Secondary 1 Mathematics Graphs Coordinate Geometry Quiz

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Secondary 1 Mathematics Quiz - Graphs Coordinate Geometry

Name: _________________________ Class: __________ Date: __________

Duration: 35 minutes Total Marks: 40 marks

Instructions:

• Answer all questions.
• Show all working clearly in the spaces provided.
• Diagrams are not drawn to scale unless stated.
• Write your answers in the simplest form.

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Section A: Plotting and Reading Coordinates (Questions 1–5, 10 marks)

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1. Write down the coordinates of point P shown on the grid below.

<image_placeholder> id: Q1-fig1 type: diagram linked_question: Q1 description: A standard Cartesian coordinate grid with axes labeled x and y, ranging from -6 to 6 on both axes. Point P is marked with a dot at coordinates (3, -2). labels: x-axis, y-axis, origin O, point P values: P at (3, -2) must_show: Grid lines, labeled axes with arrows, point P clearly marked with coordinates readable from grid intersections </image_placeholder>

[2 marks]

Answer: _________________________

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2. On the grid below, plot and label the points A(-4, 1), B(2, 5), and C(0, -3).

<image_placeholder> id: Q2-fig1 type: diagram linked_question: Q2 description: A blank Cartesian coordinate grid with axes labeled x and y, ranging from -6 to 6 on both axes, with all grid lines shown. labels: x-axis, y-axis, origin O values: axes from -6 to 6 must_show: Complete grid with labeled axes, arrows on axes, all intersection points visible for student plotting </image_placeholder>

[3 marks]

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3. Point D has coordinates (5, -2). Point E is the reflection of D in the y-axis.

(a) Write down the coordinates of E. __[1]

(b) Write down the coordinates of the point F, where F is the reflection of D in the x-axis. __[1]

Answer (a): _________________________

Answer (b): _________________________

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4. Which quadrant does the point (-7, 4) lie in? Choose from: first quadrant, second quadrant, third quadrant, or fourth quadrant.

[1 mark]

Answer: _________________________

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5. The point G(k, 2k) lies on the line y = x. Find the value of k.

[3 marks]

Answer: _________________________

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Section B: Distance and Length (Questions 6–10, 10 marks)

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6. Find the length of the line segment joining A(1, 2) and B(4, 6).

[2 marks]

Answer: _________________________ units

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7. Find the distance between C(-3, 5) and D(1, -1), giving your answer in simplified surd form where appropriate.

[3 marks]

Answer: _________________________ units

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8. The distance between E(2, p) and F(5, 9) is 5 units. Find the two possible values of p.

[3 marks]

Answer: p = _________________________ or _________________________

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9. Point H is at (6, 8). Find the length of OH, where O is the origin.

[2 marks]

Answer: _________________________ units

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10. Which point is further from the origin: J(-4, 5) or K(3, 6)? Show your working.

[2 marks]

Answer: _________________________

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Section C: Gradient and Equation of Straight Line (Questions 11–15, 10 marks)

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11. Find the gradient of the line passing through P(2, 7) and Q(6, 15).

[2 marks]

Answer: _________________________

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12. A straight line has gradient $-\frac{3}{4}$ and passes through the point (8, 1). Find the equation of the line in the form y = mx + c.

[3 marks]

Answer: y = _________________________

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13. The line $2x + 5y = 10$ meets the x-axis at A and the y-axis at B.

(a) Find the coordinates of A. __[2]

(b) Find the coordinates of B. __[1]

Answer (a): _________________________

Answer (b): _________________________

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14. Find the gradient of the line $3x - 4y + 12 = 0$.

[2 marks]

Answer: _________________________

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15. Two lines are perpendicular. The first line has gradient $\frac{2}{5}$. Find the gradient of the second line.

[2 marks]

Answer: _________________________

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Section D: Interpretation and Application (Questions 16–20, 10 marks)

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16. The graph below shows the journey of a cyclist from town A to town B.

<image_placeholder> id: Q16-fig1 type: graph linked_question: Q16 description: A distance-time graph showing a cyclist's journey from town A to town B. The x-axis is time in hours from 0 to 4, and the y-axis is distance in km from 0 to 60. The graph consists of: a straight line from (0, 0) to (1.5, 30); a horizontal line from (1.5, 30) to (2, 30); then a straight line from (2, 30) to (3.5, 60). labels: x-axis: Time (hours), y-axis: Distance (km), point labels: A at start, B at end values: coordinates as described above with time marks at 0, 1, 2, 3, 4 and distance marks at 0, 10, 20, 30, 40, 50, 60 must_show: All three segments clearly, labeled axes with units, grid lines, coordinates readable for calculating gradients </image_placeholder>

(a) Find the speed of the cyclist during the first 1.5 hours. __[2]

(b) For how many minutes did the cyclist rest? __[1]

(c) Find the average speed for the whole journey. __[2]

Answer (a): _________________________ km/h

Answer (b): _________________________ minutes

Answer (c): _________________________ km/h

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17. The table shows some values for the equation $y = 2x - 3$.

x	-2	-1	0	1	2
y

(a) Complete the table. __[2]

(b) Draw the graph of $y = 2x - 3$ for $-2 \leq x \leq 2$. __[2]

<image_placeholder> id: Q17-fig1 type: graph linked_question: Q17 description: A blank Cartesian coordinate grid with axes labeled x and y, x from -3 to 3, y from -8 to 3, with all grid lines shown. labels: x-axis, y-axis values: x-axis ticks at -3, -2, -1, 0, 1, 2, 3; y-axis ticks at -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3 must_show: Complete grid with labeled axes, sufficient range to plot all five points from the table, grid lines at integer values </image_placeholder>

Answer (a):

x	-2	-1	0	1	2
y	____	____	____	____	____

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18. The lines $y = x + 2$ and $y = 4 - x$ intersect at point P.

(a) Using algebra, find the coordinates of P. __[3]

(b) State the solution to the simultaneous equations $x - y = -2$ and $x + y = 4$. __[1]

Answer (a): P(_________________, _________________)

Answer (b): x = _________________, y = _________________

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19. The diagram shows a trapezium ABCD with vertices A(-2, 1), B(4, 1), C(6, 5), and D(0, 5).

<image_placeholder> id: Q19-fig1 type: diagram linked_question: Q19 description: A trapezium on a coordinate grid with vertices at specified coordinates. The shape has AB parallel to DC (both horizontal). labels: points A, B, C, D with coordinates labeled, x-axis, y-axis values: A(-2,1), B(4,1), C(6,5), D(0,5) must_show: Grid with axes, four labeled vertices with coordinates shown, shape ABCD drawn with AB and DC as the parallel sides </image_placeholder>

Show that AB is parallel to DC.

[3 marks]

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20. A straight line passes through (2, -3) and is parallel to the line $y = \frac{1}{2}x + 5$.

(a) Find the equation of this line. __[2]

(b) Does this line pass through the point (6, 1)? Show your working to justify your answer. __[2]

Answer (a): y = _________________________

Answer (b): _________________________

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END OF QUIZ

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Section A: 10 marks | Section B: 10 marks | Section C: 10 marks | Section D: 10 marks

TOTAL: 40 marks

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Secondary 1 Mathematics Quiz - Graphs Coordinate Geometry: Answer Key

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Section A: Plotting and Reading Coordinates

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1. [2 marks]

Answer: P(3, -2)

Working and Teaching Notes:

• Coordinates are written as (x, y), where x is the horizontal distance from the origin and y is the vertical distance.
• From the diagram, move 3 units right along the x-axis (positive x), then 2 units down (negative y).
• Marking: 1 mark for x-coordinate 3; 1 mark for y-coordinate -2.
• Common mistake: Writing (-2, 3) or (3, 2) — always remember "x before y" and check the sign of each value.

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2. [3 marks]

Expected answer: Three points correctly plotted and labeled.

Working and Teaching Notes:

• Point A(-4, 1): From origin, move 4 units left (negative x), 1 unit up (positive y).
• Point B(2, 5): From origin, move 2 units right, 5 units up.
• Point C(0, -3): On the y-axis, 3 units down — this lies on the y-axis since x = 0.
• Marking: 1 mark for each correctly plotted and labeled point.
• Common mistake: Swapping x and y coordinates; plotting C on the x-axis instead of the y-axis.

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3. [2 marks] — 1 mark each part

(a) Answer: E(-5, -2)

(b) Answer: F(5, 2)

Working and Teaching Notes:

• Reflection in the y-axis: The y-coordinate stays the same; the x-coordinate changes sign.
  • D(5, -2) → E(-5, -2)
  • Rule: (x, y) → (-x, y)
• Reflection in the x-axis: The x-coordinate stays the same; the y-coordinate changes sign.
  • D(5, -2) → F(5, 2)
  • Rule: (x, y) → (x, -y)
• Common mistake: Confusing which coordinate changes — remember "y-axis reflection changes x, x-axis reflection changes y."

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4. [1 mark]

Answer: Second quadrant

Teaching Notes:

• Quadrants are numbered counter-clockwise from the positive x-axis:
  • First quadrant: x > 0, y > 0
  • Second quadrant: x < 0, y > 0 ← (-7, 4) fits here
  • Third quadrant: x < 0, y < 0
  • Fourth quadrant: x > 0, y < 0
• For (-7, 4): x is negative, y is positive → second quadrant.

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5. [3 marks]

Answer: k = 0

Working and Teaching Notes:

• A point on the line y = x must have its x-coordinate equal to its y-coordinate.
• For G(k, 2k): the x-coordinate is k and the y-coordinate is 2k.
• Since y = x: 2k = k
• Subtract k from both sides: 2k - k = 0, so k = 0.
• Check: G(0, 0), which is the origin, and the origin lies on y = x. ✓
• Marking: 1 mark for setting up equation 2k = k; 1 mark for solving; 1 mark for verifying or stating the point.

Alternative method: Substitute into y = x: 2k = k.

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Section B: Distance and Length

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6. [2 marks]

Answer: 5 units

Working and Teaching Notes:

• Use the distance formula: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$
• Substitute A(1, 2) and B(4, 6):
  • $d = \sqrt{(4-1)^2 + (6-2)^2}$
  • $d = \sqrt{3^2 + 4^2}$
  • $d = \sqrt{9 + 16}$
  • $d = \sqrt{25}$
  • $d = 5$
• Marking: 1 mark for correct substitution; 1 mark for final answer.
• This is a classic 3-4-5 Pythagorean triple — recognizing this pattern saves time.

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7. [3 marks]

Answer: $2\sqrt{13}$ units (or $\sqrt{52}$)

Working and Teaching Notes:

• Using the distance formula with C(-3, 5) and D(1, -1):
  • $d = \sqrt{(1-(-3))^2 + ((-1)-5)^2}$
  • $d = \sqrt{(4)^2 + (-6)^2}$
  • $d = \sqrt{16 + 36}$
  • $d = \sqrt{52}$
  • $d = \sqrt{4 \times 13} = 2\sqrt{13}$
• Marking: 1 mark for correct substitution; 1 mark for $\sqrt{52}$; 1 mark for simplifying to $2\sqrt{13}$.
• Common mistake: Forgetting to square negative values properly — $(-6)^2 = 36$, not -36.

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8. [3 marks]

Answer: p = 5 or p = 13

Working and Teaching Notes:

• Using distance formula: $\sqrt{(5-2)^2 + (9-p)^2} = 5$
• Square both sides: $9 + (9-p)^2 = 25$
• $(9-p)^2 = 16$
• So $9-p = 4$ or $9-p = -4$
• If $9-p = 4$: p = 5
• If $9-p = -4$: p = 13
• Check:
  • For p = 5: E(2, 5), distance = $\sqrt{9+16} = 5$ ✓
  • For p = 13: E(2, 13), distance = $\sqrt{9+16} = 5$ ✓
• Marking: 1 mark for setting up equation; 1 mark for solving quadratic; 1 mark for both values.
• Common mistake: Forgetting the negative square root, giving only p = 5.

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9. [2 marks]

Answer: 10 units

Working and Teaching Notes:

• O is (0, 0), so this is distance from origin.
• $OH = \sqrt{(6-0)^2 + (8-0)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$
• This is another Pythagorean triple: 6-8-10 (scaled 3-4-5).
• Marking: 1 mark for method; 1 mark for answer.
• Alternatively, recognize 6-8-10 pattern directly for full credit with stated reasoning.

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10. [2 marks]

Answer: J(-4, 5) — both are at distance $\sqrt{41}$, so they are equidistant (same distance)

Working and Teaching Notes:

• Distance OJ = $\sqrt{(-4)^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41}$
• Distance OK = $\sqrt{3^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}$

Wait — let me recalculate:

• OJ = $\sqrt{16 + 25} = \sqrt{41} \approx 6.40$
• OK = $\sqrt{9 + 36} = \sqrt{45} \approx 6.71$

Corrected Answer: K(3, 6) is further from the origin.

• Marking: 1 mark for both distances calculated; 1 mark for correct comparison and conclusion.
• Common mistake: Not showing working, or only calculating one distance.

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Section C: Gradient and Equation of Straight Line

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11. [2 marks]

Answer: 2

Working and Teaching Notes:

• Gradient formula: $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{rise}}{\text{run}}$
• $m = \frac{15 - 7}{6 - 2} = \frac{8}{4} = 2$
• Marking: 1 mark for substitution; 1 mark for simplification.
• Common mistake: Reversing the formula as $\frac{x_2-x_1}{y_2-y_1}$ — remember "rise over run" (y-change over x-change).

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12. [3 marks]

Answer: $y = -\frac{3}{4}x + 7$ or equivalent

Working and Teaching Notes:

• Start with y = mx + c with m = $-\frac{3}{4}$:
  • $y = -\frac{3}{4}x + c$
• Substitute point (8, 1):
  • $1 = -\frac{3}{4}(8) + c$
  • $1 = -6 + c$
  • $c = 7$
• Final equation: $y = -\frac{3}{4}x + 7$
• Marking: 1 mark for using correct gradient; 1 mark for substituting point; 1 mark for finding c and stating equation.
• Common mistake: Arithmetic error with $-\frac{3}{4} \times 8 = -6$, or forgetting to write "y = " in final answer.

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13. [3 marks] — 2 marks (a), 1 mark (b)

(a) Answer: A(5, 0)

(b) Answer: B(0, 2)

Working and Teaching Notes:

• x-intercept (point A): Set y = 0, solve for x:
  • $2x + 5(0) = 10$
  • $2x = 10$
  • $x = 5$
  • So A(5, 0)
• y-intercept (point B): Set x = 0, solve for y:
  • $2(0) + 5y = 10$
  • $5y = 10$
  • $y = 2$
  • So B(0, 2)
• Marking (a): 1 mark for setting y = 0; 1 mark for correct coordinates.
• Marking (b): 1 mark for correct coordinates (or follow-through from method).

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14. [2 marks]

Answer: $\frac{3}{4}$

Working and Teaching Notes:

• Rearrange to y = mx + c form:
  • $3x - 4y + 12 = 0$
  • $-4y = -3x - 12$
  • $y = \frac{3}{4}x + 3$
• Gradient is the coefficient of x: $m = \frac{3}{4}$
• Marking: 1 mark for rearranging; 1 mark for identifying gradient.
• Common mistake: Stopping at $3x + 12 = 4y$ and thinking gradient is 4 or 3; sign errors when dividing by -4.

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15. [2 marks]

Answer: $-\frac{5}{2}$ (or -2.5)

Working and Teaching Notes:

• For perpendicular lines: $m_1 \times m_2 = -1$, so $m_2 = -\frac{1}{m_1}$
• Given $m_1 = \frac{2}{5}$:
  • $m_2 = -\frac{1}{\frac{2}{5}} = -\frac{5}{2}$
• Marking: 1 mark for correct formula or method; 1 mark for answer.
• Common mistake: Giving $\frac{5}{2}$ (missing negative sign) or $-\frac{2}{5}$ (reciprocal without flip).

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Section D: Interpretation and Application

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16. [5 marks] — 2 marks (a), 1 mark (b), 2 marks (c)

(a) Answer: 20 km/h

Working:

• Speed = gradient of distance-time graph = $\frac{\text{distance change}}{\text{time change}}$
• Speed = $\frac{30 - 0}{1.5 - 0} = \frac{30}{1.5} = 20$ km/h

(b) Answer: 30 minutes

Working:

• Rest period is shown by horizontal line: from t = 1.5 to t = 2 hours
• Time = 2 - 1.5 = 0.5 hours = 30 minutes

(c) Answer: $\frac{120}{7} \approx 17.1$ km/h (or exact: $17\frac{1}{7}$ km/h)

Working:

• Total distance = 60 km
• Total time = 3.5 hours
• Average speed = $\frac{60}{3.5} = \frac{120}{7} \approx 17.14$ km/h

Marking: (a) 1 mark for gradient interpretation, 1 mark for answer; (b) 1 mark; (c) 1 mark for total distance/time, 1 mark for correct calculation. Common mistake: For (c), using only moving time instead of total time.

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17. [4 marks] — 2 marks (a), 2 marks (b)

(a) Answer:

x	-2	-1	0	1	2
y	-7	-5	-3	-1	1

Working: Substitute each x into y = 2x - 3:

• x = -2: y = -4 - 3 = -7
• x = -1: y = -2 - 3 = -5
• x = 0: y = 0 - 3 = -3
• x = 1: y = 2 - 3 = -1
• x = 2: y = 4 - 3 = 1

(b) Plot all five points and join with a straight line.

Marking (a): 2 marks for all correct (deduct 1 mark for 1-2 errors). Marking (b): 1 mark for correct points plotted; 1 mark for straight line through all points. Common mistake: Sign errors with negative x values, especially x = -2: 2(-2) = -4, not 4.

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18. [4 marks] — 3 marks (a), 1 mark (b)

(a) Answer: P(1, 3)

Working:

• Set equations equal: x + 2 = 4 - x
• 2x = 2, so x = 1
• Substitute: y = 1 + 2 = 3
• Check in second equation: y = 4 - 1 = 3 ✓
• So P(1, 3)

(b) Answer: x = 1, y = 3

Teaching Notes:

• The simultaneous equations are rearrangements of the same lines:
  • y = x + 2 → x - y = -2
  • y = 4 - x → x + y = 4
• The intersection point is the solution to both equations.

Marking (a): 1 mark for eliminating one variable; 1 mark for finding x; 1 mark for finding y. Marking (b): 1 mark for stating both values (follow through from errors in part a).

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19. [3 marks]

Answer:

• Gradient of AB = $\frac{1-1}{4-(-2)} = \frac{0}{6} = 0$
• Gradient of DC = $\frac{5-5}{6-0} = \frac{0}{6} = 0$
• Both gradients equal 0, so AB // DC (parallel)

Alternatively: Both are horizontal lines (constant y-value), so both have gradient 0.

Marking: 1 mark for each gradient calculated; 1 mark for conclusion with reason. Teaching Notes: Parallel lines have equal gradients. Horizontal lines have gradient 0. Common mistake: Calculating distance instead of gradient; not stating the conclusion explicitly.

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20. [4 marks] — 2 marks each part

(a) Answer: $y = \frac{1}{2}x - 4$

Working:

• Parallel lines have equal gradient, so m = $\frac{1}{2}$
• $y = \frac{1}{2}x + c$
• Substitute (2, -3): $-3 = \frac{1}{2}(2) + c = 1 + c$
• c = -4
• Equation: $y = \frac{1}{2}x - 4$

(b) Answer: No, the line does NOT pass through (6, 1)

Working:

• Check: when x = 6, y = $\frac{1}{2}(6) - 4 = 3 - 4 = -1$
• But the point has y = 1, not -1.
• Alternatively: $1 = \frac{1}{2}(6) - 4 = -1$, and 1 ≠ -1.
• Or substitute into original: $1 = \frac{1}{2}(6) - 4$ → 1 = 3 - 4 → 1 = -1, which is false.

Marking (a): 1 mark for gradient; 1 mark for finding c and equation. Marking (b): 1 mark for substitution; 1 mark for correct conclusion with evidence. Common mistake: In (b), saying "no" without showing the check — full marks require justification.

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TOTAL MARKS: 40

Section totals: A = 10, B = 10, C = 10, D = 10