Secondary 4 Elementary Mathematics Graphs Coordinate Geometry Quiz

Secondary 4 Elementary Mathematics Quiz - Graphs Coordinate Geometry

ANSWER KEY AND MARKING SCHEME

Total Marks: 50

Section A: Basic Coordinate Geometry (Questions 1–5)

Each question carries 2 marks.

1. Gradient = (13 − 5)/(6 − 2) = 8/4 = 2 ✓✓

• M1: Correct substitution into gradient formula
• A1: Correct answer

2. Distance = √[(5 − (−3))² + (7 − 1)²] = √[8² + 6²] = √(64 + 36) = √100 = 10 ✓✓

• M1: Correct substitution into distance formula
• A1: Correct answer (accept √100 or 10)

3. Midpoint = ((4 + 10)/2, (−2 + 8)/2) = (7, 3) ✓✓

• M1: Correct substitution into midpoint formula
• A1: Correct coordinates

4. y − 4 = 3(x − 1) → y − 4 = 3x − 3 → y = 3x + 1 ✓✓

• M1: Correct use of point-gradient form
• A1: Correct equation in required form

5. Line 1: y = 2x − 5, gradient m₁ = 2 Line 2: 2y + x = 7 → 2y = −x + 7 → y = −½x + 3.5, gradient m₂ = −½ m₁ × m₂ = 2 × (−½) = −1 Therefore, the lines are perpendicular. ✓✓

• M1: Correctly finding both gradients
• A1: Correct conclusion with justification

Section B: Equations of Lines and Applications (Questions 6–10)

Each question carries 3 marks.

6. Parallel to y = 4x + 2, so m = 4 Using point (3, −1): y − (−1) = 4(x − 3) y + 1 = 4x − 12 y = 4x − 13 ✓✓✓

• M1: Identifying gradient of parallel line
• M1: Correct substitution into point-gradient form
• A1: Correct equation

7. Midpoint of AB = ((1 + 7)/2, (2 + 10)/2) = (4, 6) Gradient of AB = (10 − 2)/(7 − 1) = 8/6 = 4/3 Gradient of perpendicular bisector = −3/4 Equation: y − 6 = −¾(x − 4) y − 6 = −¾x + 3 y = −¾x + 9 ✓✓✓

• M1: Correct midpoint and gradient of AB
• M1: Correct perpendicular gradient and substitution
• A1: Correct equation

8. (a) Gradient = (−1 − 5)/(4 − (−2)) = −6/6 = −1 ✓ (b) Using point (−2, 5): y − 5 = −1(x − (−2)) y − 5 = −x − 2 y = −x + 3 x + y = 3 ✓✓

• M1 (a): Correct gradient calculation
• M1 (b): Correct substitution
• A1 (b): Correct equation in required form

9. 3x − 2y = 12 (a) x-intercept: set y = 0 → 3x = 12 → x = 4 ✓ (b) y-intercept: set x = 0 → −2y = 12 → y = −6 ✓ (c) Rearranging: −2y = −3x + 12 → y = (3/2)x − 6 → gradient = 3/2 ✓

• A1 each for correct intercepts and gradient

10. Gradient of AB = (7 − 3)/(6 − 2) = 4/4 = 1 Since A, B, C are collinear, gradient of BC = 1 (k − 7)/(8 − 6) = 1 (k − 7)/2 = 1 k − 7 = 2 k = 9 ✓✓✓

• M1: Finding gradient of AB
• M1: Setting up equation using collinearity
• A1: Correct value of k

Section C: Coordinate Geometry Problems (Questions 11–15)

Each question carries 3 marks.

11. (a) PQ = √[(5 − 1)² + (8 − 2)²] = √(16 + 36) = √52 QR = √[(9 − 5)² + (2 − 8)²] = √(16 + 36) = √52 Since PQ = QR, triangle PQR is isosceles. ✓✓ (b) Base PR = 9 − 1 = 8 units Height = 8 − 2 = 6 units Area = ½ × 8 × 6 = 24 square units ✓

• M1 (a): Correct distance calculations
• A1 (a): Correct conclusion
• M1 (b): Correct area calculation
• A1 (b): Correct area

12. (a) 2x + 1 = −x + 7 → 3x = 6 → x = 2 y = 2(2) + 1 = 5 A = (2, 5) ✓ (b) Distance OA = √[(2 − 0)² + (5 − 0)²] = √(4 + 25) = √29 (≈ 5.39) ✓✓

• M1 (a): Correctly solving simultaneous equations
• A1 (a): Correct coordinates
• M1 (b): Correct distance formula
• A1 (b): Correct distance

13. (a) Gradient = (0 − 4)/(6 − 0) = −4/6 = −2/3 y-intercept = 4 Equation: y = −⅔x + 4 ✓ (b) Triangle has base 6 (on x-axis) and height 4 (on y-axis) Area = ½ × 6 × 4 = 12 square units ✓✓

• M1 (a): Correct gradient and intercept
• A1 (a): Correct equation
• M1 (b): Correct identification of base and height
• A1 (b): Correct area

14. (a) Gradient of AB = (−1 − 3)/(2 − (−1)) = −4/3 ✓ (b) Gradient of BC = (5 − (−1))/(5 − 2) = 6/3 = 2 ✓ (c) m₁ × m₂ = (−4/3) × 2 = −8/3 ≠ −1 Therefore, AB is not perpendicular to BC. ✓

• A1 each for (a) and (b)
• M1 (c): Correct product of gradients
• A1 (c): Correct conclusion with justification

15. P is equidistant from A(2, 1) and B(8, 1), so P lies on the perpendicular bisector of AB. Midpoint of AB = (5, 1). Since A and B have same y-coordinate, perpendicular bisector is vertical line x = 5. P also lies on y = x − 2. Substituting x = 5: y = 5 − 2 = 3 P = (5, 3) ✓✓✓

• M1: Identifying perpendicular bisector as x = 5
• M1: Substituting into line equation
• A1: Correct coordinates

Section D: Graphs and Coordinate Geometry Applications (Questions 16–20)

Each question carries 4 marks.

16. (a) Gradient = (11 − 3)/(4 − 0) = 8/4 = 2 ✓ (b) y-intercept = 3, so equation is y = 2x + 3 ✓ (c) Substitute (p, 19): 19 = 2p + 3 → 2p = 16 → p = 8 ✓✓

• A1 (a): Correct gradient
• A1 (b): Correct equation
• M1 (c): Correct substitution
• A1 (c): Correct value of p

17. (a) $0.40 per 200 m =$0.40 per 0.2 km = $2.00 per km. Hence the coefficient of d is 2. ✓ **(b)** F = 3.50 + 2(5.5) = 3.50 + 11 = **$14.50** ✓ (c) 15.50 = 3.50 + 2d → 2d = 12 → d = 6 km ✓ (d) Graph: Straight line from (0, 3.50) to (8, 19.50). Axes labelled: F ($) vertical, d (km) horizontal. ✓

• A1 each for (a), (b), (c)
• A1 (d): Correct graph with labelled axes and correct line

18. (a) M = ((1 + 9)/2, (2 + 2)/2) = (5, 2) ✓ (b) Gradient of BM = (2 − 10)/(5 − 5) = −8/0 → undefined (vertical line) ✓ (c) Gradient of AC = (2 − 2)/(9 − 1) = 0/8 = 0 (horizontal line) Since BM is vertical and AC is horizontal, BM ⟂ AC. ✓ (d) AC = 9 − 1 = 8 units (base) Height = distance from B to AC = 10 − 2 = 8 units Area = ½ × 8 × 8 = 32 square units ✓

• A1 each for (a), (b), (c), (d)

19. (a) 2x + y = 8 → y = −2x + 8 → gradient = −2 ✓ (b) Perpendicular gradient = ½ ✓ (c) Using point (3, 5): y − 5 = ½(x − 3) → y − 5 = ½x − 1.5 → y = ½x + 3.5 ✓ (d) Intersection: −2x + 8 = ½x + 3.5 → −2.5x = −4.5 → x = 1.8 y = −2(1.8) + 8 = 4.4 Intersection = (1.8, 4.4) ✓

• A1 each for (a), (b), (c), (d)

20. (a) √[(10 − 2)² + (y − 1)²] = 10 √[64 + (y − 1)²] = 10 64 + (y − 1)² = 100 (y − 1)² = 36 ✓ (b) y − 1 = ±6 → y = 7 or y = −5 ✓ (c) For y = 7: gradient = (7 − 1)/(10 − 2) = 6/8 = ¾ For y = −5: gradient = (−5 − 1)/(10 − 2) = −6/8 = −¾ ✓ (d) Positive gradient occurs when y = 7 ✓

• A1 each for (a), (b), (c), (d)

END OF ANSWER KEY