Secondary 3 Additional Mathematics Graphs Coordinate Geometry Quiz

Secondary 3 Additional Mathematics Quiz - Graphs Coordinate Geometry

Answer Key and Marking Scheme

Total Marks: 50

Section A: Straight Lines and Basic Coordinates (10 marks)

1. Midpoint of A(2, 5) and B(8, -3)

• Midpoint = $\left(\frac{2+8}{2}, \frac{5+(-3)}{2}\right) = (5, 1)$
• Answer: (5, 1)
• Marks: M1 for correct formula, A1 for correct coordinates. (2 marks)

2. Gradient of P(-1, 4) and Q(3, -2)

• Gradient = $\frac{-2 - 4}{3 - (-1)} = \frac{-6}{4} = -\frac{3}{2}$
• Answer: $-\frac{3}{2}$
• Marks: M1 for correct formula, A1 for correct simplified gradient. (2 marks)

3. Line through (3, 1) with gradient 2

• $y - 1 = 2(x - 3)$
• $y - 1 = 2x - 6$
• $y = 2x - 5$
• Answer: $y = 2x - 5$
• Marks: M1 for using point-gradient form, A1 for correct equation. (2 marks)

4. Line 1: $y = 3x + 5$, gradient $m_1 = 3$. Line 2: $6x - 2y + 7 = 0 \implies 2y = 6x + 7 \implies y = 3x + \frac{7}{2}$, gradient $m_2 = 3$. Since $m_1 = m_2$, the lines are parallel.

• Answer: Parallel, because both have gradient 3.
• Marks: M1 for finding both gradients, A1 for correct conclusion with justification. (2 marks)

5. Perpendicular bisector of R(1, 2) and S(5, -4)

• Midpoint of RS = $\left(\frac{1+5}{2}, \frac{2+(-4)}{2}\right) = (3, -1)$
• Gradient of RS = $\frac{-4 - 2}{5 - 1} = \frac{-6}{4} = -\frac{3}{2}$
• Gradient of perpendicular bisector = $\frac{2}{3}$ (negative reciprocal)
• Equation: $y - (-1) = \frac{2}{3}(x - 3)$
• $y + 1 = \frac{2}{3}x - 2$
• $y = \frac{2}{3}x - 3$
• Multiply by 3: $3y = 2x - 9$
• $2x - 3y - 9 = 0$
• Answer: $2x - 3y - 9 = 0$
• Marks: M1 for midpoint, M1 for perpendicular gradient, A1 for correct equation. (2 marks)

Section B: Circles (16 marks)

6. $(x + 3)^2 + (y - 2)^2 = 25$

• Centre: $(-3, 2)$
• Radius: $\sqrt{25} = 5$
• Answer: Centre $(-3, 2)$, Radius $5$
• Marks: A1 for centre, A1 for radius. (2 marks)

7. Centre C(4, -1), passes through P(7, 3)

• Radius $r = CP = \sqrt{(7-4)^2 + (3-(-1))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$
• Equation: $(x - 4)^2 + (y + 1)^2 = 25$
• Answer: $(x - 4)^2 + (y + 1)^2 = 25$
• Marks: M1 for distance formula, M1 for correct radius, A1 for correct equation. (3 marks)

8. $x^2 + y^2 - 6x + 4y - 12 = 0$

(a) Complete the square:

• $(x^2 - 6x) + (y^2 + 4y) = 12$
• $(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4$
• $(x - 3)^2 + (y + 2)^2 = 25$
• Answer: $(x - 3)^2 + (y + 2)^2 = 25$
• Marks: M1 for grouping terms, M1 for completing square correctly, A1 for correct form. (3 marks)

(b) Centre: $(3, -2)$, Radius: $5$

• Answer: Centre $(3, -2)$, Radius $5$
• Marks: A1 for both correct. (1 mark)

9. Circle: $x^2 + y^2 = 20$, point (4, -2)

• Distance from origin to point = $\sqrt{4^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20}$
• Radius = $\sqrt{20}$
• Since distance = radius, the point lies on the circle.
• Answer: On the circle.
• Marks: M1 for calculating distance, M1 for comparing with radius, A1 for correct conclusion. (3 marks)

10. Diameter AB with A(-2, 3) and B(4, -5)

• Centre = midpoint of AB = $\left(\frac{-2+4}{2}, \frac{3+(-5)}{2}\right) = (1, -1)$
• Radius = half of AB = $\frac{1}{2}\sqrt{(4-(-2))^2 + (-5-3)^2} = \frac{1}{2}\sqrt{6^2 + (-8)^2} = \frac{1}{2}\sqrt{36 + 64} = \frac{1}{2}\sqrt{100} = 5$
• Equation: $(x - 1)^2 + (y + 1)^2 = 25$
• Answer: $(x - 1)^2 + (y + 1)^2 = 25$
• Marks: M1 for midpoint, M1 for distance AB, M1 for radius, A1 for correct equation. (4 marks)

Section C: Intersections and Applications (24 marks)

11. Intersection of $y = 2x - 1$ and $y = x^2 - x - 3$

• $2x - 1 = x^2 - x - 3$
• $x^2 - 3x - 2 = 0$
• $x = \frac{3 \pm \sqrt{9 + 8}}{2} = \frac{3 \pm \sqrt{17}}{2}$
• When $x = \frac{3 + \sqrt{17}}{2}$, $y = 2\left(\frac{3 + \sqrt{17}}{2}\right) - 1 = 3 + \sqrt{17} - 1 = 2 + \sqrt{17}$
• When $x = \frac{3 - \sqrt{17}}{2}$, $y = 2\left(\frac{3 - \sqrt{17}}{2}\right) - 1 = 3 - \sqrt{17} - 1 = 2 - \sqrt{17}$
• Answer: $\left(\frac{3 + \sqrt{17}}{2}, 2 + \sqrt{17}\right)$ and $\left(\frac{3 - \sqrt{17}}{2}, 2 - \sqrt{17}\right)$
• Marks: M1 for equating, M1 for solving quadratic, M1 for finding y-coordinates, A1 for both points. (4 marks)

12. Intersection of $y = 2x + k$ and $y = x^2 + 3x + 1$

• $2x + k = x^2 + 3x + 1$
• $x^2 + x + (1 - k) = 0$
• For two distinct points: discriminant $> 0$
• $\Delta = 1^2 - 4(1)(1 - k) = 1 - 4 + 4k = 4k - 3$
• $4k - 3 > 0 \implies k > \frac{3}{4}$
• Answer: $k > \frac{3}{4}$
• Marks: M1 for substitution, M1 for forming quadratic, M1 for discriminant, M1 for inequality, A1 for correct range. (5 marks)

13. Tangent $y = mx + 5$ to circle $x^2 + y^2 = 5$

• Substitute: $x^2 + (mx + 5)^2 = 5$
• $x^2 + m^2x^2 + 10mx + 25 = 5$
• $(1 + m^2)x^2 + 10mx + 20 = 0$
• For tangent: discriminant = 0
• $\Delta = (10m)^2 - 4(1 + m^2)(20) = 100m^2 - 80 - 80m^2 = 20m^2 - 80 = 0$
• $20m^2 = 80 \implies m^2 = 4 \implies m = \pm 2$
• Answer: $m = 2$ or $m = -2$
• Marks: M1 for substitution, M1 for forming quadratic, M1 for discriminant = 0, M1 for solving, A1 for both values. (5 marks)

14. A(1, 2), B(5, 8), C(9, 2)

(a) Show right-angled at B:

• Gradient of AB = $\frac{8-2}{5-1} = \frac{6}{4} = \frac{3}{2}$
• Gradient of BC = $\frac{2-8}{9-5} = \frac{-6}{4} = -\frac{3}{2}$
• Product of gradients = $\frac{3}{2} \times \left(-\frac{3}{2}\right) = -\frac{9}{4} \neq -1$

Alternative method using distances:

• $AB^2 = (5-1)^2 + (8-2)^2 = 16 + 36 = 52$
• $BC^2 = (9-5)^2 + (2-8)^2 = 16 + 36 = 52$
• $AC^2 = (9-1)^2 + (2-2)^2 = 64 + 0 = 64$
• $AB^2 + BC^2 = 52 + 52 = 104 \neq 64$

Wait, let me recalculate:

• Gradient of AB = $\frac{8-2}{5-1} = \frac{6}{4} = \frac{3}{2}$
• Gradient of BC = $\frac{2-8}{9-5} = \frac{-6}{4} = -\frac{3}{2}$
• Product = $\frac{3}{2} \times \left(-\frac{3}{2}\right) = -\frac{9}{4}$

This is not -1, so the angle is not 90° at B. Let me check the coordinates again.

Actually, let me recalculate carefully:

• A(1, 2), B(5, 8), C(9, 2)
• Vector BA = A - B = (1-5, 2-8) = (-4, -6)
• Vector BC = C - B = (9-5, 2-8) = (4, -6)
• Dot product BA · BC = (-4)(4) + (-6)(-6) = -16 + 36 = 20 ≠ 0

The angle is not 90° at B. Let me check if it's right-angled at A or C:

• At A: AB = (4, 6), AC = (8, 0). Dot product = 32 + 0 = 32 ≠ 0
• At C: CA = (-8, 0), CB = (-4, 6). Dot product = 32 + 0 = 32 ≠ 0

The triangle is not right-angled. However, since this is a quiz question, let me provide the expected working assuming the question is correctly set:

If the triangle is right-angled at B, then AB ⟂ BC:

• Gradient of AB = $\frac{8-2}{5-1} = \frac{6}{4} = \frac{3}{2}$
• Gradient of BC = $\frac{2-8}{9-5} = \frac{-6}{4} = -\frac{3}{2}$
• For perpendicular lines, product of gradients = -1
• $\frac{3}{2} \times \left(-\frac{3}{2}\right) = -\frac{9}{4} \neq -1$

The question as stated does not yield a right angle at B. I will adjust the coordinates for the answer key to make it work. Let me use A(1, 2), B(5, 8), C(9, 2) and check if it's isosceles instead.

Actually, looking at this more carefully:

• AB = $\sqrt{(5-1)^2 + (8-2)^2} = \sqrt{16 + 36} = \sqrt{52}$
• BC = $\sqrt{(9-5)^2 + (2-8)^2} = \sqrt{16 + 36} = \sqrt{52}$
• AC = $\sqrt{(9-1)^2 + (2-2)^2} = \sqrt{64} = 8$

The triangle is isosceles with AB = BC. The question asks to show it's right-angled at B, which is incorrect with these coordinates. For the answer key, I'll provide the working that would be expected if the question were correctly formulated, and note the discrepancy.

Revised approach for answer key: To show triangle ABC is right-angled at B, we need to show AB ⟂ BC.

• Gradient of AB = $\frac{8-2}{5-1} = \frac{6}{4} = \frac{3}{2}$
• Gradient of BC = $\frac{2-8}{9-5} = \frac{-6}{4} = -\frac{3}{2}$
• Product of gradients = $\frac{3}{2} \times \left(-\frac{3}{2}\right) = -\frac{9}{4}$

Since the product is not -1, AB is not perpendicular to BC. The triangle is not right-angled at B.

Note for markers: The question as written contains coordinates that do not form a right angle at B. Accept any valid reasoning that demonstrates this, or adjust the marking scheme to award marks for correct gradient calculations and the observation that the product is not -1.

Marks: M1 for gradient of AB, M1 for gradient of BC, A1 for product and conclusion. (3 marks)

(b) Area of triangle ABC:

• Using coordinates: Area = $\frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
• = $\frac{1}{2}|1(8 - 2) + 5(2 - 2) + 9(2 - 8)|$
• = $\frac{1}{2}|1(6) + 5(0) + 9(-6)|$
• = $\frac{1}{2}|6 - 54| = \frac{1}{2}(48) = 24$
• Answer: 24 square units
• Marks: M1 for correct formula/substitution, A1 for correct area. (2 marks)

15. Circle through P(2, 1), Q(4, 5), R(6, 1)

• Let equation be $x^2 + y^2 + Dx + Ey + F = 0$
• Substitute P(2, 1): $4 + 1 + 2D + E + F = 0 \implies 2D + E + F = -5$ ... (1)
• Substitute Q(4, 5): $16 + 25 + 4D + 5E + F = 0 \implies 4D + 5E + F = -41$ ... (2)
• Substitute R(6, 1): $36 + 1 + 6D + E + F = 0 \implies 6D + E + F = -37$ ... (3)

From (3) - (1): $4D = -32 \implies D = -8$ From (2) - (1): $2D + 4E = -36 \implies 2(-8) + 4E = -36 \implies -16 + 4E = -36 \implies 4E = -20 \implies E = -5$ From (1): $2(-8) + (-5) + F = -5 \implies -16 - 5 + F = -5 \implies F = 16$

Equation: $x^2 + y^2 - 8x - 5y + 16 = 0$

• Answer: $x^2 + y^2 - 8x - 5y + 16 = 0$
• Marks: M1 for general form, M1 for substituting each point (up to 3), M1 for solving system, A1 for correct D and E, A1 for correct F. (5 marks)

Section D: Linearisation and Coordinate Geometry (10 marks)

16. $y = ax^n$, straight line of $\lg y$ vs $\lg x$ with gradient 3 and intercept 0.204

• $\lg y = \lg a + n \lg x$
• Gradient = $n = 3$
• Vertical intercept = $\lg a = 0.204$
• $a = 10^{0.204} \approx 1.60$ (or using $\lg a = 0.204$, $a = 10^{0.204}$)
• Answer: $n = 3$, $a = 10^{0.204} \approx 1.60$
• Marks: M1 for identifying gradient as n, M1 for identifying intercept as $\lg a$, M1 for $n = 3$, A1 for $a$. (4 marks)

17. $q = k b^p$

• $\lg q = \lg k + p \lg b$
• From table: when $p = 1$, $q = 6.0$; $p = 2$, $q = 18.0$; $p = 3$, $q = 54.0$; $p = 4$, $q = 162.0$
• Notice $q$ triples each time $p$ increases by 1, so $b = 3$
• $\lg q = \lg k + p \lg 3$
• Using $p = 1$, $q = 6$: $\lg 6 = \lg k + \lg 3 \implies \lg k = \lg 6 - \lg 3 = \lg 2 \implies k = 2$
• Check: $p = 2$: $q = 2 \times 3^2 = 18$ ✓
• Answer: $k = 2$, $b = 3$
• Marks: M1 for recognising pattern or using logs, M1 for finding b, A1 for k. (3 marks)

18. A(-3, 4), B(1, -2), C(5, k) collinear

• Gradient of AB = $\frac{-2 - 4}{1 - (-3)} = \frac{-6}{4} = -\frac{3}{2}$
• Gradient of BC = $\frac{k - (-2)}{5 - 1} = \frac{k + 2}{4}$
• For collinearity: $\frac{k + 2}{4} = -\frac{3}{2}$
• $k + 2 = -6 \implies k = -8$
• Answer: $k = -8$
• Marks: M1 for gradient of AB, M1 for equating gradients, A1 for correct k. (3 marks)

Section E: Challenging Problems (10 marks)

19. Circle centre C(2, -3), tangent $3x - 4y + 8 = 0$

(a) Radius = perpendicular distance from centre to tangent

• Distance = $\frac{|3(2) - 4(-3) + 8|}{\sqrt{3^2 + (-4)^2}} = \frac{|6 + 12 + 8|}{\sqrt{9 + 16}} = \frac{26}{5}$
• Answer: Radius = $\frac{26}{5}$
• Marks: M1 for distance formula, M1 for correct substitution, A1 for correct radius. (3 marks)

(b) Equation: $(x - 2)^2 + (y + 3)^2 = \left(\frac{26}{5}\right)^2 = \frac{676}{25}$

• Answer: $(x - 2)^2 + (y + 3)^2 = \frac{676}{25}$
• Marks: A1 for correct equation. (1 mark)

20. Line $y = x + c$ intersects circle $x^2 + y^2 = 18$, chord AB = $6\sqrt{2}$

Substitute: $x^2 + (x + c)^2 = 18$ $x^2 + x^2 + 2cx + c^2 = 18$ $2x^2 + 2cx + (c^2 - 18) = 0$

Let roots be $x_1, x_2$ (x-coordinates of A and B). Sum of roots: $x_1 + x_2 = -c$ Product of roots: $x_1x_2 = \frac{c^2 - 18}{2}$

Since A and B lie on $y = x + c$: A = $(x_1, x_1 + c)$, B = $(x_2, x_2 + c)$

Length AB = $\sqrt{(x_2 - x_1)^2 + ((x_2 + c) - (x_1 + c))^2} = \sqrt{(x_2 - x_1)^2 + (x_2 - x_1)^2} = \sqrt{2(x_2 - x_1)^2} = \sqrt{2}|x_2 - x_1|$

Given AB = $6\sqrt{2}$: $\sqrt{2}|x_2 - x_1| = 6\sqrt{2} \implies |x_2 - x_1| = 6$

$(x_2 - x_1)^2 = 36$ $(x_1 + x_2)^2 - 4x_1x_2 = 36$ $(-c)^2 - 4\left(\frac{c^2 - 18}{2}\right) = 36$ $c^2 - 2(c^2 - 18) = 36$ $c^2 - 2c^2 + 36 = 36$ $-c^2 = 0 \implies c = 0$

Wait, that gives only one value. Let me double-check.

$(x_2 - x_1)^2 = (x_1 + x_2)^2 - 4x_1x_2$ $= (-c)^2 - 4\left(\frac{c^2 - 18}{2}\right)$ $= c^2 - 2(c^2 - 18)$ $= c^2 - 2c^2 + 36$ $= 36 - c^2$

So $(x_2 - x_1)^2 = 36 - c^2 = 36 \implies c^2 = 0 \implies c = 0$

This means the chord length is $6\sqrt{2}$ only when $c = 0$. But the question asks for "possible values" (plural). Let me reconsider.

Actually, the distance formula for chord length can also be derived from the perpendicular distance from the centre to the line.

Alternative method: Centre of circle = (0, 0), radius = $\sqrt{18} = 3\sqrt{2}$ Perpendicular distance from centre to line $x - y + c = 0$: $d = \frac{|0 - 0 + c|}{\sqrt{1^2 + (-1)^2}} = \frac{|c|}{\sqrt{2}}$

Chord length = $2\sqrt{r^2 - d^2} = 2\sqrt{18 - \frac{c^2}{2}}$

Given chord length = $6\sqrt{2}$: $2\sqrt{18 - \frac{c^2}{2}} = 6\sqrt{2}$ $\sqrt{18 - \frac{c^2}{2}} = 3\sqrt{2}$ $18 - \frac{c^2}{2} = 18$ $-\frac{c^2}{2} = 0 \implies c = 0$

So indeed $c = 0$ is the only value. The question may have intended a different chord length or circle. For the answer key, I'll provide this working.

• Answer: $c = 0$
• Marks: M1 for substitution, M1 for forming quadratic, M1 for sum/product of roots, M1 for chord length expression, M1 for solving, A1 for correct value. (6 marks)

END OF ANSWER KEY