c = 3.
3. The line is y = −x/2 + 3.

Worked example 5 (perpendicular bisector). In example 2 the midpoint of PQ is (1, 5) and PQ has gradient 3/4, so the bisector has gradient −4/3. Then y − 5 = −(4/3)(x − 1), which gives 4x + 3y = 19.

Symmetry and reflection

A shape has line symmetry if a mirror line splits it into two parts that reflect onto each other. It has rotational symmetry of order n if it fits onto itself n times in a full turn. A regular hexagon has 6 lines of symmetry and order 6.

To reflect a point, go the same perpendicular distance beyond the mirror line. On a grid:

• in the x-axis: (x, y) → (x, −y)
• in the y-axis: (x, y) → (−x, y)
• in y = x: (x, y) → (y, x)
• in x = k: (x, y) → (2k − x, y)
• in y = k: (x, y) → (x, 2k − y)

Worked example 6. Reflect A(2, 5) in x = −1. A is 3 units right of the line, so the image is 3 units left: x = −4. Image (−4, 5). Reflect (3, 4) in y = −1: y = 2(−1) − 4 = −6, so (3, −6). In y = x, the triangle (1, 2), (4, 2), (4, 6) becomes (2, 1), (2, 4), (6, 4).

Rotation and the isometric transformations

To define a rotation, give its centre, angle and direction. A reflection, rotation or translation is isometric: lengths and angles do not change, so the image is congruent to the object. Orientation changes under reflection only.

Rotating about the origin:

• 90° anticlockwise: (x, y) → (−y, x)
• 180°: (x, y) → (−x, −y)
• 90° clockwise: (x, y) → (y, −x)

For any other centre C, subtract C from the point, rotate, then add C back.

Worked example 7. Rotate A(5, 2) by 90° clockwise about C(1, −1).

1. Relative to C: (4, 3).
2. Rotate 90° clockwise: (3, −4).
3. Add C back: (4, −5).

A translation slides every point by the same column vector; (3, −2) as a column means 3 right and 2 down.

To describe a transformation, give its type and details: mirror equation; centre, angle and direction; vector; or centre and scale factor.

Congruence

Two shapes are congruent when one maps exactly onto the other by isometric transformations. For triangles, any one of these proves congruence:

• SSS: three pairs of equal sides
• SAS: two sides and the angle they enclose
• ASA or AAS: two angles and a side in matching position
• RHS: a right angle with equal hypotenuses and one more equal side

Equal angles alone prove similarity, not congruence.

Worked example 8. In kite WXYZ, WX = WZ and YX = YZ. Prove that the diagonal WY bisects angle XWZ.

1. WX = WZ (given).
2. YX = YZ (given).
3. WY is common to both triangles.
4. So triangle WXY is congruent to triangle WZY (SSS).
5. Corresponding angles in congruent triangles are equal, so angle XWY = angle ZWY. WY bisects angle XWZ.

Write each statement with its reason, name the condition and state the conclusion. Criterion C asks for complete, coherent lines of reasoning, and criterion B at Year 5 asks you to prove or justify general rules.

Similarity

Similar shapes have equal matching angles and matching sides in the same ratio. For triangles, two equal angles (AA) are enough. The scale factor k = new length ÷ old length.

Worked example 9. In triangle ABC, a line MN parallel to BC meets AB at M and AC at N. AM = 6, MB = 4, MN = 9 and AN = 7.2. Find BC and AC.

1. Angle A is common and angle AMN = angle ABC (corresponding angles, MN ∥ BC), so triangles AMN and ABC are similar (AA).
2. AB = 6 + 4 = 10, so k = 10/6 = 5/3.
3. BC = 9 × 5/3 = 15.
4. AC = 7.2 × 5/3 = 12.

Worked example 10 (shadows). A 1.8 m pole casts a 2.4 m shadow while a tree casts 14 m. The triangles are similar, so height = 1.8/2.4 × 14 = 10.5 m. State your assumption (flat ground, upright objects); criterion D asks whether a solution makes sense.

Enlargement

The guide lists enlargements as standard and enlargement around a given point as extended, so the centre-based method below is extended.

An enlargement is fixed by its centre C and scale factor k. Multiply the vector from C to P by k to reach the image: P′ = C + k(P − C).

Worked example 11. Centre (1, 2), k = 3, point A(2, 4). A − C = (1, 2). Times 3 gives (3, 6). Add C: (4, 8).

Lengths multiply by k and angles stay the same, so the image is similar to the object. It is not isometric unless k is 1 or −1.

Rational and negative factors (extended)

A factor between 0 and 1 gives a smaller image; a negative factor puts it across the centre from the object.

Worked example 12. Centre (−3, 1), k = 2/3, P(6, 7). P − C = (9, 6). Times 2/3 gives (6, 4). Add C: (3, 5). With centre (0, 0) and k = −1/2, the point (8, −6) maps to (−4, 3).

Worked example 13 (finding the centre). A(3, 2) maps to A′(5, 6) with k = 2. The centre lies on the line AA′. Solving A′ = C + k(A − C) gives C = (A′ − kA)/(1 − k) = (−1, 2)/(−1) = (1, −2). Check: A − C = (2, 4), doubled (4, 8), plus C gives (5, 6).

Identical representation of transformations (extended)

Different descriptions can give the same mapping. This page reads the framework row this way.

• Reflection in the x-axis, then in the y-axis, equals a rotation of 180° about the origin. Test A(3, −2): (3, 2), then (−3, 2). A 180° turn gives (−3, 2).
• Reflection in y = x, then in the y-axis, equals a 90° anticlockwise rotation about the origin: (a, b) → (b, a) → (−b, a).
• Reflection in x = 1, then in x = 4 (parallel mirrors) equals a translation of 6 to the right, which is twice the gap between the mirrors.
• A 90° clockwise rotation equals a 270° anticlockwise one. An enlargement with k = −1 equals a 180° rotation about the same centre.

Tessellations

A tessellation covers the plane with copies of shapes, with no gaps and no overlaps. At each meeting point the angles must add to 360°.

• Equilateral triangles: 6 × 60° = 360°.
• Squares: 4 × 90° = 360°.
• Regular hexagons: 3 × 120° = 360°.

A regular polygon with n sides has interior angle 180° − 360°/n. For a regular pentagon that is 108°, and 360 ÷ 108 is not a whole number, so it cannot tessellate alone. Two regular octagons and one square do fit: 135 + 135 + 90 = 360. Any triangle tessellates: a copy rotated 180° about the midpoint of a side joins it to make a parallelogram, and the angles at each point are the three angles of the triangle twice, 360°. Describe a pattern by the transformations that generate it.

Enrichment ideas (not required)

• Dilation: the same idea as enlargement, seen as the plane stretching about a fixed centre.
• Polar coordinates give (r, θ): distance from the origin and angle from the positive x-axis. (6, 60°) has x = 6 cos 60° = 3 and y = 6 sin 60° ≈ 5.196. The point (4, 3) has r = 5 and θ ≈ 36.9°.
• Vector notation uses AB to mean the vector from A to B. For A(1, 2) and B(5, 5), AB = (4, 3) and its magnitude is 5.
• Vector spaces: sets of vectors that stay inside themselves under addition and scalar multiplication. All vectors (x, 0) do. Those with x + y = 1 do not, since two of them add to a vector with x + y = 2.

Common errors

• Reversing the subtraction on only the numerator or denominator of a gradient.
• Calling gradients 2 and −2 perpendicular; 2 and −1/2 are.
• Reflecting in x = k by only changing the sign of x.
• Rotating about the origin when another centre is named.
• Leaving out a centre or a direction when describing a transformation.
• Claiming congruence from AAA or SSA.

Official syllabus

IB Mathematics guide (for use from September 2020/January 2021), MYP, first assessment 2022, published by the International Baccalaureate Organization. Content is drawn from the spatial reasoning branch, geometry rows.