Secondary 3 Elementary Mathematics Graphs Coordinate Geometry Quiz

Secondary 3 Elementary Mathematics Quiz - Graphs Coordinate Geometry

ANSWER KEY AND MARKING SCHEME

Total Marks: 50

Section A: Basic Coordinate Geometry (Questions 1–7) — 15 marks

1. Gradient = (17 − 5) / (7 − 2) = 12 / 5 = 2.4 ✓✓
(2 marks: 1 for correct substitution, 1 for correct answer)

2. Distance = √[(5 − (−3))² + (−5 − 1)²] = √[8² + (−6)²] = √(64 + 36) = √100 = 10 ✓✓
(2 marks: 1 for correct substitution, 1 for correct surd/simplified answer)

3. Midpoint = ((4 + (−6))/2, (−2 + 8)/2) = (−2/2, 6/2) = (−1, 3) ✓✓
(2 marks: 1 for each coordinate)

4. Using y − y₁ = m(x − x₁): y − 4 = −3(x − 1) → y − 4 = −3x + 3 → y = −3x + 7 ✓✓
(2 marks: 1 for correct method, 1 for correct equation)

5. Gradient AB = (8 − 2)/(3 − 1) = 6/2 = 3
Gradient BC = (14 − 8)/(5 − 3) = 6/2 = 3
Since gradient AB = gradient BC, the points are collinear. ✓✓
(2 marks: 1 for finding both gradients, 1 for correct conclusion with reasoning)

6. (a) Gradient = (6 − (−2))/(4 − 0) = 8/4 = 2 ✓
(b) y-intercept is −2, so equation is y = 2x − 2 ✓✓
(3 marks: 1 for gradient, 2 for equation; allow y − y₁ = m(x − x₁) method)

7. Parallel line has gradient 2.
Using y − 1 = 2(x − 3) → y − 1 = 2x − 6 → y = 2x − 5 ✓✓
(2 marks: 1 for recognising gradient = 2, 1 for correct equation)

Section B: Graphs of Quadratic Functions (Questions 8–14) — 20 marks

8. Vertex is (−2, −3). In y = (x + a)² + b, vertex is (−a, b).
So −a = −2 → a = 2, and b = −3. ✓✓
(2 marks: 1 for each correct value)

9. (a) x-intercepts: set y = 0 → (x − 1)(x + 5) = 0 → x = 1 or x = −5.
Coordinates: (1, 0) and (−5, 0). ✓✓
(b) Axis of symmetry: x = (1 + (−5))/2 = −2. ✓
(c) Vertex: x = −2, y = (−2 − 1)(−2 + 5) = (−3)(3) = −9. Coordinates: (−2, −9). ✓✓
(5 marks: 2 for (a), 1 for (b), 2 for (c))

10. y = −(x − 2)² + 4
Vertex: (2, 4) ✓
y-intercept: x = 0 → y = −(0 − 2)² + 4 = −4 + 4 = 0 → (0, 0) ✓
x-intercepts: set y = 0 → −(x − 2)² + 4 = 0 → (x − 2)² = 4 → x − 2 = ±2 → x = 0 or x = 4 → (0, 0) and (4, 0) ✓
Sketch: downward-opening parabola with vertex (2, 4), passing through (0, 0) and (4, 0).
(3 marks: 1 for vertex, 1 for intercepts, 1 for correct shape)

11. (a) y = x² − 4x + 1 = (x² − 4x + 4) − 4 + 1 = (x − 2)² − 3. So p = 2, q = −3. ✓✓
(b) Minimum point is (2, −3). ✓
(3 marks: 2 for completing the square, 1 for coordinates)

12. Set y = 0: 2x² + 3x − 5 = 0
(2x + 5)(x − 1) = 0 ✓
x = −5/2 or x = 1 ✓
Coordinates: (−2.5, 0) and (1, 0). ✓
(3 marks: 1 for factorisation, 1 for solving, 1 for coordinates)

13. x-intercepts at (−1, 0) and (3, 0), so y = a(x + 1)(x − 3). ✓
Passes through (0, −6): −6 = a(0 + 1)(0 − 3) = a(1)(−3) = −3a → a = 2.
Equation: y = 2(x + 1)(x − 3). ✓
(2 marks: 1 for form with a, 1 for finding a and final equation)

14. At x = 3, y = 3² − 2(3) = 9 − 6 = 3. Point is (3, 3).
Draw tangent at (3, 3). Gradient ≈ 4 (accept 3.8 to 4.2 depending on tangent drawn). ✓✓
(2 marks: 1 for correct point, 1 for reasonable gradient estimate)

Section C: Coordinate Geometry Problems (Questions 15–20) — 15 marks

15. (a) AB = √[(6 − 2)² + (4 − 1)²] = √(16 + 9) = √25 = 5 units. ✓
(b) Gradient AB = (4 − 1)/(6 − 2) = 3/4.
Gradient BC = (8 − 4)/(3 − 6) = 4/(−3) = −4/3.
Product = (3/4) × (−4/3) = −1, so AB ⟂ BC. ✓✓
(c) Area = ½ × AB × BC. BC = √[(3 − 6)² + (8 − 4)²] = √(9 + 16) = √25 = 5.
Area = ½ × 5 × 5 = 12.5 square units. ✓✓
(5 marks: 1 for (a), 2 for (b), 2 for (c))

16. (a) 3x + 4y = 12 → 4y = −3x + 12 → y = −¾x + 3. Gradient = −¾. ✓
(b) Perpendicular gradient = 4/3.
Using y − (−1) = (4/3)(x − 2) → y + 1 = (4/3)x − 8/3 → y = (4/3)x − 11/3. ✓✓
(3 marks: 1 for (a), 2 for (b))

17. (a) Midpoint of PQ = ((−2 + 4)/2, (3 + (−1))/2) = (1, 1).
Gradient PQ = (−1 − 3)/(4 − (−2)) = −4/6 = −2/3.
Perpendicular gradient = 3/2.
Equation: y − 1 = (3/2)(x − 1) → y = (3/2)x − 3/2 + 1 → y = (3/2)x − ½. ✓✓
(b) y-intercept: x = 0 → y = −½. Point is (0, −½). ✓
(3 marks: 2 for (a), 1 for (b))

18. In parallelogram ABCD, AB = DC (as vectors).
B − A = (5 − 1, 2 − 2) = (4, 0).
C − D = (7 − p, 6 − q) = (4, 0).
So 7 − p = 4 → p = 3, and 6 − q = 0 → q = 6. ✓✓
(2 marks: 1 for each coordinate; accept other valid vector methods)

19. Substitute (k, 2k) into 3x − 2y = 8:
3(k) − 2(2k) = 8 → 3k − 4k = 8 → −k = 8 → k = −8. ✓
(1 mark)

20. √[(5 − a)² + (a − 3)²] = √20
(5 − a)² + (a − 3)² = 20
(25 − 10a + a²) + (a² − 6a + 9) = 20
2a² − 16a + 34 = 20
2a² − 16a + 14 = 0
a² − 8a + 7 = 0
(a − 1)(a − 7) = 0
a = 1 or a = 7. ✓
(1 mark: award full mark for both correct values)

END OF ANSWER KEY