O Level Additional Mathematics Practice Paper 5

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

Section A

1. Gradient $m = \frac{6 - (-3)}{5 - 2} = \frac{9}{3} = 3$. Eq: $y - 6 = 3(x - 5) \implies y = 3x - 9$. (3 marks)
2. $L_1: y = \frac{2}{3}x - \frac{8}{3}$. $L_2$ gradient is $\frac{2}{3}$. $y - (-4) = \frac{2}{3}(x - 1) \implies y = \frac{2}{3}x - \frac{14}{3}$ or $2x - 3y = 14$. (3 marks)
3. $L_1$ gradient $m_1 = -\frac{3}{4}$. Perpendicular $m_2 = \frac{4}{3}$. Thus $k = \frac{4}{3}$. (2 marks)
4. Midpoint $= (\frac{-4+8}{2}, \frac{7-1}{2}) = (2, 3)$. (2 marks)
5. $m_1 = \frac{1}{2} \implies m_L = -2$. $y - 2 = -2(x - 3) \implies y = -2x + 8 \implies 2x + y - 8 = 0$. (3 marks)

Section B

6. $(x-3)^2 + (y+4)^2 = -9 + 9 + 16 = 16$. Centre $(3, -4)$, Radius $4$. (3 marks)
7. $(x-2)^2 + (y+5)^2 = (4\sqrt{2})^2 \implies (x-2)^2 + (y+5)^2 = 32$. (3 marks)
8. Centre $= (\frac{-1+5}{2}, \frac{4+2}{2}) = (2, 3)$. Radius $r = \sqrt{(2-(-1))^2 + (3-4)^2} = \sqrt{9+1} = \sqrt{10}$. Eq: $(x-2)^2 + (y-3)^2 = 10$. (4 marks)
9. $x^2 + (2x+1)^2 = 25 \implies x^2 + 4x^2 + 4x + 1 = 25 \implies 5x^2 + 4x - 24 = 0$. $(5x+12)(x-2) = 0 \implies x = 2, x = -2.4$. Points: $(2, 5)$ and $(-2.4, -3.8)$. (4 marks)
10. Substitute $y = x-2$ into circle: $(x-1)^2 + (x-2+2)^2 = 5 \implies x^2 - 2x + 1 + x^2 = 5 \implies 2x^2 - 2x - 4 = 0 \implies x^2 - x - 2 = 0$. $(x-2)(x+1) = 0$. Wait, this intersects at two points. Correction: If the question asks to "show it is a tangent", the discriminant must be 0. For $y=x-2$ and $(x-1)^2 + (y+2)^2 = 5$, we get $x=2, -1$. This is a secant. Corrected Logic for intended answer: If the line were $y = x - 1$, we would check $\Delta$. For the provided equation, the line is NOT a tangent. (Marking: Full marks if student proves $\Delta \neq 0$ and states it is not a tangent). (3 marks)
11. Centre $(3, -4)$, passes through $(0,0) \implies r^2 = 3^2 + (-4)^2 = 25$. Eq: $(x-3)^2 + (y+4)^2 = 25$. (3 marks)
12. $x^2 + (mx+1)^2 = 2 \implies x^2 + m^2x^2 + 2mx + 1 = 2 \implies (1+m^2)x^2 + 2mx - 1 = 0$. For tangency, $\Delta = 0 \implies (2m)^2 - 4(1+m^2)(-1) = 0 \implies 4m^2 + 4 + 4m^2 = 0 \implies 8m^2 = -4$ (No real $m$). Correction: If the line was $y=mx+2$, $m=0$. For $y=mx+1$, no real $m$ exists. (Marking: Full marks for showing $\Delta < 0$). (4 marks)
13. $x^2 - 4x + 3 = 2x - 5 \implies x^2 - 6x + 8 = 0 \implies (x-2)(x-4) = 0$. $x=2 \implies y=-1$; $x=4 \implies y=3$. Points: $(2, -1), (4, 3)$. (4 marks)
14. Power of point $P(10, 10) = 10^2 + 10^2 + 4(10) - 6(10) - 12 = 100 + 100 + 40 - 60 - 12 = 168$. Length $= \sqrt{168} \approx 13.0$. (4 marks)
15. Centre $(h, 2h)$. Passes through $(0,0) \implies r^2 = h^2 + (2h)^2 = 5h^2$. Passes through $(4,0) \implies (4-h)^2 + (0-2h)^2 = 5h^2 \implies 16 - 8h + h^2 + 4h^2 = 5h^2 \implies 16 - 8h = 0 \implies h = 2$. Centre $(2, 4)$, $r^2 = 5(2)^2 = 20$. Eq: $(x-2)^2 + (y-4)^2 = 20$. (5 marks)

Section C

16. Area $= \frac{1}{2} \times \text{base} \times \text{height} \implies 10 = \frac{1}{2} \times 4 \times |y| \implies |y| = 5$. Case 1: $y = 5 \implies 5 = 2x + 1 \implies x = 2$. Point $(2, 5)$. Case 2: $y = -5 \implies -5 = 2x + 1 \implies x = -3$. Point $(-3, -5)$. (5 marks)
17. $y = ax^2$ and $y = 4x - 2$. $ax^2 - 4x + 2 = 0$. For tangency, $\Delta = 0 \implies (-4)^2 - 4(a)(2) = 0 \implies 16 - 8a = 0 \implies a = 2$. $2x^2 - 4x + 2 = 0 \implies 2(x-1)^2 = 0 \implies x = 1$. $y = 4(1) - 2 = 2$. Point $(1, 2)$. (5 marks)
18. $x=0, y=5 \implies 5 = kb^0 \implies k = 5$. $x=2, y=20 \implies 20 = 5b^2 \implies b^2 = 4 \implies b = 2$ (since $b>0$ for exponential). (4 marks)
19. $y = 5(2^x) \implies \ln y = \ln 5 + x \ln 2$. Let $Y = \ln y$ and $X = x$. $Y = (\ln 2)X + \ln 5$. $M = \ln 2$ (gradient), $C = \ln 5$ (vertical intercept). (4 marks)
20. Midpoint $M = (\frac{1+3}{2}, \frac{5-1}{2}) = (2, 2)$. Gradient $AB = \frac{-1-5}{3-1} = -3$. Perpendicular gradient $= \frac{1}{3}$. Eq: $y - 2 = \frac{1}{3}(x - 2) \implies 3y - 6 = x - 2 \implies x - 3y + 4 = 0$. (4 marks)