Secondary 1 Mathematics Graphs Coordinate Geometry Quiz

Secondary 1 Mathematics Quiz - Graphs Coordinate Geometry (Answer Key)

Total Marks: 40

Section A: Multiple Choice Questions (5 × 1 = 5 marks)

1. Answer: A $(1, -1)$
Working: Substitute $x = 1$ into $y = 2x - 3$: $y = 2(1) - 3 = -1$. Point $(1, -1)$ satisfies the equation.
Check others: B: $2(2)-3=1 \neq 0$; C: $2(0)-3=-3 \neq 3$; D: $2(-1)-3=-5$ ✓ but A is also correct. Wait, D also gives -5. Let me recheck: $y = 2(-1) - 3 = -2 - 3 = -5$. So D is also on the line. But typically only one answer is correct in MCQ. Let me check the question again. The question asks "Which of the following points lies on the line". Both A and D lie on the line. This is a flaw in the question design. For the answer key, I'll note both A and D are correct, but the intended answer is likely A as it's the first correct option.
Marking note: Both A $(1, -1)$ and D $(-1, -5)$ satisfy $y = 2x - 3$. If this were a real exam, the question would be flawed. Accept either A or D. [1]

2. Answer: A $2$
Working: Gradient $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{13 - 5}{6 - 2} = \frac{8}{4} = 2$. [1]

3. Answer: B $6$
Working: Rearrange $3x + 2y = 12$ to $y = mx + c$ form: $2y = -3x + 12 \Rightarrow y = -\frac{3}{2}x + 6$. The $y$-intercept $c = 6$. Alternatively, set $x = 0$: $3(0) + 2y = 12 \Rightarrow 2y = 12 \Rightarrow y = 6$. [1]

4. Answer: B $7$
Working: Points $P(-3, 2)$ and $Q(4, 2)$ have the same $y$-coordinate, so distance $= |4 - (-3)| = 7$. [1]

5. Answer: B $y = -3x + 7$
Working: Parallel lines have the same gradient. The given line $y = -3x + 4$ has gradient $-3$. Option B has gradient $-3$. [1]

Section B: Short Answer Questions (10 × 2 = 20 marks)

6. Answer: Points plotted correctly on the grid.
Expected plot:

• $A(2, 3)$: 2 right, 3 up from origin
• $B(-1, 4)$: 1 left, 4 up from origin
• $C(-3, -2)$: 3 left, 2 down from origin
• $D(4, -1)$: 4 right, 1 down from origin
  Marking: 1 mark for all four points plotted correctly, 1 mark for correct labels. [2]

7. Answer: $-1$
Working: Gradient $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 5}{4 - (-2)} = \frac{-6}{6} = -1$. [2]

8. Answer: $y = \frac{3}{4}x - 2$
Working: Given gradient $m = \frac{3}{4}$ and $y$-intercept $c = -2$ (since line passes through $(0, -2)$). Equation: $y = \frac{3}{4}x - 2$. [2]

9. (a) Answer: $\frac{5}{2}$ or $2.5$
Working: $2y = 5x - 10 \Rightarrow y = \frac{5}{2}x - 5$. Gradient $m = \frac{5}{2}$. [1]

(b) Answer: $-5$
Working: From $y = \frac{5}{2}x - 5$, $y$-intercept $c = -5$. Or set $x = 0$: $2y = -10 \Rightarrow y = -5$. [1]

10. Answer: $(-1, 2)$
Working: Midpoint $= \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = \left(\frac{3 + (-5)}{2}, \frac{-4 + 8}{2}\right) = \left(\frac{-2}{2}, \frac{4}{2}\right) = (-1, 2)$. [2]

11. Answer: Completed table and plotted line.
Table values:

Working: Substitute each $x$ into $y = -2x + 6$:

• $x = -1$: $y = -2(-1) + 6 = 2 + 6 = 9$
• $x = 0$: $y = 6$
• $x = 1$: $y = -2(1) + 6 = 4$
• $x = 2$: $y = -2(2) + 6 = 2$
• $x = 3$: $y = -2(3) + 6 = 0$

Plotting: Points $(-1, 9)$, $(0, 6)$, $(1, 4)$, $(2, 2)$, $(3, 0)$ plotted and joined with a straight line.
Marking: 1 mark for correct table, 1 mark for correct plotting and line. [2]

12. Answer: Yes, the points are collinear.
Working: Gradient of $AB = \frac{6 - 2}{4 - 1} = \frac{4}{3}$. Gradient of $BC = \frac{10 - 6}{7 - 4} = \frac{4}{3}$. Since gradients are equal and point $B$ is common, $A$, $B$, $C$ lie on the same straight line.
Alternative: Area of triangle $= 0$ or check if $C$ satisfies equation of line through $A$ and $B$. [2]

13. Answer: $10$
Working: Distance $= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(4 - (-2))^2 + (5 - (-3))^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$. [2]

14. (a) Answer: $2$
Working: Gradient $m = \frac{11 - 3}{6 - 2} = \frac{8}{4} = 2$. [1]

(b) Answer: $y = 2x - 1$
Working: Using $y = mx + c$ with $m = 2$ and point $(2, 3)$: $3 = 2(2) + c \Rightarrow 3 = 4 + c \Rightarrow c = -1$. Equation: $y = 2x - 1$. [1]

15. Answer: $A\left(\frac{7}{4}, 0\right)$ or $(1.75, 0)$; $B(0, -7)$
Working:

• For $A$ (x-intercept): set $y = 0$: $0 = 4x - 7 \Rightarrow 4x = 7 \Rightarrow x = \frac{7}{4}$. So $A\left(\frac{7}{4}, 0\right)$.
• For $B$ (y-intercept): set $x = 0$: $y = 4(0) - 7 = -7$. So $B(0, -7)$. [1] [1]

Section C: Structured Questions (5 × 3 = 15 marks)

16. (a) Answer: $-\frac{2}{3}$
Working: Gradient $m = \frac{0 - 4}{6 - 0} = \frac{-4}{6} = -\frac{2}{3}$. [1]

(b) Answer: $y = -\frac{2}{3}x + 4$
Working: $y$-intercept is $4$ (point $P(0, 4)$). Gradient $m = -\frac{2}{3}$. Equation: $y = -\frac{2}{3}x + 4$. [1]

(c) Answer: $R\left(8, -\frac{4}{3}\right)$ or $\left(8, -1\frac{1}{3}\right)$
Working: Substitute $x = 8$ into $y = -\frac{2}{3}x + 4$: $y = -\frac{2}{3}(8) + 4 = -\frac{16}{3} + 4 = -\frac{16}{3} + \frac{12}{3} = -\frac{4}{3}$. [1]

17. (a) Answer: $\frac{1}{2}$
Working: The given line is $y = \frac{1}{2}x + 1$, so gradient $m = \frac{1}{2}$. [1]

(b) Answer: $-2$
Working: Perpendicular gradients satisfy $m_1 \times m_2 = -1$. So $m_2 = -\frac{1}{m_1} = -\frac{1}{1/2} = -2$. [1]

(c) Answer: $y = -2x + 1$
Working: Perpendicular line has gradient $-2$ and passes through $(2, -3)$. Substitute: $-3 = -2(2) + c \Rightarrow -3 = -4 + c \Rightarrow c = 1$. Equation: $y = -2x + 1$. [1]

18. (a) Answer: $2\sqrt{13}$ or $\sqrt{52}$
Working: $AB = \sqrt{(4 - (-2))^2 + (5 - 1)^2} = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$. [1]

(b) Answer: $(5, 1)$
Working: Midpoint of $BC = \left(\frac{4 + 6}{2}, \frac{5 + (-3)}{2}\right) = \left(\frac{10}{2}, \frac{2}{2}\right) = (5, 1)$. [1]

(c) Answer: No, triangle $ABC$ is not right-angled.
Working: Check if any two sides are perpendicular using gradients or Pythagoras.
Gradients: $m_{AB} = \frac{4}{6} = \frac{2}{3}$, $m_{BC} = \frac{-8}{2} = -4$, $m_{AC} = \frac{-4}{8} = -\frac{1}{2}$.
Products: $m_{AB} \times m_{BC} = \frac{2}{3} \times (-4) = -\frac{8}{3} \neq -1$; $m_{BC} \times m_{AC} = -4 \times (-\frac{1}{2}) = 2 \neq -1$; $m_{AC} \times m_{AB} = -\frac{1}{2} \times \frac{2}{3} = -\frac{1}{3} \neq -1$. No perpendicular sides.
Alternatively, check Pythagoras: $AB^2 = 52$, $BC^2 = 68$, $AC^2 = 80$. No sum of two equals the third ($52 + 68 = 120 \neq 80$, etc.). [1]

19. (a) Answer: $-\frac{2}{3}$
Working: $2x + 3y = 12 \Rightarrow 3y = -2x + 12 \Rightarrow y = -\frac{2}{3}x + 4$. Gradient $m = -\frac{2}{3}$. [1]

(b) Answer: $-\frac{2}{3}$
Working: Parallel lines have equal gradients. [1]

(c) Answer: $y = -\frac{2}{3}x + 2$
Working: $L_2$ has gradient $-\frac{2}{3}$ and passes through $(-3, 4)$. Substitute: $4 = -\frac{2}{3}(-3) + c \Rightarrow 4 = 2 + c \Rightarrow c = 2$. Equation: $y = -\frac{2}{3}x + 2$. [1]

20. (a) Answer: $\frac{3}{5}$
Working: Gradient of $PQ = \frac{5 - 2}{3 - (-2)} = \frac{3}{5}$. [1]

(b) Answer: $\frac{3}{5}$
Working: Gradient of $RS = \frac{-2 - 1}{1 - 6} = \frac{-3}{-5} = \frac{3}{5}$. [1]

(c) Answer: $PQ$ and $RS$ are parallel (equal gradients). Quadrilateral $PQRS$ is a trapezium (or trapezoid).
Working: Since $m_{PQ} = m_{RS} = \frac{3}{5}$, $PQ \parallel RS$. A quadrilateral with one pair of parallel sides is a trapezium.
Check other pair: $m_{QR} = \frac{1 - 5}{6 - 3} = -\frac{4}{3}$, $m_{SP} = \frac{2 - (-2)}{-2 - 1} = \frac{4}{-3} = -\frac{4}{3}$. So $QR \parallel SP$ as well! Both pairs of opposite sides are parallel, so $PQRS$ is actually a parallelogram.
Correction: The question asks "What can you conclude about lines PQ and RS? Hence, state the special name of quadrilateral PQRS." Since both pairs of opposite sides are parallel, it's a parallelogram. But the question leads students to first notice $PQ \parallel RS$, then check the other pair. The special name is parallelogram. [1]

Marking Notes for Teachers:

• Q1: Both A and D are mathematically correct. Award mark for either.
• Q11: Allow follow-through marks if table has arithmetic errors but plotting is consistent with their table.
• Q12: Accept gradient method, area method, or equation substitution method.
• Q18(c): Accept either gradient product method or Pythagoras method. Must show working for the mark.
• Q20(c): The quadrilateral is a parallelogram (both pairs of opposite sides parallel). If student only states trapezium based on $PQ \parallel RS$ without checking the other pair, award partial credit (0.5/1) but full mark requires parallelogram.
• For all coordinate geometry questions: Award method marks for correct formula substitution even if arithmetic error occurs.