Secondary 4 Additional Mathematics Practice Paper 1

Answer Key - Additional Mathematics Secondary 4

Topic: Graphs & Coordinate Geometry
Version: 1 of 5

Section A: Short Answer Questions

1.
Step 1: Find gradient of $L_1$.
$m_{L1} = \frac{-3 - 5}{6 - 2} = \frac{-8}{4} = -2$.
Step 2: Find gradient of $L_2$ (perpendicular).
$m_{L2} = -\frac{1}{m_{L1}} = -\frac{1}{-2} = \frac{1}{2}$.
Step 3: Find midpoint of $AB$.
$M = \left(\frac{2+6}{2}, \frac{5+(-3)}{2}\right) = (4, 1)$.
Step 4: Equation of $L_2$.
$y - 1 = \frac{1}{2}(x - 4) \Rightarrow 2(y - 1) = x - 4 \Rightarrow 2y - 2 = x - 4$.
$x - 2y - 2 = 0$ or $y = \frac{1}{2}x + 1$.
Answer: $x - 2y - 2 = 0$ [3]

2.
Step 1: Equate $y$ values.
$x^2 - 4x + 3 = x - 1$
$x^2 - 5x + 4 = 0$
Step 2: Solve for $x$.
$(x - 4)(x - 1) = 0 \Rightarrow x = 1$ or $x = 4$.
Step 3: Find corresponding $y$.
If $x = 1, y = 1 - 1 = 0$. Point $(1, 0)$.
If $x = 4, y = 4 - 1 = 3$. Point $(4, 3)$.
Answer: $(1, 0)$ and $(4, 3)$ [3]

3.
Step 1: Complete the square for $x$ and $y$.
$x^2 - 6x + y^2 + 4y = 12$
$(x - 3)^2 - 9 + (y + 2)^2 - 4 = 12$
$(x - 3)^2 + (y + 2)^2 = 25$
Step 2: Identify centre and radius.
Centre $(3, -2)$, Radius $\sqrt{25} = 5$.
Answer: Centre $(3, -2)$, Radius $5$ [3]

4.
Step 1: Set up intersection equation.
$kx + 2 = x^2 - 3x + 5$
$x^2 - (3 + k)x + 3 = 0$
Step 2: Condition for no intersection is discriminant $< 0$.
$b^2 - 4ac < 0$
$(-(3+k))^2 - 4(1)(3) < 0$
$(3+k)^2 - 12 < 0$
Step 3: Solve inequality.
$(3+k)^2 < 12$
$-\sqrt{12} < 3+k < \sqrt{12}$
$-2\sqrt{3} - 3 < k < 2\sqrt{3} - 3$
Answer: $-3 - 2\sqrt{3} < k < -3 + 2\sqrt{3}$ [3]

5.
Step 1: Calculate side lengths.
$PQ = \sqrt{(5-1)^2 + (6-2)^2} = \sqrt{16+16} = \sqrt{32}$
$QR = \sqrt{(9-5)^2 + (2-6)^2} = \sqrt{16+16} = \sqrt{32}$
$PR = \sqrt{(9-1)^2 + (2-2)^2} = \sqrt{64} = 8$
Since $PQ = QR$, it is isosceles.
Step 2: Find area.
Base $PR$ is horizontal, length $8$.
Height is vertical distance from $Q(5,6)$ to line $y=2$ (line PR). Height $= 6 - 2 = 4$.
Area $= \frac{1}{2} \times 8 \times 4 = 16$.
Answer: Isosceles shown, Area $= 16$ [4]

6.
Step 1: Midpoint of $AB$.
$M = (\frac{3+7}{2}, \frac{-1+5}{2}) = (5, 2)$.
Step 2: Gradient of $AB$.
$m_{AB} = \frac{5 - (-1)}{7 - 3} = \frac{6}{4} = \frac{3}{2}$.
Step 3: Gradient of perpendicular bisector.
$m_{\perp} = -\frac{2}{3}$.
Step 4: Equation.
$y - 2 = -\frac{2}{3}(x - 5)$
$3(y - 2) = -2(x - 5)$
$3y - 6 = -2x + 10$
$2x + 3y - 16 = 0$.
Answer: $2x + 3y - 16 = 0$ [4]

7.
Step 1: Find point on curve.
$x = 3 \Rightarrow y = \frac{12}{3} = 4$. Point $(3, 4)$.
Step 2: Find gradient of tangent.
$y = 12x^{-1} \Rightarrow \frac{dy}{dx} = -12x^{-2} = -\frac{12}{x^2}$.
At $x = 3, m = -\frac{12}{9} = -\frac{4}{3}$.
Step 3: Equation of tangent.
$y - 4 = -\frac{4}{3}(x - 3)$
$3(y - 4) = -4(x - 3)$
$3y - 12 = -4x + 12$
$4x + 3y = 24$.
Step 4: Find intercepts.
x-intercept ($y=0$): $4x = 24 \Rightarrow x = 6$. $A(6, 0)$.
y-intercept ($x=0$): $3y = 24 \Rightarrow y = 8$. $B(0, 8)$.
Answer: $A(6, 0), B(0, 8)$ [4]

8.
Step 1: Radius is distance from centre $(4, -3)$ to origin $(0,0)$.
$r^2 = (4-0)^2 + (-3-0)^2 = 16 + 9 = 25$.
Step 2: Equation.
$(x - 4)^2 + (y + 3)^2 = 25$
$x^2 - 8x + 16 + y^2 + 6y + 9 = 25$
$x^2 + y^2 - 8x + 6y = 0$.
Answer: $x^2 + y^2 - 8x + 6y = 0$ [3]

9.
Step 1: Find intersection $P$.
$2x + 1 = -x + 7 \Rightarrow 3x = 6 \Rightarrow x = 2$.
$y = 2(2) + 1 = 5$. $P(2, 5)$.
Step 2: Distance from origin.
$OP = \sqrt{2^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29}$.
Answer: $\sqrt{29}$ [3]

Section B: Structured Questions

10.
(a) Gradient $AC = \frac{7 - 1}{3 - 1} = \frac{6}{2} = 3$. [1]
(b) Gradient of altitude from $B$ is $-\frac{1}{3}$.
Passes through $B(5, 3)$.
$y - 3 = -\frac{1}{3}(x - 5) \Rightarrow 3(y - 3) = -(x - 5) \Rightarrow 3y - 9 = -x + 5$.
$x + 3y - 14 = 0$. [3]
(c) Solve simultaneous equations for foot of perpendicular ($F$).
Line $AC$: $y - 1 = 3(x - 1) \Rightarrow y = 3x - 2$.
Substitute into altitude eq: $x + 3(3x - 2) - 14 = 0 \Rightarrow x + 9x - 6 - 14 = 0 \Rightarrow 10x = 20 \Rightarrow x = 2$.
$y = 3(2) - 2 = 4$.
Foot is $(2, 4)$. [3]
(d) Base $AC = \sqrt{(3-1)^2 + (7-1)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}$.
Height $BF = \sqrt{(5-2)^2 + (3-4)^2} = \sqrt{9 + 1} = \sqrt{10}$.
Area $= \frac{1}{2} \times 2\sqrt{10} \times \sqrt{10} = 10$. [3]
(Alternative: Shoelace formula or box method yields same result)

11.
(a) $x^2 - 2x - 3 = -x^2 + 4x + 1 \Rightarrow 2x^2 - 6x - 4 = 0 \Rightarrow x^2 - 3x - 2 = 0$.
$x = \frac{3 \pm \sqrt{9 - 4(1)(-2)}}{2} = \frac{3 \pm \sqrt{17}}{2}$.
Let $x_1 = \frac{3 + \sqrt{17}}{2}, x_2 = \frac{3 - \sqrt{17}}{2}$.
$y_1 = x_1 - 1$? No, use linear eq from subtraction?
Actually, subtracting the two curve equations gives the line through intersections directly (see part b).
Let's find y using $y = x^2 - 2x - 3$.
This is messy. Better to find the line first? No, question asks for coordinates.
$y = \frac{3 \pm \sqrt{17}}{2} - 1$? Wait, is $y=x-1$ the line?
Subtracting $C_1$ from $C_2$: $2x^2 - 6x - 4 = 0$ is not a line.
Wait, $C_1: y = x^2 - 2x - 3$, $C_2: y = -x^2 + 4x + 1$.
Intersection: $2x^2 - 6x - 4 = 0 \implies x^2 - 3x - 2 = 0$.
Roots are irrational.
$y = x^2 - 2x - 3$. Since $x^2 = 3x + 2$, $y = (3x + 2) - 2x - 3 = x - 1$.
So $y_1 = x_1 - 1 = \frac{3 + \sqrt{17}}{2} - \frac{2}{2} = \frac{1 + \sqrt{17}}{2}$.
$y_2 = x_2 - 1 = \frac{3 - \sqrt{17}}{2} - \frac{2}{2} = \frac{1 - \sqrt{17}}{2}$.
Points: $\left(\frac{3 + \sqrt{17}}{2}, \frac{1 + \sqrt{17}}{2}\right)$ and $\left(\frac{3 - \sqrt{17}}{2}, \frac{1 - \sqrt{17}}{2}\right)$. [4]
(b) The line passing through intersections is found by subtracting the equations?
Actually, we found $y = x - 1$ during substitution.
Equation: $y = x - 1$ or $x - y - 1 = 0$. [2]
(c) Gradient of $y = x - 1$ is $1$. Gradient of $y = 3x$ is $3$.
$1 \neq 3$, so not parallel. [1]

12.
(a) $(x - 2)^2 + (y - 1)^2 = 25$. [1]
(b) Substitute $y = 2x + k$ into circle eq.
$(x - 2)^2 + (2x + k - 1)^2 = 25$
$x^2 - 4x + 4 + 4x^2 + 4x(k - 1) + (k - 1)^2 = 25$
$5x^2 + x[-4 + 4k - 4] + [4 + k^2 - 2k + 1 - 25] = 0$
$5x^2 + (4k - 8)x + (k^2 - 2k - 20) = 0$.
For tangent, discriminant $= 0$.
$(4k - 8)^2 - 4(5)(k^2 - 2k - 20) = 0$
$16(k - 2)^2 - 20(k^2 - 2k - 20) = 0$
Divide by 4: $4(k^2 - 4k + 4) - 5(k^2 - 2k - 20) = 0$
$4k^2 - 16k + 16 - 5k^2 + 10k + 100 = 0$
$-k^2 - 6k + 116 = 0 \Rightarrow k^2 + 6k - 116 = 0$.
$k = \frac{-6 \pm \sqrt{36 - 4(1)(-116)}}{2} = \frac{-6 \pm \sqrt{36 + 464}}{2} = \frac{-6 \pm \sqrt{500}}{2} = \frac{-6 \pm 10\sqrt{5}}{2} = -3 \pm 5\sqrt{5}$. [5]
(c) $k > 0 \Rightarrow k = -3 + 5\sqrt{5}$.
Solve for $x$ using $x = \frac{-b}{2a}$ from quadratic formula (since disc=0).
$x = \frac{-(4k - 8)}{10} = \frac{8 - 4k}{10} = \frac{4 - 2k}{5}$.
Substitute $k$: $x = \frac{4 - 2(-3 + 5\sqrt{5})}{5} = \frac{4 + 6 - 10\sqrt{5}}{5} = \frac{10 - 10\sqrt{5}}{5} = 2 - 2\sqrt{5}$.
$y = 2x + k = 2(2 - 2\sqrt{5}) + (-3 + 5\sqrt{5}) = 4 - 4\sqrt{5} - 3 + 5\sqrt{5} = 1 + \sqrt{5}$.
Point of contact: $(2 - 2\sqrt{5}, 1 + \sqrt{5})$. [4]

13.
(a) Midpoint of $AB$: $(\frac{-2+4}{2}, \frac{0+6}{2}) = (1, 3)$.
Gradient $AB$: $\frac{6-0}{4-(-2)} = \frac{6}{6} = 1$.
Gradient perp bisector: $-1$.
Eq: $y - 3 = -1(x - 1) \Rightarrow y = -x + 4$. [4]
(b) Centre lies on $y = x$ and $y = -x + 4$.
$x = -x + 4 \Rightarrow 2x = 4 \Rightarrow x = 2$.
$y = 2$. Centre $(2, 2)$. [2]
(c) Radius squared $r^2 = (2 - (-2))^2 + (2 - 0)^2 = 4^2 + 2^2 = 20$.
Eq: $(x - 2)^2 + (y - 2)^2 = 20$ or $x^2 + y^2 - 4x - 4y - 12 = 0$. [2]
(d) Distance $CD^2 = (6 - 2)^2 + (2 - 2)^2 = 16 + 0 = 16$.
$r^2 = 20$. Since $16 < 20$, $D$ is inside the circle. [2]

Section C: Problem Solving

14.
(a) Gradient $AB = 2$. Since $ABCD$ is a rectangle, $BC \perp AB$.
Gradient $BC = -\frac{1}{2}$.
Passes through $C(7, 8)$.
$y - 8 = -\frac{1}{2}(x - 7) \Rightarrow 2(y - 8) = -(x - 7) \Rightarrow 2y - 16 = -x + 7$.
$x + 2y - 23 = 0$. [3]
(b) $B$ is intersection of $y = 2x$ and $x + 2y - 23 = 0$.
$x + 2(2x) - 23 = 0 \Rightarrow 5x = 23 \Rightarrow x = 4.6$.
$y = 2(4.6) = 9.2$.
$B(4.6, 9.2)$. [4]
(c) Midpoint of $AC$ is same as midpoint of $BD$.
$M_{AC} = (\frac{1+7}{2}, \frac{2+8}{2}) = (4, 5)$.
Let $D(x, y)$. $\frac{x + 4.6}{2} = 4 \Rightarrow x + 4.6 = 8 \Rightarrow x = 3.4$.
$\frac{y + 9.2}{2} = 5 \Rightarrow y + 9.2 = 10 \Rightarrow y = 0.8$.
$D(3.4, 0.8)$. [2]
(d) Length $AB = \sqrt{(4.6 - 1)^2 + (9.2 - 2)^2} = \sqrt{3.6^2 + 7.2^2} = \sqrt{12.96 + 51.84} = \sqrt{64.8}$.
Length $BC = \sqrt{(7 - 4.6)^2 + (8 - 9.2)^2} = \sqrt{2.4^2 + (-1.2)^2} = \sqrt{5.76 + 1.44} = \sqrt{7.2}$.
Area $= \sqrt{64.8} \times \sqrt{7.2} = \sqrt{466.56} = 21.6$. [3]

15.
(a) $C_1: (x - 2)^2 - 4 + (y - 3)^2 - 9 - 12 = 0 \Rightarrow (x - 2)^2 + (y - 3)^2 = 25$.
Centre $O_1(2, 3)$, $r_1 = 5$.
$C_2: (x + 1)^2 - 1 + (y - 4)^2 - 16 + 13 = 0 \Rightarrow (x + 1)^2 + (y - 4)^2 = 4$.
Centre $O_2(-1, 4)$, $r_2 = 2$. [4]
(b) Distance $O_1O_2 = \sqrt{(-1 - 2)^2 + (4 - 3)^2} = \sqrt{9 + 1} = \sqrt{10} \approx 3.16$.
Sum of radii $= 5 + 2 = 7$. Diff of radii $= 5 - 2 = 3$.
Since $3 < \sqrt{10} < 7$, the circles intersect at two distinct points. [3]
(c) Subtract equations:
$(x^2 + y^2 - 4x - 6y - 12) - (x^2 + y^2 + 2x - 8y + 13) = 0$
$-6x + 2y - 25 = 0 \Rightarrow 6x - 2y + 25 = 0$. [3]
(d) Distance from $O_1(2, 3)$ to line $6x - 2y + 25 = 0$.
$d = \frac{|6(2) - 2(3) + 25|}{\sqrt{6^2 + (-2)^2}} = \frac{|12 - 6 + 25|}{\sqrt{40}} = \frac{31}{\sqrt{40}}$.
Half-chord length $h = \sqrt{r_1^2 - d^2} = \sqrt{25 - \frac{961}{40}} = \sqrt{\frac{1000 - 961}{40}} = \sqrt{\frac{39}{40}}$.
Total length $= 2h = 2\sqrt{\frac{39}{40}} = \sqrt{\frac{156}{40}} = \sqrt{3.9} \approx 1.97$. [4]

16.
(a) Let $P(x, y)$. $PA = 2 PB$.
$PA^2 = 4 PB^2$.
$x^2 + (y - 4)^2 = 4[(x - 3)^2 + y^2]$.
$x^2 + y^2 - 8y + 16 = 4[x^2 - 6x + 9 + y^2]$.
$x^2 + y^2 - 8y + 16 = 4x^2 - 24x + 36 + 4y^2$.
$3x^2 + 3y^2 - 24x + 8y + 20 = 0$.
$x^2 + y^2 - 8x + \frac{8}{3}y + \frac{20}{3} = 0$. [5]
(b) Complete square:
$(x - 4)^2 - 16 + (y + \frac{4}{3})^2 - \frac{16}{9} + \frac{20}{3} = 0$.
$(x - 4)^2 + (y + \frac{4}{3})^2 = 16 + \frac{16}{9} - \frac{60}{9} = 16 - \frac{44}{9} = \frac{144 - 44}{9} = \frac{100}{9}$.
Circle, Centre $(4, -\frac{4}{3})$, Radius $\frac{10}{3}$. [3]
(c) Distance from Origin to Centre $(4, -4/3)$:
$d^2 = 16 + \frac{16}{9} = \frac{160}{9}$.
Radius squared $r^2 = \frac{100}{9}$.
Since $\frac{160}{9} > \frac{100}{9}$, Origin is outside. [2]

17.
(a) Substitute $y = mx + c$ into $x^2 + y^2 = 25$.
$x^2 + (mx + c)^2 = 25 \Rightarrow (1 + m^2)x^2 + 2mcx + c^2 - 25 = 0$.
Tangent $\Rightarrow$ Discriminant $= 0$.
$(2mc)^2 - 4(1 + m^2)(c^2 - 25) = 0$.
$4m^2c^2 - 4(c^2 - 25 + m^2c^2 - 25m^2) = 0$.
$m^2c^2 - c^2 + 25 - m^2c^2 + 25m^2 = 0$.
$-c^2 + 25 + 25m^2 = 0 \Rightarrow c^2 = 25(1 + m^2)$. [4]
(b) Line passes through $(7, 1) \Rightarrow 1 = 7m + c \Rightarrow c = 1 - 7m$.
Substitute into (a):
$(1 - 7m)^2 = 25(1 + m^2)$.
$1 - 14m + 49m^2 = 25 + 25m^2$.
$24m^2 - 14m - 24 = 0$.
$12m^2 - 7m - 12 = 0$.
$(4m + 3)(3m - 4) = 0$.
$m = -\frac{3}{4}$ or $m = \frac{4}{3}$. [4]
(c) If $m = -\frac{3}{4}, c = 1 - 7(-\frac{3}{4}) = 1 + \frac{21}{4} = \frac{25}{4}$.
Eq: $y = -\frac{3}{4}x + \frac{25}{4} \Rightarrow 3x + 4y - 25 = 0$.
If $m = \frac{4}{3}, c = 1 - 7(\frac{4}{3}) = 1 - \frac{28}{3} = -\frac{25}{3}$.
Eq: $y = \frac{4}{3}x - \frac{25}{3} \Rightarrow 4x - 3y - 25 = 0$. [2]