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 1–5)

1. Gradient (m = \frac{17 - 5}{8 - 2} = \frac{12}{6} = 2)
[M1] for correct substitution into gradient formula
[A1] for correct answer
Answer: 2

2. Distance (d = \sqrt{(5 - (-3))^2 + (7 - 1)^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10)
[M1] for correct substitution into distance formula
[A1] for correct answer in surd form (or simplified to 10)
Answer: 10 (or (\sqrt{100}))

3. Midpoint (M = \left(\frac{4 + 10}{2}, \frac{-2 + 6}{2}\right) = (7, 2))
[M1] for correct midpoint formula
[A1] for both coordinates correct
Answer: (7, 2)

4. Rearranging: (3y = 6x - 9 \Rightarrow y = 2x - 3)
Gradient (m = 2)
(y)-intercept: when (x = 0), (y = -3), so ((0, -3))
[A1] for correct gradient
[A1] for correct (y)-intercept coordinates
Answers: Gradient: 2; (y)-intercept: (0, –3)

5. Using (y - y_1 = m(x - x_1)):
(y - 1 = -\frac{1}{2}(x - 4))
(y - 1 = -\frac{1}{2}x + 2)
(y = -\frac{1}{2}x + 3)
[M1] for correct substitution into point-gradient formula
[A1] for correct equation in required form
Answer: (y = -\frac{1}{2}x + 3)

Section B: Equations of Lines and Parallel/Perpendicular Lines (Questions 6–10)

6. Gradient (m = \frac{12 - 4}{5 - 1} = \frac{8}{4} = 2)
Using point ((1, 4)): (y - 4 = 2(x - 1))
(y - 4 = 2x - 2)
(y = 2x + 2)
In form (ax + by = c): (2x - y = -2) (or (-2x + y = 2))
[M1] for correct gradient
[M1] for correct equation in (y = mx + c) form
[A1] for correct equation in required form (accept equivalent forms)
Answer: (2x - y = -2) (or equivalent)

7. Parallel lines have equal gradients, so (m = 3).
Using point ((2, 5)): (y - 5 = 3(x - 2))
(y - 5 = 3x - 6)
(y = 3x - 1)
[M1] for identifying gradient of parallel line as 3
[A1] for correct equation
Answer: (y = 3x - 1)

8. Rearranging (2x + 3y = 6):
(3y = -2x + 6)
(y = -\frac{2}{3}x + 2)
Gradient of (L_3) is (-\frac{2}{3}).
Gradient of perpendicular line (= \frac{3}{2}) (negative reciprocal).
[M1] for finding gradient of given line
[A1] for correct perpendicular gradient
Answer: (\frac{3}{2})

9. Midpoint of (GH): (M = \left(\frac{2 + 8}{2}, \frac{3 + 11}{2}\right) = (5, 7))
Gradient of (GH): (m = \frac{11 - 3}{8 - 2} = \frac{8}{6} = \frac{4}{3})
Gradient of perpendicular bisector: (m_{\perp} = -\frac{3}{4})
Equation: (y - 7 = -\frac{3}{4}(x - 5))
(y - 7 = -\frac{3}{4}x + \frac{15}{4})
(y = -\frac{3}{4}x + \frac{15}{4} + \frac{28}{4})
(y = -\frac{3}{4}x + \frac{43}{4})
[M1] for correct midpoint
[M1] for correct perpendicular gradient
[A1] for correct equation (accept any equivalent form)
Answer: (y = -\frac{3}{4}x + \frac{43}{4}) (or equivalent)

10. Gradient (AB = \frac{8 - 2}{3 - 1} = \frac{6}{2} = 3)
Gradient (BC = \frac{14 - 8}{5 - 3} = \frac{6}{2} = 3)
Since gradient (AB =) gradient (BC) and they share point (B), the three points are collinear.
[M1] for calculating both gradients
[A1] for conclusion with reasoning
Answer: Gradients are equal (both 3), therefore points are collinear.

Section C: Graphs of Quadratic Functions (Questions 11–15)

11. (y = (x - 3)^2 + 2)
(a) Turning point: ((3, 2))
[A1] for correct coordinates

(b) Minimum (coefficient of ((x - 3)^2) is positive)
[A1] for correct nature

(c) When (x = 0): (y = (0 - 3)^2 + 2 = 9 + 2 = 11), so ((0, 11))
[A1] for correct (y)-intercept

Answers:
(a) (3, 2)
(b) Minimum
(c) (0, 11)

12. (y = -(x + 1)^2 + 9)
(a) Axis of symmetry: (x = -1)
[A1] for correct equation

(b) For (x)-intercepts, set (y = 0):
(0 = -(x + 1)^2 + 9)
((x + 1)^2 = 9)
(x + 1 = \pm 3)
(x = -1 + 3 = 2) or (x = -1 - 3 = -4)
Intercepts: ((2, 0)) and ((-4, 0))
[M1] for setting (y = 0) and solving
[A1] for both correct coordinates

Answers:
(a) (x = -1)
(b) ((2, 0)) and ((-4, 0))

13. (y = (x - 2)(x + 4))
(a) (x)-intercepts: when (y = 0), (x = 2) or (x = -4), so ((2, 0)) and ((-4, 0))
[A1] for both correct coordinates

(b) Axis of symmetry: (x = \frac{2 + (-4)}{2} = -1)
When (x = -1): (y = (-1 - 2)(-1 + 4) = (-3)(3) = -9)
Turning point: ((-1, -9))
[M1] for finding axis of symmetry
[A1] for correct turning point

Answers:
(a) ((2, 0)) and ((-4, 0))
(b) ((-1, -9))

14. (y = x^2 - 6x + 5)
(y = (x^2 - 6x + 9) - 9 + 5)
(y = (x - 3)^2 - 4)
Minimum value of (y = -4) when (x = 3).
[M1] for completing the square correctly
[A1] for correct vertex form
[A1] for correct minimum value and corresponding (x)

Answers:
(y = (x - 3)^2 - 4)
Minimum value: (-4)
Value of (x): (3)

15. Substitute ((1, 8)) into (y = x^2 + 4x + c):
(8 = 1^2 + 4(1) + c)
(8 = 1 + 4 + c)
(8 = 5 + c)
(c = 3)
[M1] for correct substitution
[A1] for correct value of (c)

Answer: (c = 3)

Section D: Graphical Solutions and Applications (Questions 16–20)

16. (y = x^2 - 2x - 3)
(a) (x^2 - 2x - 3 = 0)
((x - 3)(x + 1) = 0)
(x = 3) or (x = -1)
[A1] for both correct solutions

(b) (x^2 - 2x - 3 = x + 1)
(x^2 - 3x - 4 = 0)
((x - 4)(x + 1) = 0)
(x = 4) or (x = -1)
[M1] for setting up equation correctly
[M1] for solving quadratic
[A1] for both correct solutions

Answers:
(a) (x = -1) or (x = 3)
(b) (x = -1) or (x = 4)

17. Graph drawing and gradient estimation:

• Correct axes with specified scales [1 mark]
• Correct plotting of all 7 points [1 mark]
• Smooth curve through points [1 mark]

At (x = 4), (y = 4^2 - 4(4) + 3 = 16 - 16 + 3 = 3). Point is ((4, 3)).
Draw tangent at ((4, 3)).
Using points on tangent, e.g., ((3, 1)) and ((5, 5)):
Gradient (= \frac{5 - 1}{5 - 3} = \frac{4}{2} = 2)
(Allow answers between 1.8 and 2.2 depending on tangent accuracy.)
[A1] for reasonable gradient estimate

Answer: Approximately 2 (accept 1.8 to 2.2)

18. Gradient of line through ((1, 5)) and ((4, 11)):
(m = \frac{11 - 5}{4 - 1} = \frac{6}{3} = 2)
Equation: (y - 5 = 2(x - 1) \Rightarrow y = 2x + 3)
Intersection with (y = 2x - 1):
(2x + 3 = 2x - 1)
(3 = -1) (contradiction)
The lines are parallel (both have gradient 2) and distinct, so they do not intersect.
[M1] for finding equation of first line
[M1] for setting up intersection equation
[A1] for concluding no intersection (or stating lines are parallel)

Answer: No intersection (lines are parallel)

19. (a) (AB = \sqrt{(4 - (-2))^2 + (5 - 1)^2} = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13})
[A1] for correct length (accept (\sqrt{52}) or (2\sqrt{13}))

(b) (AC = \sqrt{(6 - (-2))^2 + (1 - 1)^2} = \sqrt{8^2 + 0^2} = \sqrt{64} = 8)
(BC = \sqrt{(6 - 4)^2 + (1 - 5)^2} = \sqrt{2^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5})
Since no two sides are equal, the triangle is NOT isosceles.
(Note: This is a correction—the question asks to "show that triangle ABC is isosceles" but the coordinates given produce a scalene triangle. Accept correct working showing sides are different, with conclusion that it is not isosceles.)
[M1] for calculating at least one other side length
[A1] for correct conclusion based on working

Answers:
(a) (AB = \sqrt{52}) (or (2\sqrt{13}))
(b) (AC = 8), (BC = \sqrt{20}). All three sides have different lengths, so the triangle is not isosceles.

20. (a) Substitute ((3, 10)) into (y = 2x + k):
(10 = 2(3) + k)
(10 = 6 + k)
(k = 4)
[A1] for correct value of (k)

(b) Line: (y = 2x + 4)
Intersection with (y = x^2 + 1):
(x^2 + 1 = 2x + 4)
(x^2 - 2x - 3 = 0)
((x - 3)(x + 1) = 0)
(x = 3) or (x = -1)
When (x = 3): (y = 2(3) + 4 = 10), point ((3, 10))
When (x = -1): (y = 2(-1) + 4 = 2), point ((-1, 2))
[M1] for setting up equation
[M1] for solving quadratic
[A1] for both correct coordinates

Answers:
(a) (k = 4)
(b) ((3, 10)) and ((-1, 2))

END OF ANSWER KEY