Secondary 4 Additional Mathematics Graphs Coordinate Geometry Quiz

Secondary 4 Additional Mathematics Quiz - Graphs Coordinate Geometry

Answer Key

Question 1 (3 marks)

(a) For $A$ (on $x$-axis, $y = 0$): $2x - 0 = 6 \Rightarrow x = 3$, so $A = (3, 0)$. [1]

For $B$ (on $y$-axis, $x = 0$): $0 - 3y = 6 \Rightarrow y = -2$, so $B = (0, -2)$. [½]

(b) $AB = \sqrt{(3-0)^2 + (0-(-2))^2} = \sqrt{9 + 4} = \sqrt{13}$ units. [1½]

Common mistake: Forgetting to set $y = 0$ for $x$-intercept and $x = 0$ for $y$-intercept.

Question 2 (3 marks)

(a) Midpoint of $PQ = \left(\dfrac{2+8}{2}, \dfrac{5+(-3)}{2}\right) = (5, 1)$. [1]

(b) Gradient of $PQ = \dfrac{-3 - 5}{8 - 2} = \dfrac{-8}{6} = -\dfrac{4}{3}$. [1]

(c) Gradient of perpendicular line $= \dfrac{3}{4}$ (negative reciprocal of $-\dfrac{4}{3}$).

Equation: $y - 4 = \dfrac{3}{4}(x - 1)$, so $4y - 16 = 3x - 3$, giving $3x - 4y + 13 = 0$. [1]

Common mistake: Confusing perpendicular gradient ($m_1 \cdot m_2 = -1$) with parallel gradient ($m_1 = m_2$).

Question 3 (3 marks)

(a) Completing the square:

$x^2 - 6x + y^2 + 4y = 12$

$(x - 3)^2 - 9 + (y + 2)^2 - 4 = 12$

$(x - 3)^2 + (y + 2)^2 = 25$ [2]

(b) Centre: $(3, -2)$, Radius: $5$ units. [1]

Common mistake: Forgetting to subtract the constants added during completing the square.

Question 4 (3 marks)

(a) For $A$ and $B$ ($y = 0$): $x^2 - 6x + 5 = 0 \Rightarrow (x-1)(x-5) = 0$, so $x = 1$ or $x = 5$.

$A = (1, 0)$, $B = (5, 0)$. [1]

For $C$ ($x = 0$): $y = 0 - 0 + 5 = 5$, so $C = (0, 5)$. [½]

(b) Vertex: $x = -\dfrac{-6}{2(1)} = 3$, $y = 9 - 18 + 5 = -4$.

Vertex $= (3, -4)$. [1½]

Common mistake: Using $x = \dfrac{b}{2a}$ instead of $x = -\dfrac{b}{2a}$.

Question 5 (3 marks)

Substitute $y = mx + 7$ into $x^2 + y^2 = 25$:

$x^2 + (mx + 7)^2 = 25$

$x^2 + m^2x^2 + 14mx + 49 = 25$

$(1 + m^2)x^2 + 14mx + 24 = 0$ [1]

For tangency, discriminant $= 0$:

$(14m)^2 - 4(1 + m^2)(24) = 0$

$196m^2 - 96(1 + m^2) = 0$

$196m^2 - 96 - 96m^2 = 0$

$100m^2 = 96$

$m^2 = \dfrac{24}{25}$

$m = \pm\dfrac{2\sqrt{6}}{5}$ [2]

Common mistake: Forgetting to set discriminant $= 0$ for tangency condition.

Question 6 (6 marks)

(a) Gradient of $AB = \dfrac{4 - 2}{7 - 1} = \dfrac{2}{6} = \dfrac{1}{3}$. [1]

(b) Midpoint of $AB = \left(\dfrac{1+7}{2}, \dfrac{2+4}{2}\right) = (4, 3)$.

Gradient of perpendicular bisector $= -3$ (negative reciprocal of $\dfrac{1}{3}$).

Equation: $y - 3 = -3(x - 4) \Rightarrow y - 3 = -3x + 12 \Rightarrow 3x + y = 15$.

Note: The question states $3x + y = 17$, but the correct calculation gives $3x + y = 15$. The answer key uses the correct mathematical result. [2]

(c) Gradient of $BC = \dfrac{8 - 4}{3 - 7} = \dfrac{4}{-4} = -1$.

Midpoint of $BC = \left(\dfrac{7+3}{2}, \dfrac{4+8}{2}\right) = (5, 6)$.

Gradient of perpendicular bisector of $BC = 1$ (negative reciprocal of $-1$).

Equation: $y - 6 = 1(x - 5) \Rightarrow y = x + 1$. [2]

(d) Solving $3x + y = 15$ and $y = x + 1$:

$3x + x + 1 = 15 \Rightarrow 4x = 14 \Rightarrow x = \dfrac{7}{2}$

$y = \dfrac{7}{2} + 1 = \dfrac{9}{2}$

Circumcentre $= \left(\dfrac{7}{2}, \dfrac{9}{2}\right)$. [1]

Question 7 (7 marks)

(a) Completing the square:

$x^2 + 4x + y^2 - 8y = -k$

$(x + 2)^2 - 4 + (y - 4)^2 - 16 = -k$

$(x + 2)^2 + (y - 4)^2 = 20 - k$ [2]

Centre $C = (-2, 4)$. [½]

(b) Radius $= 3$, so $r^2 = 9 = 20 - k \Rightarrow k = 11$. [1½]

(c) Substitute $y = 2x + 1$ into $(x + 2)^2 + (y - 4)^2 = 9$:

$(x + 2)^2 + (2x + 1 - 4)^2 = 9$

$(x + 2)^2 + (2x - 3)^2 = 9$

$x^2 + 4x + 4 + 4x^2 - 12x + 9 = 9$

$5x^2 - 8x + 4 = 0$ [1]

$(5x - 2)(x - 2) = 0 \Rightarrow x = \dfrac{2}{5}$ or $x = 2$

When $x = \dfrac{2}{5}$: $y = \dfrac{4}{5} + 1 = \dfrac{9}{5}$, so $P = \left(\dfrac{2}{5}, \dfrac{9}{5}\right)$.

When $x = 2$: $y = 4 + 1 = 5$, so $Q = (2, 5)$. [2]

Question 8 (7 marks)

(a) At intersection: $x^2 - 4x + 7 = x + 3$

$x^2 - 5x + 4 = 0 \Rightarrow (x - 1)(x - 4) = 0$

$x = 1 \Rightarrow y = 4$, so $A = (1, 4)$.

$x = 4 \Rightarrow y = 7$, so $B = (4, 7)$. [2]

(b) $\dfrac{dy}{dx} = 2x - 4$. At $x = 1$: gradient $= 2(1) - 4 = -2$.

Tangent at $A$: $y - 4 = -2(x - 1) \Rightarrow y = -2x + 6$. [2]

(c) At $D$ ($x = 0$): $y = 6$, so $D = (0, 6)$. [1]

(d) Area $= \int_1^4 [(x + 3) - (x^2 - 4x + 7)] \, dx = \int_1^4 (-x^2 + 5x - 4) \, dx$

$= \left[-\dfrac{x^3}{3} + \dfrac{5x^2}{2} - 4x\right]_1^4$

At $x = 4$: $-\dfrac{64}{3} + 40 - 16 = -\dfrac{64}{3} + 24 = \dfrac{8}{3}$

At $x = 1$: $-\dfrac{1}{3} + \dfrac{5}{2} - 4 = -\dfrac{1}{3} - \dfrac{3}{2} = -\dfrac{11}{6}$

Area $= \dfrac{8}{3} - (-\dfrac{11}{6}) = \dfrac{16}{6} + \dfrac{11}{6} = \dfrac{27}{6} = \dfrac{9}{2}$ square units. [2]

Question 9 (5 marks)

(a) At $(0, 5)$: $c = 5$. [½]

At $(1, 0)$: $a + b + c = 0 \Rightarrow a + b = -5$. [½]

At $(3, 8)$: $9a + 3b + c = 8 \Rightarrow 9a + 3b = 3 \Rightarrow 3a + b = 1$. [½]

(b) From $a + b = -5$ and $3a + b = 1$: subtracting gives $2a = 6 \Rightarrow a = 3$.

Then $b = -8$. With $c = 5$.

Equation: $y = 3x^2 - 8x + 5$. [2]

(c) Minimum at $x = -\dfrac{-8}{2(3)} = \dfrac{4}{3}$.

$y = 3\left(\dfrac{16}{9}\right) - 8\left(\dfrac{4}{3}\right) + 5 = \dfrac{48}{9} - \dfrac{32}{3} + 5 = \dfrac{48 - 96 + 45}{9} = -\dfrac{3}{9} = -\dfrac{1}{3}$.

Minimum point $= \left(\dfrac{4}{3}, -\dfrac{1}{3}\right)$. [1½]

Question 10 (5 marks)

(a) Circle 1: Centre $(0, 0)$, radius $4$. Circle 2: Centre $(6, 0)$, radius $2$. [1]

(b) Distance between centres $= \sqrt{(6-0)^2 + (0-0)^2} = 6$.

Sum of radii $= 4 + 2 = 6$. Since distance $=$ sum of radii, the circles touch externally. [1½]

(c) The point of contact lies on the line joining the centres (the $x$-axis), at distance $4$ from $(0, 0)$ towards $(6, 0)$.

Point of contact $= (4, 0)$. [1]

(d) The common tangent at the point of contact is perpendicular to the $x$-axis at $x = 4$.

Equation: $x = 4$. [1½]

Question 11 (2 marks)

Line $4x + 2y = 7$ has gradient $-2$.

Parallel line through $(-1, 3)$: $y - 3 = -2(x + 1) \Rightarrow y = -2x + 1$.

Or $2x + y = 1$. [2]

Question 12 (2 marks)

$r^2 = (4-1)^2 + (-2-3)^2 = 9 + 25 = 34$.

Equation: $(x - 1)^2 + (y - 3)^2 = 34$. [2]

Question 13 (2 marks)

Vertex on $y$-axis means $x = 0$ at vertex, so $-\dfrac{b}{2(2)} = 0 \Rightarrow b = 0$. [2]

Question 14 (2 marks)

Substitute $y = 3x - 1$ into $y = x^2 + kx + 5$:

$3x - 1 = x^2 + kx + 5 \Rightarrow x^2 + (k-3)x + 6 = 0$

For tangency, discriminant $= 0$:

$(k-3)^2 - 24 = 0 \Rightarrow (k-3)^2 = 24 \Rightarrow k - 3 = \pm 2\sqrt{6}$

$k = 3 \pm 2\sqrt{6}$ [2]

Question 15 (2 marks)

Using section formula: $\left(\dfrac{3(-5) + 1(3)}{3+1}, \dfrac{3(7) + 1(-1)}{3+1}\right) = \left(\dfrac{-15+3}{4}, \dfrac{21-1}{4}\right) = (-3, 5)$. [2]

Question 16 (2 marks)

$x^2 + y^2 - 10x + 6y + 30 = 0$

$(x - 5)^2 - 25 + (y + 3)^2 - 9 + 30 = 0$

$(x - 5)^2 + (y + 3)^2 = 4$

Centre $(5, -3)$, radius $2$. [1]

Length of tangent from $(7, -1)$:

$PT^2 = (7-5)^2 + (-1+3)^2 - 4 = 4 + 4 - 4 = 4$

$PT = 2$ units. [1]

Question 17 (2 marks)

Passes through $(2, 0)$: $4 + 2p + q = 0 \Rightarrow 2p + q = -4$. [½]

Minimum value: $x = -\dfrac{p}{2}$, $y = \dfrac{p^2}{4} - \dfrac{p^2}{2} + q = -\dfrac{p^2}{4} + q = -9$. [½]

So $q = -9 + \dfrac{p^2}{4}$. Substituting: $2p + (-9 + \dfrac{p^2}{4}) = -4$

$\dfrac{p^2}{4} + 2p - 5 = 0 \Rightarrow p^2 + 8p - 20 = 0 \Rightarrow (p+10)(p-2) = 0$

$p = 2$ or $p = -10$.

If $p = 2$: $q = -8$. If $p = -10$: $q = -34$.

Check: For $p = 2$, vertex at $x = -1$, $y = 1 - 2 + q = -1 + q = -9 \Rightarrow q = -8$ ✓

For $p = -10$, vertex at $x = 5$, $y = 25 - 50 + q = -25 + q = -9 \Rightarrow q = -34$ ✓

Both solutions valid: $(p, q) = (2, -8)$ or $(-10, -34)$. [1½]

Question 18 (2 marks)

$\dfrac{dy}{dx} = 2x - 2$. At $x = 2$: gradient $= 2$. So $m = 2$. [1]

Line passes through $(2, 3)$: $3 = 2(2) + c \Rightarrow c = -1$. [1]

Question 19 (2 marks)

Let equation be $x^2 + y^2 + Dx + Ey + F = 0$.

At $(0, 0)$: $F = 0$. [½]

At $(4, 0)$: $16 + 4D = 0 \Rightarrow D = -4$. [½]

At $(0, 3)$: $9 + 3E = 0 \Rightarrow E = -3$. [½]

Equation: $x^2 + y^2 - 4x - 3y = 0$.

Or $(x - 2)^2 + (y - \dfrac{3}{2})^2 = \dfrac{25}{4}$. [½]

Question 20 (2 marks)

Axis of symmetry $x = 1$: $-\dfrac{b}{2a} = 1 \Rightarrow b = -2a$. [½]

At $(0, -2)$: $c = -2$. [½]

At $(3, 7)$: $9a + 3b + c = 7 \Rightarrow 9a + 3(-2a) - 2 = 7 \Rightarrow 3a = 9 \Rightarrow a = 3$.

Then $b = -6$, $c = -2$.

Equation: $y = 3x^2 - 6x - 2$. [1]

Mark Summary

Note: Total marks = 60 (adjusted from stated 45 to reflect actual allocation).

Common Mistakes Summary

1. Sign errors in completing the square — always verify by expanding back.
2. Confusing perpendicular and parallel gradients — remember $m_1 \cdot m_2 = -1$ for perpendicular lines.
3. Forgetting discriminant = 0 for tangency conditions.
4. Arithmetic errors when substituting negative coordinates into distance or section formulas.
5. Incorrect vertex formula — use $x = -\dfrac{b}{2a}$, not $x = \dfrac{b}{2a}$.