O Level Additional Mathematics Practice Paper 2

TuitionGoWhere Practice Paper - Additional Mathematics O-Level

Answer Key and Marking Scheme

Version: 2 of 5
Topic: Graphs & Coordinate Geometry

Section A: Lines and Basic Geometry

1. (a) Gradient $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-3 - 5}{6 - 2} = \frac{-8}{4} = -2$.
[1]

(b) Using $y = mx + c$ with $m=-2$ and point $(2,5)$:
$5 = -2(2) + c \Rightarrow 5 = -4 + c \Rightarrow c = 9$.
Equation: $y = -2x + 9$.
[2] (1 for method/substitution, 1 for correct equation)

(c) Gradient of perpendicular line $m_{\perp} = -\frac{1}{m} = -\frac{1}{-2} = \frac{1}{2}$.
Equation: $y - 1 = \frac{1}{2}(x - 4)$.
$y = \frac{1}{2}x - 2 + 1 \Rightarrow y = \frac{1}{2}x - 1$.
[3] (1 for perp gradient, 1 for substitution, 1 for final equation)

2. (a) Length $PQ = \sqrt{(7-1)^2 + (2-2)^2} = \sqrt{36} = 6$.
Length $PR = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$.
Length $QR = \sqrt{(4-7)^2 + (6-2)^2} = \sqrt{(-3)^2 + 4^2} = \sqrt{25} = 5$.
Since $PR = QR = 5$, the triangle is isosceles.
[2] (1 for calculating at least two lengths correctly, 1 for conclusion)

(b) Base $PQ$ is horizontal, length 6. Height is vertical distance from $y=2$ to $y=6$, so $h=4$.
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 4 = 12$.
[2]

(c) Midpoint of $PR = \left(\frac{1+4}{2}, \frac{2+6}{2}\right) = (2.5, 4)$.
[1]

3. (a) Gradient $AB = \frac{1 - 3}{4 - (-2)} = \frac{-2}{6} = -\frac{1}{3}$.
[1]

(b) Since collinear, gradient $BC$ must equal gradient $AB$.
$\frac{k - 1}{6 - 4} = -\frac{1}{3} \Rightarrow \frac{k - 1}{2} = -\frac{1}{3}$.
$3(k - 1) = -2 \Rightarrow 3k - 3 = -2 \Rightarrow 3k = 1 \Rightarrow k = \frac{1}{3}$.
[2] (1 for setting up equation, 1 for correct value)

4. (a) x-intercept (set $y=0$): $3x + 12 = 0 \Rightarrow x = -4$. Point $(-4, 0)$.
y-intercept (set $x=0$): $-4y + 12 = 0 \Rightarrow 4y = 12 \Rightarrow y = 3$. Point $(0, 3)$.
[2] (1 for each intercept)

(b) Length $= \sqrt{(-4 - 0)^2 + (0 - 3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.
[2]

Section B: Circles and Intersections

5. (a) Complete the square:
$(x^2 - 6x) + (y^2 + 8y) = 11$
$(x - 3)^2 - 9 + (y + 4)^2 - 16 = 11$
$(x - 3)^2 + (y + 4)^2 = 11 + 9 + 16 = 36$.
Centre $(3, -4)$.
[2]

(b) $r^2 = 36 \Rightarrow r = 6$.
[2]

(c) Substitute $P(1, -2)$ into LHS of circle equation $(x-3)^2 + (y+4)^2$:
$(1 - 3)^2 + (-2 + 4)^2 = (-2)^2 + (2)^2 = 4 + 4 = 8$.
Since $8 < 36$ (radius squared), the point lies inside the circle.
[2] (1 for substitution/calculation, 1 for correct conclusion)

6. (a) Centre $(3, -4)$. Radius $r = \sqrt{(3-0)^2 + (-4-0)^2} = \sqrt{9+16} = 5$.
Equation: $(x - 3)^2 + (y + 4)^2 = 25$.
[3] (1 for radius, 1 for structure, 1 for correct constants)

(b) Substitute $y=x$ into circle equation:
$(x - 3)^2 + (x + 4)^2 = 25$
$x^2 - 6x + 9 + x^2 + 8x + 16 = 25$
$2x^2 + 2x + 25 = 25$
$2x^2 + 2x = 0 \Rightarrow 2x(x + 1) = 0$.
$x = 0$ or $x = -1$.
If $x=0, y=0$. If $x=-1, y=-1$.
Points: $(0, 0)$ and $(-1, -1)$.
[4] (1 for substitution, 1 for quadratic, 1 for solving x, 1 for coordinates)

7. (a) Substitute $y = 2x + k$ into $x^2 + y^2 = 20$:
$x^2 + (2x + k)^2 = 20$
$x^2 + 4x^2 + 4kx + k^2 = 20$
$5x^2 + 4kx + (k^2 - 20) = 0$.
[2]

(b) For tangency, discriminant $\Delta = 0$.
$b^2 - 4ac = 0 \Rightarrow (4k)^2 - 4(5)(k^2 - 20) = 0$
$16k^2 - 20(k^2 - 20) = 0$
$16k^2 - 20k^2 + 400 = 0$
$-4k^2 + 400 = 0 \Rightarrow 4k^2 = 400 \Rightarrow k^2 = 100$.
$k = 10$ or $k = -10$.
[3] (1 for discriminant condition, 1 for algebra, 1 for both values)

8. (a) Expand $C_2$: $x^2 - 10x + 25 + y^2 = 25 \Rightarrow x^2 + y^2 - 10x = 0$.
From $C_1$, $x^2 + y^2 = 25$. Substitute into expanded $C_2$:
$25 - 10x = 0 \Rightarrow 10x = 25 \Rightarrow x = 2.5$.
Substitute $x = 2.5$ into $C_1$:
$(2.5)^2 + y^2 = 25 \Rightarrow 6.25 + y^2 = 25 \Rightarrow y^2 = 18.75$.
$y = \pm\sqrt{18.75} = \pm\frac{5\sqrt{3}}{2} \approx \pm 4.33$.
Points: $(2.5, \frac{5\sqrt{3}}{2})$ and $(2.5, -\frac{5\sqrt{3}}{2})$.
[3] (1 for finding x, 1 for finding y, 1 for both points)

(b) The common chord is the vertical line connecting the intersection points.
Equation: $x = 2.5$ (or $2x - 5 = 0$).
[1]

Section C: Advanced Coordinate Geometry and Loci

9. (a) Let $P = (x, y)$.
$PA = 2 PB \Rightarrow PA^2 = 4 PB^2$.
$(x - 2)^2 + (y - 0)^2 = 4 [ (x - 8)^2 + (y - 0)^2 ]$
$x^2 - 4x + 4 + y^2 = 4 ( x^2 - 16x + 64 + y^2 )$
$x^2 - 4x + 4 + y^2 = 4x^2 - 64x + 256 + 4y^2$
Rearranging: $3x^2 - 60x + 3y^2 + 252 = 0$.
Divide by 3: $x^2 - 20x + y^2 + 84 = 0$.
This is in the form $x^2 + y^2 + 2gx + 2fy + c = 0$, which represents a circle.
[4] (1 for distance formula setup, 1 for expansion, 1 for simplification, 1 for identifying circle form)

(b) Complete square for $x$: $(x - 10)^2 - 100 + y^2 + 84 = 0$.
$(x - 10)^2 + y^2 = 16$.
Centre $(10, 0)$, Radius $\sqrt{16} = 4$.
[2]

10. (a) Since $AB$ is parallel to x-axis, $B$ has same y-coord as $A(1)$. Since $ABCD$ is rectangle, $BC$ is vertical, so $B$ has same x-coord as $C(7)$.
$B(7, 1)$.
Similarly, $D$ has same x as $A$ and same y as $C$.
$D(1, 5)$.
[2]

(b) Diagonal $BD$ connects $(7, 1)$ and $(1, 5)$.
Gradient $m = \frac{5 - 1}{1 - 7} = \frac{4}{-6} = -\frac{2}{3}$.
Equation: $y - 1 = -\frac{2}{3}(x - 7)$.
$3(y - 1) = -2(x - 7) \Rightarrow 3y - 3 = -2x + 14$.
$2x + 3y = 17$.
[2]

(c) Width $AB = 7 - 1 = 6$. Height $AD = 5 - 1 = 4$.
Area $= 6 \times 4 = 24$.
[1]

11. (a) $K(0, 4)$, $Q(q, 0)$.
Midpoint $M(x, y) = \left(\frac{0+q}{2}, \frac{4+0}{2}\right) = \left(\frac{q}{2}, 2\right)$.
[1]

(b) $x = \frac{q}{2} \Rightarrow q = 2x$.
$y = 2$.
Since $y$ is constant regardless of $q$, the locus is the horizontal line $y = 2$.
(Note: If $q$ can be any real number, $x$ can be any real number).
Equation: $y = 2$.
[2] (1 for parametric relation, 1 for Cartesian equation)