IB MYP Mathematics: Coordinate geometry and transformations (MYP Mathematics) – Revision Notes

These revision notes condense the spatial reasoning geometry rows of the IB Mathematics guide (for use from September 2020/January 2021), first assessment 2022: coordinates, lines and gradients, the coordinate formulae, the transformations, congruence, similarity and tessellations. Items from the extended MYP 4-5 column are marked “extended” and items from the enrichment list are marked “enrichment”. Everything else is MYP 1-3 or MYP 4-5 (standard). For full explanations read the study guide, then test yourself with the practice set. The course hub, the printable checklist and the free diagnostics point you to weak topics.

Key formulae

Idea	Rule
Line	y = mx + c; m is the gradient, c the y-intercept
Gradient	(y₂ − y₁)/(x₂ − x₁)
Parallel lines	Same gradient
Perpendicular lines (extended)	m₁ × m₂ = −1, so the partner of m is −1/m
Distance	√[(x₂ − x₁)² + (y₂ − y₁)²]
Midpoint	((x₁ + x₂)/2, (y₁ + y₂)/2)
Enlargement (extended), centre C, factor k	P′ = C + k(P − C)
Reflection in x = k	(x, y) → (2k − x, y)
Reflection in y = k	(x, y) → (x, 2k − y)

Methods in steps

Equation of a line through two points

1. Work out m from the two points.
2. Put one point into y = mx + c to get c.
3. Test the other point.

Perpendicular bisector (extended)

1. Find the midpoint of the segment.
2. Find its gradient, then flip and negate it.
3. Use the midpoint and the new gradient to find c.

Rotation about a centre C

1. Subtract C from the point.
2. Rotate about the origin.
3. Add C back.

Finding an enlargement centre (extended)

1. Join two object points to their images; the lines meet at the centre.
2. Scale factor = image length ÷ object length, negative if the image is on the other side of the centre.

Proving triangles congruent

1. List each equal pair with the reason: given, common side, vertically opposite, alternate, corresponding.
2. Name the condition: SSS, SAS, ASA, AAS or RHS.
3. State the conclusion and, if needed, the consequence (for example, corresponding angles are equal).

Rotation rules about the origin

Turn	Mapping
90° anticlockwise	(x, y) → (−y, x)
180°	(x, y) → (−x, −y)
90° clockwise	(x, y) → (y, −x)

Congruence and similarity conditions

Test	Needs	Proves
SSS	Three pairs of equal sides	Congruent
SAS	Two sides and the angle they enclose	Congruent
ASA or AAS	Two angles and a side in matching position	Congruent
RHS	A right angle, equal hypotenuses, one more equal side	Congruent
AA	Two pairs of equal angles	Similar
Ratio	All three pairs of sides in the same ratio	Similar

Short proof. Triangle XYZ has XY = XZ, and W is the midpoint of YZ. Given XY = XZ and YW = ZW, with XW common, triangles XYW and XZW are congruent by SSS. So angle XYW = angle XZW: the base angles of an isosceles triangle are equal.

Small worked reminders

Gradient and equation. Through (1, 7) and (5, −1): m = −8/4 = −2. Then 7 = −2 + c, so c = 9 and y = −2x + 9.

Distance and midpoint. (−2, 1) and (4, 9): differences 6 and 8, distance 10, midpoint (1, 5).

Rotation. (4, 1) turned 90° anticlockwise about (2, 2): relative (2, −1) becomes (1, 2); add (2, 2) to get (3, 4).

Perpendicular line (extended). Perpendicular to y = −3x + 2 through (6, 5): m = 1/3, then 5 = 2 + c, so c = 3 and y = x/3 + 3.

Rational factor (extended). Centre (4, −2), k = 3/4, point (8, 10): the offset (4, 12) becomes (3, 9), so the image is (7, 7).

Similar triangles. In triangle PQR, S is on PQ and ST is parallel to QR, with T on PR. PS = 4, SQ = 2, ST = 6. Then PQ = 6, k = 6/4 = 1.5 and QR = 1.5 × 6 = 9.

Symmetry. A non-square rectangle has 2 lines of symmetry and order 2. An equilateral triangle has 3 lines and order 3. A parallelogram that is not a rhombus has no lines of symmetry but order 2.

Similarity. Two triangles with sides 5, 7, 9 and 10, 14, 18 have the same ratio 2, so they are similar but not congruent.

Tessellation test. A regular polygon tessellates alone if 360 ÷ (interior angle) is a whole number. Triangle 6, square 4, hexagon 3. Pentagon: 360/108 is not whole.

Mixed tessellations. Angles at a point must total 360°. Three triangles and two squares give 3 × 60 + 2 × 90 = 360. Two hexagons and two triangles give 2 × 120 + 2 × 60 = 360.

Must-know distinctions

• Congruent or similar. Congruent shapes match exactly. Similar shapes have the same form and a constant scale factor. All congruent shapes are similar, with k = 1.
• Isometric or not. Reflection, rotation and translation keep length and angle. Enlargement changes length unless k is 1 or −1.
• Parallel or perpendicular. Equal gradients are parallel. A product of −1 is perpendicular (extended).
• Enlargement factor. Between 0 and 1 gives a smaller image. Negative puts the image on the far side of the centre (extended).
• Gradient of a vertical line. It has no gradient; its equation is x = k. A horizontal line has gradient 0.

Describing a transformation fully

Type	Say
Reflection	The equation of the mirror line
Rotation	The centre, the angle and the direction
Translation	The column vector
Enlargement	The centre and the scale factor

Isometric transformations give a congruent image. Check one vertex and one side length to confirm your description before you write it down. For criterion C, name the transformation first and give the details second.

Identical representations (extended)

Description 1	Description 2
Reflect in the x-axis, then the y-axis	Rotate 180° about the origin
Reflect in y = x, then in the y-axis	Rotate 90° anticlockwise about the origin
Reflect in two parallel mirrors	Translate by twice the gap, perpendicular to the mirrors
Enlarge with k = −1 about C	Rotate 180° about C
Rotate 90° clockwise	Rotate 270° anticlockwise

Enrichment in brief

• Dilation: the enlargement idea about a fixed centre.
• Polar coordinates: (r, θ) converts to x = r cos θ, y = r sin θ.
• Vector notation: AB is the vector from A to B, found as B − A; its magnitude is the distance AB.
• Vector spaces: a set closed under vector addition and scalar multiplication.

Quick self-test

1. Give the gradient and y-intercept of 2y = 8x − 10.
2. Find the line parallel to that one through (2, 9).
3. Find the distance and midpoint for (−1, −2) and (4, 10).
4. A line has gradient 2/3. What is the gradient of a perpendicular line? (extended)
5. Reflect (−3, 6) in y = x, then reflect (−3, 6) in x = 2, then reflect (−3, 6) in y = −2.
6. Rotate (2, −5) by 90° anticlockwise and by 180°, both about the origin.
7. Are triangles with sides 5, 7, 9 and 10, 14, 18 congruent, similar, both or neither?
8. Enlarge (3, 1) with centre (2, 2) and scale factor −2. (extended)
9. Explain why regular octagons cannot tessellate on their own.
10. Reflections in y = 1 and then y = 5 equal which single transformation? (extended)
11. A(−1, 3) and B(2, −1). Write AB as a column and find its magnitude. (enrichment)
12. Give the Cartesian coordinates of the polar point (8, 30°), to 3 significant figures. (enrichment)

Answers

1. Halving gives y = 4x − 5: gradient 4, y-intercept −5.
2. m = 4, so 9 = 8 + c, giving c = 1 and y = 4x + 1.
3. Differences 5 and 12 give distance 13. Midpoint (1.5, 4).
4. −3/2.
5. (6, −3); (7, 6); (−3, −10).
6. (5, 2) and (−2, 5).
7. Similar only. The sides are in ratio 2, so they are not the same size.
8. (3, 1) − (2, 2) = (1, −1). Times −2 gives (−2, 2). Add the centre: (0, 4).
9. The interior angle is 135°. 360 ÷ 135 is about 2.67, not a whole number, so the angles cannot fit round a point.
10. A translation of 8 up, since 2 × 4 = 8. Test: y = 0 goes to 2, then to 8.
11. AB = (3, −4) as a column; magnitude √(9 + 16) = 5.
12. x = 8 cos 30° = 4√3 ≈ 6.93 and y = 8 sin 30° = 4, so (6.93, 4).

Where marks are usually lost

• Using (y₁ − y₂)/(x₂ − x₁), mixing the order of subtraction.
• Finding c but leaving the equation as y = mx or giving only m.
• Calling a quadrilateral a square or rectangle after checking sides only. Check right angles with gradients.
• Forgetting that a perpendicular gradient needs both a flip and a sign change.
• Rotating about the origin when the centre is elsewhere.
• Describing a transformation without all its details, such as a rotation with no direction.
• Naming congruence without listing the equal pairs and the reason for each.
• Matching the wrong sides when working out a scale factor; put matching sides in the same order.
• Using AAA or SSA to claim congruence.
• Claiming a shape tessellates because it “looks” like it fits, without the 360° angle argument.

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.