O Level Additional Mathematics Graphs Coordinate Geometry Quiz

O-Level Additional Mathematics Quiz - Graphs Coordinate Geometry (Answer Key)

1. (a) Gradient $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 5}{8 - 2} = \frac{-6}{6} = -1$. [1] (b) Midpoint $= \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) = \left(\frac{2+8}{2}, \frac{5+(-1)}{2}\right) = (5, 2)$. [2]

2. (a) $3x - 2y + 6 = 0 \Rightarrow 2y = 3x + 6 \Rightarrow y = \frac{3}{2}x + 3$. Gradient $m_1 = \frac{3}{2}$. [1] (b) Gradient of perpendicular line $m_2 = -\frac{1}{m_1} = -\frac{2}{3}$. Equation: $y - 1 = -\frac{2}{3}(x - 4)$. $3(y - 1) = -2(x - 4)$ $3y - 3 = -2x + 8$ $2x + 3y - 11 = 0$. [3]

3. (a) Gradient $PQ = \frac{6-2}{3-(-1)} = \frac{4}{4} = 1$. Gradient $QR = \frac{-2-6}{5-3} = \frac{-8}{2} = -4$. Product of gradients $1 \times (-4) = -4 \neq -1$. Correction in logic for student check: Let's re-calculate coordinates. $P(-1, 2), Q(3, 6) \rightarrow m_{PQ} = 1$. $Q(3, 6), R(5, -2) \rightarrow m_{QR} = -4$. Wait, the question asks to show it is right-angled at Q. Let's check lengths: $PQ^2 = (3 - -1)^2 + (6 - 2)^2 = 16 + 16 = 32$. $QR^2 = (5 - 3)^2 + (-2 - 6)^2 = 4 + 64 = 68$. $PR^2 = (5 - -1)^2 + (-2 - 2)^2 = 36 + 16 = 52$. $32 + 52 \neq 68$. It is not right-angled at Q with these coordinates. Self-Correction for Answer Key based on generated question: Let's re-evaluate the coordinates in Q3. $P(-1, 2), Q(3, 6), R(5, -2)$. $m_{PQ} = 1$. $m_{QR} = -4$. $m_{PR} = \frac{-2-2}{5-(-1)} = \frac{-4}{6} = -\frac{2}{3}$. None of the products are -1. The question as generated in the prompt text contains a flaw in the specific numbers chosen for a "Show that" question. Adjustment for valid Answer Key: Let's assume the question intended $R(7, 2)$. $m_{PQ} = 1$. $m_{QR} = \frac{2-6}{7-3} = -1$. Product = -1. However, I must answer the question as written or note the error. Let's check $P(-1,2), Q(3,6), R(5, -2)$ again. Vector $QP = (-4, -4)$. Vector $QR = (2, -8)$. Dot product $-8 + 32 = 24 \neq 0$. Vector $PQ = (4, 4)$. Vector $PR = (6, -4)$. Dot product $24 - 16 = 8 \neq 0$. Vector $RQ = (-2, 8)$. Vector $RP = (-6, 4)$. Dot product $12 + 32 = 44 \neq 0$. The triangle is not right-angled. Note to User: In a real exam generation, the numbers would be verified. For this practice key, I will provide the method for a valid right-angled triangle, e.g., if $R$ was $(7,2)$. Revised Answer for Q3 (assuming typo in question generation for R): If we assume the question meant to ask for the area regardless, or if we adjust R to $(7,2)$: (a) $m_{PQ}=1, m_{QR}=-1 \Rightarrow$ Perpendicular. (b) Area $= \frac{1}{2} \times PQ \times QR = \frac{1}{2} \sqrt{32} \sqrt{32} = 16$. Given the strict constraint to answer the generated text: I will provide the calculation for the area using the "Shoelace" or determinant formula for the coordinates given, and note that it is not right-angled, but answer the area part. (b) Area $= \frac{1}{2} |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|$ $= \frac{1}{2} |-1(6 - (-2)) + 3(-2 - 2) + 5(2 - 6)|$ $= \frac{1}{2} |-1(8) + 3(-4) + 5(-4)|$ $= \frac{1}{2} |-8 - 12 - 20| = \frac{1}{2} |-40| = 20$. [2] (Note: Part (a) "Show that" fails with these numbers. In a live exam, students would likely find the gradients and show they are not perpendicular, or the question would be flawed. For the purpose of this key, we provide the Area calculation which is robust.)

4. Gradient $AB = \frac{7-3}{4-1} = \frac{4}{3}$. Gradient $BC = \frac{11-7}{k-4} = \frac{4}{k-4}$. Since collinear, gradients are equal: $\frac{4}{3} = \frac{4}{k-4}$. $3 = k - 4 \Rightarrow k = 7$. [2]

5. $m = \frac{-6 - 4}{3 - (-2)} = \frac{-10}{5} = -2$. $y = -2x + c$. Substitute $(-2, 4)$: $4 = -2(-2) + c \Rightarrow 4 = 4 + c \Rightarrow c = 0$. $m = -2, c = 0$. [3]

6. (a) $(x - 3)^2 + (y + 2)^2 = 25$. [1] (b) $x^2 - 6x + 9 + y^2 + 4y + 4 = 25$. $x^2 + y^2 - 6x + 4y + 13 - 25 = 0$. $x^2 + y^2 - 6x + 4y - 12 = 0$. [2]

7. (a) Centre $(h, k)$ from $x^2 + y^2 + 2gx + 2fy + c = 0$. $2g = -6 \Rightarrow g = -3 \Rightarrow h = -g = 3$. $2f = 8 \Rightarrow f = 4 \Rightarrow k = -f = -4$. Centre: $(3, -4)$. [2] (b) Radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-3)^2 + 4^2 - (-11)} = \sqrt{9 + 16 + 11} = \sqrt{36} = 6$. [2]

8. Substitute $y = 2x + k$ into $x^2 + y^2 = 20$: $x^2 + (2x + k)^2 = 20$ $x^2 + 4x^2 + 4kx + k^2 - 20 = 0$ $5x^2 + 4kx + (k^2 - 20) = 0$. For tangent, discriminant $\Delta = 0$. $b^2 - 4ac = 0$ $(4k)^2 - 4(5)(k^2 - 20) = 0$ $16k^2 - 20k^2 + 400 = 0$ $-4k^2 + 400 = 0$ $k^2 = 100 \Rightarrow k = \pm 10$. [4]

9. (a) Centre is midpoint of $AB$: $(\frac{1+5}{2}, \frac{2+6}{2}) = (3, 4)$. [1] (b) Radius squared $r^2 = (3-1)^2 + (4-2)^2 = 2^2 + 2^2 = 8$. Equation: $(x - 3)^2 + (y - 4)^2 = 8$. [3]

10. (a) Centre $C(2, -1)$. Point $P(5, 3)$. Gradient $CP = \frac{3 - (-1)}{5 - 2} = \frac{4}{3}$. [2] (b) Gradient of tangent $m_T = -\frac{1}{m_{CP}} = -\frac{3}{4}$. Equation: $y - 3 = -\frac{3}{4}(x - 5)$. $4(y - 3) = -3(x - 5)$ $4y - 12 = -3x + 15$ $3x + 4y - 27 = 0$. [3]

11. Substitute $y = x + 1$ into $x^2 + y^2 = 1$: $x^2 + (x + 1)^2 = 1$ $x^2 + x^2 + 2x + 1 = 1$ $2x^2 + 2x = 0$ $2x(x + 1) = 0$. Discriminant of $2x^2 + 2x + 0 = 0$: $\Delta = b^2 - 4ac = 2^2 - 4(2)(0) = 4$. Since $\Delta > 0$, there are two distinct real roots. Therefore, the line intersects the circle at two points. [4]

12. General equation: $x^2 + y^2 + 2gx + 2fy + c = 0$. Passes through $(0,0) \Rightarrow c = 0$. Passes through $(4,0) \Rightarrow 16 + 0 + 8g + 0 + 0 = 0 \Rightarrow 8g = -16 \Rightarrow g = -2$. Passes through $(0,3) \Rightarrow 0 + 9 + 0 + 6f + 0 = 0 \Rightarrow 6f = -9 \Rightarrow f = -1.5$. Equation: $x^2 + y^2 - 4x - 3y = 0$. [4]

13. $x^2 - 4 = x + 2$ $x^2 - x - 6 = 0$ $(x - 3)(x + 2) = 0$ $x = 3$ or $x = -2$. If $x = 3, y = 3 + 2 = 5 \Rightarrow (3, 5)$. If $x = -2, y = -2 + 2 = 0 \Rightarrow (-2, 0)$. Coordinates: $(3, 5)$ and $(-2, 0)$. [4]

14. $\frac{6}{x} = 7 - x$ $6 = 7x - x^2$ $x^2 - 7x + 6 = 0$ $(x - 6)(x - 1) = 0$ $x = 6$ or $x = 1$. If $x = 6, y = 7 - 6 = 1 \Rightarrow (6, 1)$. If $x = 1, y = 7 - 1 = 6 \Rightarrow (1, 6)$. Coordinates: $(6, 1)$ and $(1, 6)$. [4]

15. (a) Midpoint of $AC = (\frac{1+9}{2}, \frac{1+5}{2}) = (5, 3)$. Midpoint of $BD = (\frac{4+6}{2}, \frac{5+1}{2}) = (5, 3)$. Since diagonals bisect each other, $ABCD$ is a parallelogram. [2] (b) Base $AD$ is horizontal? No. Vector $AB = (3, 4)$. Vector $AD = (5, 0)$. Area using determinant/cross product magnitude: $|x_A(y_B - y_D) + x_B(y_D - y_A) + x_D(y_A - y_B)|$? No, simpler: Base $AD$ length $= 6 - 1 = 5$ (Horizontal segment? No, $A(1,1), D(6,1)$ is horizontal). Height of $B$ from $AD$ (line $y=1$) is $5 - 1 = 4$. Area $= \text{Base} \times \text{Height} = 5 \times 4 = 20$. [2]

16. (a) Vertical: $y$, Horizontal: $x^2$. [1] (b) Equation of line $Y = mX + c$ where $Y=y, X=x^2$. $m = \frac{28 - 10}{5 - 2} = \frac{18}{3} = 6$. So $a = 6$. $10 = 6(2) + b \Rightarrow 10 = 12 + b \Rightarrow b = -2$. $a = 6, b = -2$. [3]

17. (a) Gradient $m = \frac{1.1 - 0.3}{4 - 0} = \frac{0.8}{4} = 0.2$. [1] (b) Equation: $\log_{10} y = 0.2x + c$. At $x=0, \log_{10} y = 0.3 \Rightarrow c = 0.3$. $\log_{10} y = 0.2x + 0.3$. $y = 10^{0.2x + 0.3} = 10^{0.3} \cdot (10^{0.2})^x$. $A = 10^{0.3} \approx 2.00$ (or exactly $10^{0.3}$). $b = 10^{0.2} \approx 1.58$ (or exactly $10^{0.2}$). $A = 10^{0.3}, b = 10^{0.2}$. [3]

18. (a) $y = 10 - 2x$. Substitute into curve: $10 - 2x = x^2 - 2x + 4$ $10 = x^2 + 4$ $x^2 = 6$. Wait, $x^2 - 6 = 0$ has real roots. They do intersect. Let's re-read the generated question. Line: $2x + y = 10 \Rightarrow y = 10 - 2x$. Curve: $y = x^2 - 2x + 4$. $10 - 2x = x^2 - 2x + 4$ $10 = x^2 + 4$ $x^2 = 6 \Rightarrow x = \pm\sqrt{6}$. The question asks to "Show that the line L does not intersect". Error in Question Generation: The line does intersect. Correction for Key: I will provide the working showing they do intersect, and note the discrepancy. However, usually "Show that" implies the premise is true. Let's check the vertex of the parabola: $x = -b/2a = 1$. $y = 1 - 2 + 4 = 3$. Vertex $(1,3)$. Distance from vertex to line $2x+y-10=0$: $d = \frac{|2(1) + 3 - 10|}{\sqrt{2^2+1^2}} = \frac{|-5|}{\sqrt{5}} = \sqrt{5} \approx 2.23$. Since the parabola opens upward and the line is above the vertex? At $x=1$, Line $y = 8$. Curve $y = 3$. Line is above. At $x=\sqrt{6} \approx 2.45$, Line $y = 10 - 4.9 = 5.1$. Curve $y = 6 - 4.9 + 4 = 5.1$. They touch/cross. Actually, $x^2 = 6$ means 2 intersection points. Key Answer: The premise in Q18(a) is mathematically incorrect for the given equations. Alternative Interpretation: Did I copy the question right? "Show that the line L does not intersect". If the line was $2x + y = 1$, then $1 - 2x = x^2 - 2x + 4 \Rightarrow x^2 = -3$ (No solution). Given the constraint to answer the provided text, I will state: "Substitution leads to $x^2 = 6$, which has real solutions. Thus, the line does intersect the curve. The question statement appears to contain an error." (b) Shortest distance is 0 since they intersect.

19. (a) Midpoint $AB = (1, 2.5)$. Gradient $AB = \frac{4-1}{3-(-2)} = \frac{3}{5}$. Gradient perp $= -\frac{5}{3}$. Eq: $y - 2.5 = -\frac{5}{3}(x - 1)$. $3y - 7.5 = -5x + 5$ $5x + 3y - 12.5 = 0$ or $10x + 6y - 25 = 0$. [3] (b) Solve system:

1. $10x + 6y = 25$
2. $2x - 7y = 13 \Rightarrow 10x - 35y = 65$ Subtract (2) from (1): $(6y - (-35y)) = 25 - 65$ $41y = -40 \Rightarrow y = -\frac{40}{41}$. $2x = 13 + 7(-\frac{40}{41}) = \frac{533 - 280}{41} = \frac{253}{41}$. $x = \frac{253}{82}$. Circumcentre: $(\frac{253}{82}, -\frac{40}{41})$. [3]

20. (a) Since it touches the x-axis at $(4,0)$, the centre's x-coordinate is 4. $h = 4$. [1] (b) Centre is $(4, k)$. Radius $r = |k|$ (since it touches x-axis). Equation: $(x - 4)^2 + (y - k)^2 = k^2$. Passes through $(2, 2)$: $(2 - 4)^2 + (2 - k)^2 = k^2$ $4 + 4 - 4k + k^2 = k^2$ $8 - 4k = 0 \Rightarrow 4k = 8 \Rightarrow k = 2$. Radius $r = 2$. $k = 2, r = 2$. [4]