IB MYP Mathematics: Coordinate geometry and transformations (MYP Mathematics) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions follow the IB Mathematics guide (for use from September 2020/January 2021), first assessment 2022, spatial reasoning geometry rows: lines and gradients, coordinate formulae, symmetry, transformations, congruence, similarity and tessellations. MYP work is judged against criteria rather than by marks, so the bracketed numbers only show how much working each part deserves. Questions 4, 9 and 10 use the MYP 4-5 (extended) column and are labelled. Start with the study guide or the revision notes if you need support. See the course hub and the printable checklist.

Questions

1. Write down the gradient and y-intercept of 7x + 3y = 21. [2]

2. A line passes through (2, −3) and (6, 5).

(a) Find the equation of the line, giving it as y = mx + c. [3] (b) Find the equation of the line parallel to it that passes through (1, 4). [2]

3. M is the point (−2, 3) and N is the point (3, 15).

(a) Calculate the length MN. [2] (b) Find the midpoint of MN. [1] (c) Find the gradient of MN. [1] (d) K(3, 1) is the midpoint of AB, where A is (−1, 4). Find the coordinates of B. [2]

4. (Extended) The points are P(1, 1), Q(5, −1), R(6, 1) and S(2, 3).

(a) Find the gradients of PQ, QR, RS and SP. [2] (b) Explain why PQRS is a rectangle. [2] (c) Show that PQRS is not a square. [2]

5. (a) How many lines of symmetry does a rectangle that is not a square have, and what is its order of rotational symmetry? [2] (b) Find the image of A(2, −3) in the y-axis, of B(−4, 5) in the line y = x, and of C(5, 2) in the line x = 3. [3] (c) A point (2, 5) is reflected to (2, −1). Give the equation of the mirror line. [1]

6. Point A is (−2, 5) and C is (3, 1).

(a) Rotate A by 90° anticlockwise about C and give the image. [3] (b) Rotate A by 180° about C and give the image. [1]

7. In triangle ABC, AB = AC. D is the point on BC where AD is perpendicular to BC. Prove that triangle ABD is congruent to triangle ACD, and state what this shows about D. [4]

8. (a) Triangle ABC has AB = 8, BC = 12 and CA = 10. Triangle DEF is similar to it, with DE corresponding to AB and DE = 6. Find EF and FD. [3] (b) A person 1.6 m tall casts a shadow 2.5 m long. A flagpole casts a shadow 11.25 m long at the same moment. Find the height of the flagpole and state one assumption you made. [3]

9. (a) (Extended) Enlarge P(3, 2) with centre (2, 1) and scale factor 3. [2] (b) (Extended) Enlarge Q(5, 9) with centre (−1, 3) and scale factor 1/3. [3] (c) (Extended) An enlargement with centre the origin maps A(4, 6) to A′(−2, −3). Find the scale factor. [2]

10. (Extended) Show that reflecting in the line y = x and then in the x-axis is the same as a rotation of 90° clockwise about the origin. Test it on (3, 5), then show it for a general point (a, b). [4]

11. (a) Show that regular dodecagons (12 sides) cannot tessellate on their own. [2] (b) Show that a regular dodecagon, a regular hexagon and a square can fit together round a point. [2]

Answers

1. 3y = −7x + 21 [1], so y = −(7/3)x + 7. The gradient is −7/3 and the y-intercept is 7 [1]. Examiner insight: Rearranging first and stating both values clearly is the safest way to show the reading is deliberate; quoting 7x/3 as the gradient, or the intercept as a point, blurs the answer.

2. (a) m = (5 − (−3))/(6 − 2) = 8/4 = 2 [1]. Using (2, −3): −3 = 4 + c [1], so c = −7 and the line is y = 2x − 7 [1]. (b) Parallel lines have the same gradient, m = 2 [1]. Then 4 = 2 + c, so c = 2 and y = 2x + 2 [1]. Examiner insight: Checking the second point in your equation takes seconds and catches sign slips; show the substitution rather than just quoting c.

3. (a) The differences are 5 and 12, so MN² = 25 + 144 = 169 [1], MN = 13 [1]. (b) Midpoint = ((−2 + 3)/2, (3 + 15)/2) = (0.5, 9) [1]. (c) Gradient = 12/5 = 2.4 [1]. (d) (−1 + x)/2 = 3 gives x = 7 [1]; (4 + y)/2 = 1 gives y = −2 [1]. B is (7, −2). Examiner insight: Work in the middle: B is the reflection of A through K, so adding the same step from K that A is from K is a quick check.

4. (a) PQ: (−1 − 1)/(5 − 1) = −1/2; RS: (3 − 1)/(2 − 6) = −1/2 [1]. QR: (1 − (−1))/(6 − 5) = 2; SP: (1 − 3)/(1 − 2) = 2 [1]. (b) Opposite sides have equal gradients, so they are parallel and PQRS is a parallelogram [1]. Adjacent gradients multiply to (−1/2) × 2 = −1, so there is a right angle; a parallelogram with a right angle is a rectangle [1]. (c) PQ = √(16 + 4) = 2√5 and QR = √(1 + 4) = √5 [1]. Adjacent sides are unequal, so it is not a square [1]. Examiner insight: Give the product of gradients as evidence for a right angle; saying it “looks” perpendicular rarely persuades. Lengths decide square versus rectangle.

5. (a) 2 lines of symmetry [1] and rotational symmetry of order 2 [1]. (b) A′ = (−2, −3) [1]. B′ = (5, −4) [1]. C is 2 right of x = 3, so C′ is 2 left: (1, 2) [1]. (c) The x-coordinate is unchanged and the y-coordinates are 5 and −1, so the mirror is halfway: y = 2 [1]. Examiner insight: A diagonal of a non-square rectangle is not a line of symmetry; this is the usual trap in part (a). For part (c) give the equation, not just the number 2.

6. (a) Relative to C: A − C = (−5, 4) [1]. Rotate 90° anticlockwise with (x, y) → (−y, x): (−4, −5) [1]. Add C: (−1, −4) [1]. (b) (−5, 4) → (5, −4); add C: (8, −3) [1]. Examiner insight: Show the subtract, rotate, add steps. A single wrong sign gives a point that looks reasonable, and only visible working lets a slip be pinpointed.

7. Angle ADB = angle ADC = 90° [1]. AB = AC (given), and these are the hypotenuses [1]. AD is common to both [1]. So the triangles are congruent by RHS, and therefore BD = CD: D is the midpoint of BC [1]. Examiner insight: Write each statement with its reason and state which condition you use. Stopping at “so they are congruent” leaves the final inference, that BD = CD, unproven.

8. (a) k = 6/8 = 3/4 [1]. EF = 12 × 3/4 = 9 [1]; FD = 10 × 3/4 = 7.5 [1]. (b) Height ÷ shadow = 1.6/2.5 = 0.64 [1], so the flagpole is 0.64 × 11.25 = 7.2 m [1]. Assumption: the ground is level and both the person and the flagpole are upright, so the sun’s angle gives similar triangles [1]. Examiner insight: Match corresponding sides in order before dividing. For a real-life context, a one-line justification of the model is part of an answer that makes sense in context.

9. (a) P − C = (1, 1) [1]; times 3 and add C: (5, 4) [1]. (b) Q − C = (6, 6) [1]; times 1/3 gives (2, 2) [1]; add C: (1, 5) [1]. (c) Ratio of coordinates: −2/4 = −1/2 and −3/6 = −1/2 [1], so k = −1/2 (the image lies across the centre from A) [1]. Examiner insight: A smaller image still counts as an enlargement. Quote a negative factor when the image is on the other side of the centre; a positive 1/2 would be a different point.

10. Take (3, 5). In y = x it becomes (5, 3) [1]. In the x-axis it becomes (5, −3) [1]. A 90° clockwise turn sends (x, y) to (y, −x), so (3, 5) goes to (5, −3), the same point [1]. In general (a, b) → (b, a) → (b, −a), and the clockwise rule gives (b, −a) [1]. Examiner insight: One test point is evidence, not proof. The general (a, b) step is what turns the check into a justification.

11. (a) Interior angle = 180° − 360°/12 = 150° [1]. 360 ÷ 150 = 2.4 is not a whole number, so copies cannot fill the angle round a point [1]. (b) The hexagon has interior angle 120° and the square has 90° [1]. 150 + 120 + 90 = 360, so they fit with no gap [1]. Examiner insight: Give the angle sum and say it equals 360°; saying the shapes “fit nicely” does not show a tessellation.

Where marks are usually lost

• Reversing the subtraction order on only one of the two differences in a gradient.
• Giving only the gradient when the question asks for the equation.
• Using the distance formula without squaring both differences before adding.
• Checking sides only before calling a quadrilateral a rectangle or square.
• Forgetting that perpendicular gradients need a flip and a sign change (extended).
• Rotating about the origin when a different centre is given, or ignoring the direction.
• Reflecting in x = k by changing only the sign of the x-coordinate.
• Starting a congruence proof with the conclusion, or not stating the condition used.
• Matching the wrong sides in similar triangles.
• Describing a tessellation without the 360° argument.

Next steps

• Recap with the revision notes.
• Re-read the study guide for the topics you missed.
• Tick off topics on the printable checklist and browse the course hub.
• Try all free 10-minute diagnostics: diagnostics.
• Book a free trial class: trial class.

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.