Secondary 2 Mathematics Graphs Coordinate Geometry Quiz

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Secondary 2 Mathematics Quiz - Graphs Coordinate Geometry

Name: ___________________________
Class: ___________________________
Date: ___________________________
Score: _______ / 40

Duration: 50 minutes
Total Marks: 40

Instructions:

• Answer all 20 questions.
• Show your working clearly where required.
• Use the provided spaces to write your answers.
• Calculators may be used.

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Section A (Questions 1–8) — Basic Coordinate Geometry (2 marks each)

1. Plot the point $A(3, -2)$ on a Cartesian plane. What are the coordinates of the point $B$ which is 4 units to the left of $A$ and 5 units above $A$?

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2. Find the gradient of the line passing through the points $(2, 5)$ and $(6, 13)$.

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3. Write down the equation of a line with gradient $3$ that passes through the point $(0, -1)$.

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4. The line $y = -2x + 7$ cuts the $y$-axis at point $P$. State the coordinates of $P$.

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5. Two points are $C(-1, 4)$ and $D(5, 4)$. What is the length of the line segment $CD$?

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6. A line has gradient $0$ and passes through $(2, 9)$. Write its equation.

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7. Find the $x$-intercept of the line $y = 4x - 8$.

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8. The coordinates of three vertices of a rectangle are $(1, 1)$, $(1, 5)$, and $(4, 1)$. What are the coordinates of the fourth vertex?

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Section B (Questions 9–14) — Graphs of Linear Functions (3 marks each)

9. Draw the graph of $y = 2x - 3$ for $x = -1, 0, 1, 2, 3$ on the grid below. Use your graph to find the value of $y$ when $x = 2.5$.

Image pending generation: graph for Q9.

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10. Find the gradient and $y$-intercept of the line $3y = 6x + 9$.

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11. A taxi company charges a flat fee of \3$plus$$2$per km. Write an equation connecting cost$C$(in$$`) and distance $d$ (in km). Hence find the cost of a 7 km ride.

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12. The points $E(0, 2)$ and $F(4, 0)$ lie on a straight line. Find the equation of the line $EF$ in the form $y = mx + c$.

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13. Two lines have equations $y = 2x + 1$ and $y = -x + 7$. Find the coordinates of their point of intersection.

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14. A line passes through $(1, 3)$ and is parallel to the line $y = 4x - 2$. Find its equation.

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Section C (Questions 15–20) — Application and Interpretation (4 marks each)

15. The table below shows the distance travelled by a cyclist over time.

Time (min)	0	5	10	15	20
Distance (km)	0	2	4	6	8

Plot the points on the grid and draw the best straight line. Find the gradient and explain what it represents.

Image pending generation: graph for Q15.

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16. A line has equation $2x + 5y = 10$. Find the $x$- and $y$-intercepts and use them to sketch the line.

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17. Triangle $PQR$ has vertices $P(1, 2)$, $Q(1, 6)$, $R(5, 2)$. Show that $PQR$ is a right-angled triangle by using gradients.

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18. The cost $C$ of printing $n$ copies is given by $C = 50 + 0.20n$. Draw the graph for $n = 0, 100, 200, 300$ and find the cost of 250 copies from your graph.

Image pending generation: graph for Q18.

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19. Two points $M(-2, 3)$ and $N(4, -1)$ are given. Find the midpoint of $MN$ and the gradient of the line $MN$.

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20. A mobile plan costs \20$monthly plus$$0.10$per SMS. Another plan costs$$30$ monthly with unlimited SMS. Write equations for both and find the number of SMS where both cost the same.

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

Total Marks: 40
Topic: Graphs & Coordinate Geometry (Sec 2 G3 syllabus: N6 Functions and Graphs)

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Q1. $B(-1, 3)$
Marks: 2
Teaching note: Starting at $A(3, -2)$, moving 4 left means subtract 4 from x: $3 - 4 = -1$. Moving 5 up means add 5 to y: $-2 + 5 = 3$. So $B(-1, 3)$. Common mistake: mixing up x (horizontal) and y (vertical) directions.

Q2. Gradient $= 2$
Marks: 2
Teaching note: Gradient formula $m = \frac{y_2 - y_1}{x_2 - x_1}$. Using $(2,5)$ and $(6,13)$: $m = \frac{13 - 5}{6 - 2} = \frac{8}{4} = 2$. The gradient tells us the line rises 2 units for every 1 unit right.

Q3. $y = 3x - 1$
Marks: 2
Teaching note: Equation of line: $y = mx + c$, where $m$ is gradient, $c$ is $y$-intercept. Given $m = 3$ and point $(0, -1)$ means $c = -1$. So $y = 3x - 1$.

Q4. $P(0, 7)$
Marks: 2
Teaching note: $y$-intercept occurs when $x = 0$. Substitute: $y = -2(0) + 7 = 7$. So $P(0, 7)$.

Q5. Length $= 6$ units
Marks: 2
Teaching note: Both points have same $y$-coordinate (4), so distance is difference in x: $|5 - (-1)| = 6$. Formula: length $= \sqrt{(5+1)^2 + (4-4)^2} = 6$.

Q6. $y = 9$
Marks: 2
Teaching note: Gradient 0 means horizontal line. It passes through $(2,9)$ so $y$ is always 9. Equation: $y = 9$.

Q7. $x = 2$
Marks: 2
Teaching note: $x$-intercept is where $y = 0$. $0 = 4x - 8 \Rightarrow 4x = 8 \Rightarrow x = 2$.

Q8. $(4, 5)$
Marks: 2
Teaching note: Rectangle has opposite sides equal and parallel. Given $(1,1)$, $(1,5)$ vertical side, $(1,1)$ to $(4,1)$ horizontal. Fourth vertex shares x with $(4,1)$ and y with $(1,5)$: $(4,5)$.

Q9. Points: $(-1,-5), (0,-3), (1,-1), (2,1), (3,3)$; at $x=2.5$, $y = 2(2.5)-3 = 2$
Marks: 3
Teaching note: Substitute each x into $y=2x-3$. Plot and join. For $x=2.5$: $y=5-3=2$. Marking: 1 for correct points, 1 for line, 1 for $y=2$.

Q10. Gradient $= 2$, $y$-intercept $= 3$
Marks: 3
Teaching note: Divide equation by 3: $y = 2x + 3$. So $m=2$, $c=3$. Marking: 1 for rearranging, 1 for gradient, 1 for intercept.

Q11. $C = 3 + 2d$; for 7 km: C = 3 + 14 = \17$*Marks: 3* Teaching note: Flat fee is constant (3), per km is variable (2d). Equation$C = 3 + 2d$. At$d=7$:$C = 3+2(7)=17$. Marking: 1 eqn, 1 sub, 1 answer.

Q12. $y = -\frac{1}{2}x + 2$
Marks: 3
Teaching note: Gradient $m = \frac{0-2}{4-0} = -\frac{1}{2}$. Intercept $c = 2$ (point E). So $y = -\frac{1}{2}x + 2$. Marking: 1 grad, 1 intercept, 1 eqn.

Q13. $(2, 5)$
Marks: 3
Teaching note: Set equal: $2x+1 = -x+7 \Rightarrow 3x = 6 \Rightarrow x=2$. Then $y = 2(2)+1 = 5$. Intersection $(2,5)$. Marking: 1 eqn, 1 x, 1 y.

Q14. $y = 4x - 1$
Marks: 3
Teaching note: Parallel lines have same gradient $m=4$. Through $(1,3)$: $3 = 4(1) + c \Rightarrow c = -1$. So $y = 4x -1$. Marking: 1 parallel grad, 1 find c, 1 eqn.

Q15. Gradient $= 0.4$ km/min (or 24 km/h); represents speed
Marks: 4
Teaching note: Plot points, line through them. Gradient $= \frac{8-0}{20-0} = 0.4$ km per min. This is the cyclist's constant speed. Marking: 1 plot, 1 line, 1 grad, 1 meaning.

Q16. $x$-int $= 5$, $y$-int $= 2$
Marks: 4
Teaching note: For $x$-int set $y=0$: $2x=10 \Rightarrow x=5$. For $y$-int set $x=0$: $5y=10 \Rightarrow y=2$. Sketch line through $(5,0)$ and $(0,2)$. Marking: 1 each intercept, 2 for sketch.

Q17. Grad $PQ = \frac{6-2}{1-1}$ undefined (vertical); grad $PR = \frac{2-2}{5-1}=0$ (horizontal); perpendicular $\Rightarrow$ right angle at P
Marks: 4
Teaching note: $PQ$ vertical, $PR$ horizontal, so angle at P is 90°. Marking: 2 for gradients, 2 for conclusion.

Q18. Points: $(0,50),(100,70),(200,90),(300,110)$; at $n=250$, $C = 50+0.2(250)=100$
Marks: 4
Teaching note: Substitute n. From graph or calc: $C=100$. Marking: 1 pts, 1 line, 1 read, 1 answer.

Q19. Midpoint $= (1, 1)$, gradient $= -\frac{2}{3}$
Marks: 4
Teaching note: Midpoint $= (\frac{-2+4}{2}, \frac{3+(-1)}{2}) = (1,1)$. Grad $= \frac{-1-3}{4-(-2)} = -\frac{4}{6} = -\frac{2}{3}$. Marking: 2 midpoint, 2 gradient.

Q20. Plan A: $C = 20 + 0.10s$; Plan B: $C = 30$; equal at $s = 100$ SMS
Marks: 4
Teaching note: Set $20+0.10s = 30 \Rightarrow 0.10s = 10 \Rightarrow s=100$. Marking: 1 eqn A, 1 eqn B, 2 solve.