Secondary 4 Elementary Mathematics Graphs Coordinate Geometry Quiz

<!-- TuitionGoWhere generation metadata: stage=3-0; model=google/gemma-4-31b-it; model_label=Gemma 4 31B; generated=2026-05-31; Sources: Stage 2-1 real exam-derived templates and Stage 2-2 exam-enriched syllabus. -->

Secondary 4 Elementary Mathematics Quiz - Graphs Coordinate Geometry

Name: ____________________ Class: __________ Date: __________ Score: ________ / 50

Duration: 75 Minutes
Total Marks: 50
Instructions: Answer all questions. Show all necessary working. Use a calculator where appropriate.

---

Section A: Basic Coordinate Geometry (Questions 1-8)

Focus: Gradient, Distance, and Midpoints

1. Find the gradient of the straight line passing through the points $A(-3, 5)$ and $B(2, -1)$. [2]
   
   Answer: ____________________

2. Calculate the length of the line segment joining $P(4, -2)$ and $Q(-1, 6)$. Give your answer to 3 significant figures. [2]
   
   Answer: ____________________

3. The midpoint of the line segment $RS$ is $M(2, 3)$. Given that the coordinates of $R$ are $(-1, 7)$, find the coordinates of $S$. [2]
   
   Answer: ____________________

4. A straight line $L$ has a gradient of $-\frac{2}{3}$ and passes through the point $(6, 1)$. Find the equation of line $L$ in the form $y = mx + c$. [2]
   
   Answer: ____________________

5. Line $A$ has the equation $y = 4x - 5$. Line $B$ is parallel to Line $A$ and passes through $(0, 7)$. Find the equation of Line $B$. [2]
   
   Answer: ____________________

6. Line $C$ has the equation $2y - 3x = 12$. Find the gradient of a line perpendicular to Line $C$. [2]
   
   Answer: ____________________

7. Find the coordinates of the point where the line $y = 3x + 2$ intersects the x-axis. [2]
   
   Answer: ____________________

8. Point $K(2, 5)$ is equidistant from point $L(0, 1)$ and point $M(4, 1)$. Verify this by calculating the lengths $KL$ and $KM$. [3]
   
   Answer: ____________________

---

Section B: Graphical Interpretation & Functions (Questions 9-15)

Focus: Linear Graphs, Piecewise Functions, and Power Functions

9. A mobile data plan charges a fixed monthly fee of \10$and an additional$$2$for every GB of data used. Express the total monthly cost$C$in terms of data used$g$. [2]
   
   Answer: ____________________

10. A courier service charges \5$for the first 2 kg of a parcel and$$3$per kg for every additional kg thereafter. (a) Find the cost of sending a 5 kg parcel. [2] (b) If the total cost is$$23$, find the weight of the parcel. [2]
    
    Answer (a): ____________________ (b): ____________________

11. A graph of the function $y = ax^2$ passes through the point $(3, 18)$. Find the value of $a$. [2]
    
    Answer: ____________________

12. Given the function $y = \frac{k}{x}$, the graph passes through $(4, 5)$. Find the coordinates of the point on the graph where $y = 2$. [2]
    
    Answer: ____________________

13. The relationship between two variables $u$ and $v$ is given by $u = kv^2$. If $v$ is increased by $20\%$, calculate the percentage increase in $u$. [3]
    
    Answer: ____________________

14. A distance-time graph shows a car accelerating uniformly from rest to a speed of $20\text{ m/s}$ in $10\text{ s}$. Describe the shape of the distance-time graph for this period. [2]
    
    Answer: ____________________

15. A function is defined by $y = 12 - \frac{24}{x}$. Find the x-intercept and y-intercept of this graph. [3]
    
    Answer: ____________________

---

Section C: Advanced Coordinate Problems (Questions 16-20)

Focus: Intersections, Perpendicular Bisectors, and Geometric Proofs

16. Find the coordinates of the point of intersection of the lines $y = 2x - 3$ and $y = -x + 6$. [3]
    
    Answer: ____________________

17. Points $A(1, 2)$ and $B(5, 4)$ are the endpoints of a diameter of a circle. Find the coordinates of the center of the circle. [2]
    
    Answer: ____________________

18. Find the equation of the perpendicular bisector of the line segment joining $P(-2, 4)$ and $Q(4, 6)$. [4]
    
    Answer: ____________________

19. A triangle has vertices $X(0, 0)$, $Y(6, 0)$, and $Z(3, 3\sqrt{3})$. Show that triangle $XYZ$ is an equilateral triangle. [4]
    
    Answer: ____________________

20. Line $L_1$ passes through $(2, 3)$ and $(4, 7)$. Line $L_2$ is perpendicular to $L_1$ and passes through the midpoint of the segment joining $(2, 3)$ and $(4, 7)$. Find the equation of $L_2$. [4]
    
    Answer: ____________________

<!-- TuitionGoWhere generation metadata: stage=3-0; model=google/gemma-4-31b-it; model_label=Gemma 4 31B; generated=2026-05-31; Sources: Stage 2-1 real exam-derived templates and Stage 2-2 exam-enriched syllabus. -->

Secondary 4 Elementary Mathematics Quiz - Answers

Topic: Graphs Coordinate Geometry

1. $m = \frac{-1 - 5}{2 - (-3)} = \frac{-6}{5} = -1.2$

2. $d = \sqrt{(-1-4)^2 + (6-(-2))^2} = \sqrt{(-5)^2 + 8^2} = \sqrt{25 + 64} = \sqrt{89} \approx 9.43$

3. $M = (\frac{x_R + x_S}{2}, \frac{y_R + y_S}{2}) \Rightarrow 2 = \frac{-1 + x_S}{2} \Rightarrow x_S = 5$; $3 = \frac{7 + y_S}{2} \Rightarrow y_S = -1$. Point $S(5, -1)$.

4. $y - 1 = -\frac{2}{3}(x - 6) \Rightarrow y = -\frac{2}{3}x + 4 + 1 \Rightarrow y = -\frac{2}{3}x + 5$.

5. Parallel means $m = 4$. $y = 4x + 7$.

6. $2y = 3x + 12 \Rightarrow y = \frac{3}{2}x + 6$. Gradient $m_1 = \frac{3}{2}$. Perpendicular gradient $m_2 = -\frac{2}{3}$.

7. $0 = 3x + 2 \Rightarrow 3x = -2 \Rightarrow x = -\frac{2}{3}$. Point $(-\frac{2}{3}, 0)$.

8. $KL = \sqrt{(0-2)^2 + (1-5)^2} = \sqrt{4 + 16} = \sqrt{20}$. $KM = \sqrt{(4-2)^2 + (1-5)^2} = \sqrt{4 + 16} = \sqrt{20}$. Since $KL = KM$, $K$ is equidistant.

9. $C = 10 + 2g$

10. (a) First 2kg = \5$. Remaining 3kg =$3 \times 3 = $9$. Total =$5 + 9 = $14$. (b)$23 = 5 + 3(w - 2) \Rightarrow 18 = 3(w - 2) \Rightarrow 6 = w - 2 \Rightarrow w = 8\text{ kg}$.

11. $18 = a(3)^2 \Rightarrow 18 = 9a \Rightarrow a = 2$.

12. $5 = \frac{k}{4} \Rightarrow k = 20$. If $y = 2$, $2 = \frac{20}{x} \Rightarrow x = 10$. Point $(10, 2)$.

13. $u_{old} = kv^2$. $v_{new} = 1.2v$. $u_{new} = k(1.2v)^2 = 1.44kv^2 = 1.44u_{old}$. Increase = $44\%$.

14. A curve (parabola) starting from $(0,0)$ and concave upwards, as distance increases quadratically with time during uniform acceleration.

15. X-intercept: $0 = 12 - \frac{24}{x} \Rightarrow \frac{24}{x} = 12 \Rightarrow x = 2$. Point $(2, 0)$. Y-intercept: $x=0$ is undefined (asymptote). No y-intercept.

16. $2x - 3 = -x + 6 \Rightarrow 3x = 9 \Rightarrow x = 3$. $y = 2(3) - 3 = 3$. Point $(3, 3)$.

17. Midpoint = $(\frac{1+5}{2}, \frac{2+4}{2}) = (3, 3)$.

18. Midpoint $M = (\frac{-2+4}{2}, \frac{4+6}{2}) = (1, 5)$. Gradient $PQ = \frac{6-4}{4-(-2)} = \frac{2}{6} = \frac{1}{3}$. Perpendicular gradient = $-3$. Eq: $y - 5 = -3(x - 1) \Rightarrow y = -3x + 8$.

19. $XY = \sqrt{(6-0)^2 + (0-0)^2} = 6$. $YZ = \sqrt{(3-6)^2 + (3\sqrt{3}-0)^2} = \sqrt{9 + 27} = \sqrt{36} = 6$. $XZ = \sqrt{(3-0)^2 + (3\sqrt{3}-0)^2} = \sqrt{9 + 27} = \sqrt{36} = 6$. All sides equal, therefore equilateral.

20. Gradient $L_1 = \frac{7-3}{4-2} = 2$. Perpendicular gradient = $-\frac{1}{2}$. Midpoint = $(3, 5)$. Eq: $y - 5 = -\frac{1}{2}(x - 3) \Rightarrow y = -\frac{1}{2}x + 1.5 + 5 \Rightarrow y = -\frac{1}{2}x + 6.5$.