Secondary 2 Mathematics Graphs Coordinate Geometry Quiz

Secondary 2 Mathematics Quiz - Graphs Coordinate Geometry (Answer Key)

Total Marks: 40

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Section A: Gradient and Straight Line Equations (Questions 1–5) [10 marks]

1. [2 marks]

Answer: $-\frac{4}{3}$ or $-1\frac{1}{3}$

Working: Gradient $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - (-3)}{-4 - 2} = \frac{8}{-6} = -\frac{4}{3}$

Marking notes:

• 1 mark for correct substitution into gradient formula
• 1 mark for correct simplification to $-\frac{4}{3}$
• Common mistake: Sign error when subtracting negative coordinates

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2. [2 marks]

Answer: $y = -\frac{3}{2}x + \frac{11}{2}$ or $y = -1.5x + 5.5$

Working: Gradient $m = \frac{-2 - 4}{5 - 1} = \frac{-6}{4} = -\frac{3}{2}$ Using point $(1, 4)$: $y - 4 = -\frac{3}{2}(x - 1)$ $y - 4 = -\frac{3}{2}x + \frac{3}{2}$ $y = -\frac{3}{2}x + \frac{3}{2} + 4 = -\frac{3}{2}x + \frac{11}{2}$

Marking notes:

• 1 mark for correct gradient
• 1 mark for correct equation in $y = mx + c$ form
• Accept equivalent forms (decimals or fractions)

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3. [2 marks]

Answer: $3x + 4y = 32$

Working: Using $y - y_1 = m(x - x_1)$ with $m = -\frac{3}{4}$ and $(8, 2)$: $y - 2 = -\frac{3}{4}(x - 8)$ $y - 2 = -\frac{3}{4}x + 6$ $y = -\frac{3}{4}x + 8$ Multiply by 4: $4y = -3x + 32$ $3x + 4y = 32$

Marking notes:

• 1 mark for correct equation in any form (e.g., $y = -\frac{3}{4}x + 8$)
• 1 mark for correct rearrangement to $ax + by = c$ with integer coefficients
• Common mistake: Not multiplying through to clear fractions

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4. [2 marks]

Answer: (a) $\frac{3}{2}$ or $1.5$ (b) $(0, -6)$

Working: (a) $3x - 2y = 12 \Rightarrow -2y = -3x + 12 \Rightarrow y = \frac{3}{2}x - 6$, so gradient $= \frac{3}{2}$ (b) $y$-intercept: set $x = 0$, $y = -6$, so $(0, -6)$

Marking notes:

• 1 mark each part
• For (a), accept gradient from rearranging or using $-\frac{a}{b}$ from $ax + by = c$
• For (b), coordinates must be in $(x, y)$ form

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5. [2 marks]

Answer: Gradient of $PQ = -1$, Gradient of $QR = -1$. Since both gradients are equal, the points are collinear.

Working: Gradient $PQ = \frac{-1 - 5}{4 - (-2)} = \frac{-6}{6} = -1$ Gradient $QR = \frac{-7 - (-1)}{10 - 4} = \frac{-6}{6} = -1$ Since gradient $PQ =$ gradient $QR$, and $Q$ is a common point, $P$, $Q$, $R$ are collinear.

Marking notes:

• 1 mark for both gradients correctly calculated as $-1$
• 1 mark for conclusion that equal gradients with common point implies collinearity
• Must show both gradient calculations

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Section B: Graphs of Linear Functions and Intercepts (Questions 6–10) [10 marks]

6. [2 marks]

Answer: Graph of $y = 2x - 3$ for $-2 \le x \le 4$

Expected graph features:

• Straight line passing through $(-2, -7)$, $(0, -3)$, $(2, 1)$, $(4, 5)$
• Line drawn only for $-2 \le x \le 4$ (endpoints marked)
• Axes labelled, scale consistent

Marking notes:

• 1 mark for at least 3 correct points plotted
• 1 mark for correct straight line through points with correct domain
• Deduct if line extends beyond $x = -2$ or $x = 4$ without indication

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7. [2 marks]

Answer: $y = -x + 2$

Working: Gradient $m = \frac{-2 - 4}{4 - (-2)} = \frac{-6}{6} = -1$ $y$-intercept: from graph, line crosses $y$-axis at $(0, 2)$, so $c = 2$ Equation: $y = -x + 2$

Marking notes:

• 1 mark for correct gradient ($-1$)
• 1 mark for correct equation
• Can also use point-gradient form with either given point

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8. [2 marks]

Answer: (a) $-\frac{1}{2}$ (b) $(0, 4)$

Working: (a) Equation is in $y = mx + c$ form, so gradient $m = -\frac{1}{2}$ (b) $y$-intercept is $c = 4$, so coordinates are $(0, 4)$

Marking notes:

• 1 mark each part
• Direct reading from $y = mx + c$ form
• For (b), must give coordinates, not just $4$

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9. [2 marks]

Answer: $k = 5$

Working: Substitute $(2, 11)$ into $y = 3x + k$: $11 = 3(2) + k$ $11 = 6 + k$ $k = 5$

Marking notes:

• 1 mark for correct substitution
• 1 mark for correct value of $k$
• Common mistake: Arithmetic error ($11 - 6 = 5$)

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10. [2 marks]

Answer: $m = \frac{1}{2}$, $c = -3$

Working: $y$-intercept is $(0, -3)$, so $c = -3$ Gradient $m = \frac{0 - (-3)}{6 - 0} = \frac{3}{6} = \frac{1}{2}$ (Alternatively: $x$-intercept is $(6, 0)$, so $0 = m(6) - 3 \Rightarrow 6m = 3 \Rightarrow m = \frac{1}{2}$)

Marking notes:

• 1 mark for $c = -3$ (direct from $y$-intercept)
• 1 mark for $m = \frac{1}{2}$ (using intercepts or gradient formula)
• Accept $m = 0.5$

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Section C: Applications and Problem Solving (Questions 11–15) [10 marks]

11. [2 marks]

Answer: (a) $23$ (b) $10$ km

Working: (a) $C = 2.5(8) + 3 = 20 + 3 = 23$ (b) $28 = 2.5x + 3 \Rightarrow 2.5x = 25 \Rightarrow x = 10$

Marking notes:

• 1 mark each part
• For (b), must show equation setup and solving
• Units: dollars for (a), km for (b)

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12. [2 marks]

Answer: (a) $210$ (b) $40$

Working: (a) $C = 8(20) + 50 = 160 + 50 = 210$ (b) $370 = 8n + 50 \Rightarrow 8n = 320 \Rightarrow n = 40$

Marking notes:

• 1 mark each part
• For (b), correct algebraic manipulation required
• Context: $n$ must be positive integer

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13. [2 marks]

Answer: (a) Gradients: $2$ and $-\frac{1}{2}$ (b) The product of the gradients is $2 \times (-\frac{1}{2}) = -1$, so the lines are perpendicular.

Working: (a) From $y = 2x + 1$, gradient $= 2$. From $y = -\frac{1}{2}x + 7$, gradient $= -\frac{1}{2}$. (b) $m_1 \times m_2 = 2 \times (-\frac{1}{2}) = -1$. For perpendicular lines, $m_1 m_2 = -1$.

Marking notes:

• 1 mark for both gradients correctly identified
• 1 mark for correct explanation using $m_1 m_2 = -1$
• Must explicitly state the product equals $-1$

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14. [2 marks]

Answer: $y = -\frac{3}{4}x + \frac{13}{2}$ or $y = -0.75x + 6.5$

Working: Gradient of $L_1 = \frac{0 - 3}{4 - 0} = -\frac{3}{4}$ Since $L_2 \parallel L_1$, gradient of $L_2 = -\frac{3}{4}$ $L_2$ passes through $(2, 5)$: $y - 5 = -\frac{3}{4}(x - 2)$ $y - 5 = -\frac{3}{4}x + \frac{3}{2}$ $y = -\frac{3}{4}x + \frac{3}{2} + 5 = -\frac{3}{4}x + \frac{13}{2}$

Marking notes:

• 1 mark for correct gradient ($-\frac{3}{4}$)
• 1 mark for correct equation using point-gradient form
• Common mistake: Using wrong gradient or arithmetic error in finding $c$

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15. [2 marks]

Answer: Gradient $= 3$, Equation: $y = 3x - a$

Working: Gradient $= \frac{8a - 2a}{3a - a} = \frac{6a}{2a} = 3$ (since $a \neq 0$) Using point $(a, 2a)$: $y - 2a = 3(x - a)$ $y - 2a = 3x - 3a$ $y = 3x - a$

Marking notes:

• 1 mark for correct gradient ($3$, independent of $a$)
• 1 mark for correct equation in terms of $a$
• Key concept: $a$ cancels in gradient calculation

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Section D: Extended Questions (Questions 16–20) [10 marks]

16. [2 marks]

Answer: (a) $-\frac{3}{2}$ (b) $y = -\frac{3}{2}x + 3$

Working: (a) Gradient $= \frac{-3 - 6}{4 - (-2)} = \frac{-9}{6} = -\frac{3}{2}$ (b) Using $A(-2, 6)$: $y - 6 = -\frac{3}{2}(x + 2)$ $y - 6 = -\frac{3}{2}x - 3$ $y = -\frac{3}{2}x + 3$ Check with $B(4, -3)$: $-3 = -\frac{3}{2}(4) + 3 = -6 + 3 = -3$ ✓

Marking notes:

• 1 mark for correct gradient
• 1 mark for correct equation
• Can verify with second point

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17. [2 marks]

Answer: (a) $P(\frac{5}{4}, 0)$, $Q(0, -5)$ (b) $\frac{5\sqrt{17}}{4}$ (so $k = \frac{425}{16}$ — but wait, let's recalculate)

Correction for (b): $P(\frac{5}{4}, 0)$, $Q(0, -5)$ $PQ = \sqrt{(\frac{5}{4} - 0)^2 + (0 - (-5))^2} = \sqrt{\frac{25}{16} + 25} = \sqrt{\frac{25}{16} + \frac{400}{16}} = \sqrt{\frac{425}{16}} = \frac{\sqrt{425}}{4} = \frac{5\sqrt{17}}{4}$

Wait — the question asks for $\sqrt{k}$ where $k$ is an integer. Let me re-check the question design. The length is $\frac{5\sqrt{17}}{4}$, which is not of the form $\sqrt{k}$ with integer $k$. This is a flaw in the question design. Let me provide the correct answer based on the actual calculation.

Revised Answer: (a) $P(\frac{5}{4}, 0)$, $Q(0, -5)$ (b) $\frac{5\sqrt{17}}{4}$

Working: (a) $x$-intercept: $0 = 4x - 5 \Rightarrow x = \frac{5}{4}$, so $P(\frac{5}{4}, 0)$ $y$-intercept: $y = 4(0) - 5 = -5$, so $Q(0, -5)$ (b) $PQ = \sqrt{(\frac{5}{4})^2 + 5^2} = \sqrt{\frac{25}{16} + 25} = \sqrt{\frac{425}{16}} = \frac{\sqrt{425}}{4} = \frac{5\sqrt{17}}{4}$

Marking notes:

• 1 mark for both intercepts correct
• 1 mark for correct distance formula application and simplification
• Note: The form $\sqrt{k}$ with integer $k$ is not achievable with these numbers; accept simplified surd form

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18. [2 marks]

Answer: (a) $x$-axis: $(6, 0)$, $y$-axis: $(0, 4)$ (b) $12$ square units

Working: (a) $x$-intercept: $2x + 3(0) = 12 \Rightarrow x = 6$, so $(6, 0)$ $y$-intercept: $2(0) + 3y = 12 \Rightarrow y = 4$, so $(0, 4)$ (b) Triangle $OAB$ has base $OA = 6$ and height $OB = 4$ Area $= \frac{1}{2} \times 6 \times 4 = 12$

Marking notes:

• 1 mark for both intercepts correct
• 1 mark for correct area calculation
• Must identify base and height correctly (intercepts on axes)

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19. [2 marks]

Answer: (a) $1$ (b) $y = x - 5$

Working: (a) Gradient $AB = \frac{6 - 2}{5 - 1} = \frac{4}{4} = 1$ (b) Line through $C(7, 2)$ parallel to $AB$ has gradient $1$: $y - 2 = 1(x - 7)$ $y - 2 = x - 7$ $y = x - 5$

Marking notes:

• 1 mark for correct gradient of $AB$
• 1 mark for correct equation of parallel line through $C$
• Parallel lines have equal gradients

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20. [2 marks]

Answer: (a) $-\frac{1}{3}$ (b) $x + 3y = 12$

Working: (a) $L_1: y = 3x - 4$, gradient $= 3$ $L_2 \perp L_1$, so gradient $m_2 = -\frac{1}{3}$ (since $3 \times (-\frac{1}{3}) = -1$) (b) $L_2$ passes through $(6, 2)$ with gradient $-\frac{1}{3}$: $y - 2 = -\frac{1}{3}(x - 6)$ $y - 2 = -\frac{1}{3}x + 2$ $y = -\frac{1}{3}x + 4$ Multiply by 3: $3y = -x + 12$ $x + 3y = 12$

Marking notes:

• 1 mark for correct perpendicular gradient ($-\frac{1}{3}$)
• 1 mark for correct equation in required form $ax + by = c$ with integer coefficients
• Common mistake: Forgetting to multiply through to clear fractions

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End of Answer Key