Secondary 4 Additional Mathematics Practice Paper 5

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TuitionGoWhere Practice Paper - Additional Mathematics Secondary 4

TuitionGoWhere Practice Paper (AI)

Subject: Additional Mathematics (4049/4047) Level: Secondary 4 Paper: Practice Paper – Graphs & Coordinate Geometry Version: 5 of 5 Duration: 1 hour 30 minutes Total Marks: 80

Name: _________________________ Class: _________________________ Date: _________________________

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Instructions to Candidates

1. This paper consists of 20 questions on the topic of Graphs and Coordinate Geometry.
2. Answer all questions.
3. Write your answers in the spaces provided.
4. Unless otherwise stated, give non-exact answers correct to three significant figures.
5. Solutions by accurate drawing will not be accepted.
6. You are expected to use a scientific calculator where appropriate.
7. The number of marks is given in brackets [ ] at the end of each question or part question.
8. You are reminded of the need for clear presentation in your answers.

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Section A: Straight Lines and Linear Relations (20 marks)

Answer all questions in this section.

1. The points A(–2, 3) and B(4, –1) are given.

(a) Find the gradient of the line AB. [1]

Answer: _________________________

(b) Find the equation of the perpendicular bisector of AB. Give your answer in the form $ax + by + c = 0$, where $a$, $b$, and $c$ are integers. [3]

Answer: _________________________

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2. The line $L_1$ has equation $2x - 3y + 6 = 0$. The line $L_2$ passes through the point P(5, 1) and is parallel to $L_1$.

(a) Find the equation of $L_2$. [2]

Answer: _________________________

(b) Find the coordinates of the point where $L_2$ meets the $x$-axis. [1]

Answer: _________________________

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3. The line $L$ has equation $y = 2x - 1$. The point Q has coordinates $(k, 3k + 1)$.

(a) Find the value of $k$ for which Q lies on $L$. [2]

Answer: _________________________

(b) For the value of $k$ found in part (a), find the perpendicular distance from Q to the $y$-axis. [1]

Answer: _________________________

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4. The points R(1, 4), S(5, 2), and T(3, –2) are the vertices of a triangle.

(a) Show that the triangle RST is right-angled at S. [2]

Answer: _________________________

(b) Find the area of triangle RST. [2]

Answer: _________________________

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5. The line $L_1$ passes through the points A(–1, 2) and B(3, 6). The line $L_2$ is perpendicular to $L_1$ and passes through the midpoint of AB.

Find the equation of $L_2$. Give your answer in the form $y = mx + c$. [3]

Answer: _________________________

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Section B: Quadratic Curves and Intersections (20 marks)

Answer all questions in this section.

6. The curve $C$ has equation $y = x^2 - 4x + 7$.

(a) Express $x^2 - 4x + 7$ in the form $(x - p)^2 + q$, where $p$ and $q$ are constants. [2]

Answer: _________________________

(b) Hence state the coordinates of the minimum point of $C$. [1]

Answer: _________________________

(c) Find the set of values of $k$ for which the line $y = 2x + k$ does not intersect the curve $C$. [3]

Answer: _________________________

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7. The line $y = mx + 2$ is a tangent to the curve $y = x^2 + 3x + c$ at the point where $x = 1$.

(a) Find the value of $m$ and the value of $c$. [4]

Answer: _________________________

(b) Find the coordinates of the point of tangency. [1]

Answer: _________________________

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8. The curve $y = 2x^2 + bx + 8$ has its minimum point on the $x$-axis.

(a) Find the possible values of $b$. [3]

Answer: _________________________

(b) For the positive value of $b$, find the coordinates of the minimum point. [2]

Answer: _________________________

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9. The quadratic function $f(x) = ax^2 + bx + c$ has a maximum value of 9 when $x = 2$, and $f(0) = 5$.

Find the values of $a$, $b$, and $c$. [4]

Answer: _________________________

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Section C: Circles (20 marks)

Answer all questions in this section.

10. A circle $C_1$ has equation $x^2 + y^2 - 6x + 4y - 12 = 0$.

(a) Find the coordinates of the centre and the radius of $C_1$. [3]

Answer: _________________________

(b) The point P(7, –5) lies on $C_1$. Find the equation of the tangent to $C_1$ at P. Give your answer in the form $ax + by + c = 0$, where $a$, $b$, and $c$ are integers. [3]

Answer: _________________________

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11. A circle passes through the points A(2, 1) and B(6, 5). The centre of the circle lies on the line $y = x - 2$.

Find the equation of the circle. Give your answer in the form $(x - a)^2 + (y - b)^2 = r^2$. [5]

Answer: _________________________

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12. The circle $C$ has centre (3, –1) and radius 5.

(a) Write down the equation of $C$ in standard form. [1]

Answer: _________________________

(b) Find the coordinates of the points where $C$ meets the $x$-axis. [3]

Answer: _________________________

(c) Determine whether the point (8, 2) lies inside, on, or outside the circle $C$. Show your working clearly. [2]

Answer: _________________________

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13. A circle has equation $x^2 + y^2 + 2gx + 2fy + c = 0$. The circle passes through the origin and has its centre on the line $x + y = 0$. The radius of the circle is $\sqrt{8}$.

Find the possible equations of the circle. [4]

Answer: _________________________

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Section D: Coordinate Geometry Applications (20 marks)

Answer all questions in this section.

14. The variables $x$ and $y$ are related by the equation $y = ax^n$, where $a$ and $n$ are constants.

(a) Explain how a straight line graph may be drawn to represent this relationship, stating clearly the variables that should be plotted and what the gradient and intercept represent. [3]

Answer: _________________________

(b) The table shows experimental values of $x$ and $y$.

$x$	2	4	6	8	10
$y$	3.2	12.8	28.8	51.2	80.0

Using a suitable straight line graph, estimate the values of $a$ and $n$. [4]

Answer: _________________________

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15. The curve $C$ has equation $y = \frac{4}{x - 2}$, for $x > 2$.

(a) State the equation of the vertical asymptote of $C$. [1]

Answer: _________________________

(b) State the equation of the horizontal asymptote of $C$. [1]

Answer: _________________________

(c) The line $y = x + k$ intersects $C$ at two distinct points. Find the set of possible values of $k$. [4]

Answer: _________________________

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16. The points P(–3, 1), Q(1, 5), and R(5, 1) are given.

(a) Show that PQ = QR. [2]

Answer: _________________________

(b) Find the area of triangle PQR. [2]

Answer: _________________________

(c) The point S is such that PQRS is a rhombus. Find the coordinates of S. [2]

Answer: _________________________

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17. The line $L$ has equation $3x + 4y = 24$. The line $L$ meets the $x$-axis at A and the $y$-axis at B.

(a) Find the coordinates of A and B. [2]

Answer: _________________________

(b) Find the area of triangle OAB, where O is the origin. [1]

Answer: _________________________

(c) A circle has AB as its diameter. Find the equation of this circle. [3]

Answer: _________________________

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18. The curve $y = x^2 - 2x - 3$ and the line $y = x + 1$ intersect at points P and Q.

(a) Find the coordinates of P and Q. [3]

Answer: _________________________

(b) Find the length of the line segment PQ. [2]

Answer: _________________________

(c) Find the equation of the perpendicular bisector of PQ. [2]

Answer: _________________________

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19. A curve has equation $y = \frac{2x + 1}{x - 1}$, for $x \neq 1$.

(a) Express $\frac{2x + 1}{x - 1}$ in the form $A + \frac{B}{x - 1}$, where $A$ and $B$ are constants. [2]

Answer: _________________________

(b) Hence state the equations of the asymptotes of the curve. [2]

Answer: _________________________

(c) Find the coordinates of the points where the curve meets the axes. [2]

Answer: _________________________

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20. The diagram shows a quadrilateral ABCD with vertices A(–2, 1), B(4, 3), C(6, –1), and D(0, –3).

[Note: No diagram is provided; use the coordinates given.]

(a) Show that ABCD is a parallelogram. [2]

Answer: _________________________

(b) Find the coordinates of the point of intersection of the diagonals AC and BD. [2]

Answer: _________________________

(c) Determine whether ABCD is a rhombus. Justify your answer. [2]

Answer: _________________________

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END OF PAPER

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Check your work carefully. Ensure all answers are in the spaces provided.

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TuitionGoWhere Practice Paper - Additional Mathematics Secondary 4

Answer Key and Marking Scheme – Graphs & Coordinate Geometry (Version 5)

Total Marks: 80

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Section A: Straight Lines and Linear Relations (20 marks)

1. A(–2, 3), B(4, –1)

(a) Gradient of AB [ m = \frac{-1 - 3}{4 - (-2)} = \frac{-4}{6} = -\frac{2}{3} ] Answer: $-\frac{2}{3}$ [1]

(b) Midpoint of AB: $\left(\frac{-2+4}{2}, \frac{3+(-1)}{2}\right) = (1, 1)$ [1]

Gradient of perpendicular bisector: $m_{\perp} = \frac{3}{2}$ (since $m \cdot m_{\perp} = -1$) [1]

Equation: $y - 1 = \frac{3}{2}(x - 1)$ $2y - 2 = 3x - 3$ $3x - 2y - 1 = 0$ [1]

Answer: $3x - 2y - 1 = 0$ [3]

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2. $L_1: 2x - 3y + 6 = 0$

(a) Gradient of $L_1$: $2x - 3y + 6 = 0 \implies 3y = 2x + 6 \implies y = \frac{2}{3}x + 2$, so $m = \frac{2}{3}$ [1]

$L_2$ is parallel, so $m = \frac{2}{3}$. Passes through P(5, 1): $y - 1 = \frac{2}{3}(x - 5)$ $3y - 3 = 2x - 10$ $2x - 3y - 7 = 0$ [1]

Answer: $2x - 3y - 7 = 0$ [2]

(b) Meets $x$-axis where $y = 0$: $2x - 3(0) - 7 = 0 \implies 2x = 7 \implies x = \frac{7}{2}$ Answer: $\left(\frac{7}{2}, 0\right)$ [1]

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3. $L: y = 2x - 1$, Q$(k, 3k + 1)$

(a) Q lies on $L$: $3k + 1 = 2k - 1$ [1] $3k - 2k = -1 - 1 \implies k = -2$ [1]

Answer: $k = -2$ [2]

(b) When $k = -2$, Q = $(-2, 3(-2) + 1) = (-2, -5)$. Perpendicular distance from Q to $y$-axis is $|x\text{-coordinate}| = |-2| = 2$. Answer: 2 units [1]

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4. R(1, 4), S(5, 2), T(3, –2)

(a) Gradient of RS: $\frac{2 - 4}{5 - 1} = \frac{-2}{4} = -\frac{1}{2}$ [0.5] Gradient of ST: $\frac{-2 - 2}{3 - 5} = \frac{-4}{-2} = 2$ [0.5] Product of gradients: $-\frac{1}{2} \times 2 = -1$ [0.5] Since product = –1, RS $\perp$ ST, so triangle is right-angled at S. [0.5]

Answer: Shown [2]

(b) Area = $\frac{1}{2} \times \text{RS} \times \text{ST}$ RS = $\sqrt{(5-1)^2 + (2-4)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$ [0.5] ST = $\sqrt{(3-5)^2 + (-2-2)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$ [0.5] Area = $\frac{1}{2} \times 2\sqrt{5} \times 2\sqrt{5} = \frac{1}{2} \times 20 = 10$ [1]

Answer: 10 square units [2]

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5. A(–1, 2), B(3, 6)

Gradient of $L_1$: $\frac{6 - 2}{3 - (-1)} = \frac{4}{4} = 1$ [0.5] Midpoint of AB: $\left(\frac{-1+3}{2}, \frac{2+6}{2}\right) = (1, 4)$ [0.5] Gradient of $L_2$ (perpendicular): $m_{\perp} = -1$ [0.5] Equation: $y - 4 = -1(x - 1)$ $y - 4 = -x + 1$ $y = -x + 5$ [1.5]

Answer: $y = -x + 5$ [3]

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Section B: Quadratic Curves and Intersections (20 marks)

6. $C: y = x^2 - 4x + 7$

(a) $x^2 - 4x + 7 = (x^2 - 4x + 4) + 3 = (x - 2)^2 + 3$ [2] Answer: $(x - 2)^2 + 3$ [2]

(b) Minimum point: $(2, 3)$ [1] Answer: $(2, 3)$ [1]

(c) Intersection: $x^2 - 4x + 7 = 2x + k$ $x^2 - 6x + (7 - k) = 0$ [1] For no intersection, discriminant < 0: $(-6)^2 - 4(1)(7 - k) < 0$ $36 - 28 + 4k < 0$ $8 + 4k < 0$ $k < -2$ [2]

Answer: $k < -2$ [3]

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7. $y = mx + 2$, $y = x^2 + 3x + c$, tangent at $x = 1$

(a) At $x = 1$, point on curve: $y = 1^2 + 3(1) + c = 4 + c$ [0.5] At $x = 1$, point on line: $y = m(1) + 2 = m + 2$ [0.5] Since they meet: $4 + c = m + 2 \implies c = m - 2$ ... (1) [0.5]

For tangency, the line and curve have equal gradients at $x = 1$: Curve gradient: $\frac{dy}{dx} = 2x + 3$, at $x = 1$: $2(1) + 3 = 5$ [0.5] Line gradient: $m$ [0.5] So $m = 5$ [0.5] From (1): $c = 5 - 2 = 3$ [0.5]

Answer: $m = 5$, $c = 3$ [4]

(b) Point of tangency: $x = 1$, $y = 5(1) + 2 = 7$ (or $y = 1^2 + 3(1) + 3 = 7$) Answer: $(1, 7)$ [1]

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8. $y = 2x^2 + bx + 8$, minimum on $x$-axis

(a) Minimum on $x$-axis means the minimum value is 0 and the discriminant = 0 (since the vertex touches the $x$-axis). [1] Complete the square: $2x^2 + bx + 8 = 2\left(x^2 + \frac{b}{2}x\right) + 8 = 2\left(x + \frac{b}{4}\right)^2 - 2\left(\frac{b}{4}\right)^2 + 8$ $= 2\left(x + \frac{b}{4}\right)^2 + 8 - \frac{b^2}{8}$ [1] Minimum value = $8 - \frac{b^2}{8} = 0 \implies \frac{b^2}{8} = 8 \implies b^2 = 64 \implies b = \pm 8$ [1]

Answer: $b = 8$ or $b = -8$ [3]

(b) For $b = 8$: $y = 2x^2 + 8x + 8 = 2(x^2 + 4x) + 8 = 2(x + 2)^2 - 8 + 8 = 2(x + 2)^2$ Minimum at $(-2, 0)$. [2]

Answer: $(-2, 0)$ [2]

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9. $f(x) = ax^2 + bx + c$, max value 9 at $x = 2$, $f(0) = 5$

Maximum at $x = 2$: $f(x) = a(x - 2)^2 + 9$ (since $a < 0$ for maximum) [1] $f(0) = 5$: $a(0 - 2)^2 + 9 = 5 \implies 4a + 9 = 5 \implies 4a = -4 \implies a = -1$ [1] So $f(x) = -(x - 2)^2 + 9 = -(x^2 - 4x + 4) + 9 = -x^2 + 4x - 4 + 9 = -x^2 + 4x + 5$ [1] Thus $a = -1$, $b = 4$, $c = 5$. [1]

Answer: $a = -1$, $b = 4$, $c = 5$ [4]

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Section C: Circles (20 marks)

10. $C_1: x^2 + y^2 - 6x + 4y - 12 = 0$

(a) Complete 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] Centre: $(3, -2)$, radius: $\sqrt{25} = 5$ [1]

Answer: Centre $(3, -2)$, radius 5 [3]

(b) P(7, –5). Gradient of radius CP: $\frac{-5 - (-2)}{7 - 3} = \frac{-3}{4} = -\frac{3}{4}$ [1] Gradient of tangent: $m_{\perp} = \frac{4}{3}$ [0.5] Equation: $y - (-5) = \frac{4}{3}(x - 7)$ $y + 5 = \frac{4}{3}x - \frac{28}{3}$ $3y + 15 = 4x - 28$ $4x - 3y - 43 = 0$ [1.5]

Answer: $4x - 3y - 43 = 0$ [3]

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11. A(2, 1), B(6, 5), centre on $y = x - 2$

Let centre be $(a, a - 2)$. [0.5] Distance to A = distance to B: $(a - 2)^2 + (a - 2 - 1)^2 = (a - 6)^2 + (a - 2 - 5)^2$ [1] $(a - 2)^2 + (a - 3)^2 = (a - 6)^2 + (a - 7)^2$ $(a^2 - 4a + 4) + (a^2 - 6a + 9) = (a^2 - 12a + 36) + (a^2 - 14a + 49)$ $2a^2 - 10a + 13 = 2a^2 - 26a + 85$ $-10a + 13 = -26a + 85$ $16a = 72 \implies a = 4.5$ [1.5] Centre: $(4.5, 2.5)$ [0.5] Radius: $\sqrt{(4.5 - 2)^2 + (2.5 - 1)^2} = \sqrt{2.5^2 + 1.5^2} = \sqrt{6.25 + 2.25} = \sqrt{8.5}$ [1] Equation: $(x - 4.5)^2 + (y - 2.5)^2 = 8.5$ [0.5]

Answer: $(x - 4.5)^2 + (y - 2.5)^2 = 8.5$ [5]

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12. Centre (3, –1), radius 5

(a) $(x - 3)^2 + (y + 1)^2 = 25$ [1] Answer: $(x - 3)^2 + (y + 1)^2 = 25$ [1]

(b) Meets $x$-axis: $y = 0$ $(x - 3)^2 + (0 + 1)^2 = 25$ $(x - 3)^2 + 1 = 25$ $(x - 3)^2 = 24$ $x - 3 = \pm \sqrt{24} = \pm 2\sqrt{6}$ $x = 3 \pm 2\sqrt{6}$ [2] Points: $(3 + 2\sqrt{6}, 0)$ and $(3 - 2\sqrt{6}, 0)$ [1]

Answer: $(3 + 2\sqrt{6}, 0)$ and $(3 - 2\sqrt{6}, 0)$ [3]

(c) Distance from (8, 2) to centre (3, –1): $\sqrt{(8 - 3)^2 + (2 - (-1))^2} = \sqrt{25 + 9} = \sqrt{34}$ [1] Since $\sqrt{34} \approx 5.83 > 5$, the point lies outside the circle. [1]

Answer: Outside the circle (distance $\sqrt{34} > 5$) [2]

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13. $x^2 + y^2 + 2gx + 2fy + c = 0$, passes through origin, centre on $x + y = 0$, radius $\sqrt{8}$

Centre: $(-g, -f)$. On $x + y = 0$: $-g - f = 0 \implies f = -g$ [1] Passes through (0, 0): $0 + 0 + 0 + 0 + c = 0 \implies c = 0$ [0.5] Radius: $\sqrt{g^2 + f^2 - c} = \sqrt{g^2 + (-g)^2 - 0} = \sqrt{2g^2} = \sqrt{8}$ [1] $2g^2 = 8 \implies g^2 = 4 \implies g = \pm 2$ [0.5] If $g = 2$, $f = -2$: equation $x^2 + y^2 + 4x - 4y = 0$ [0.5] If $g = -2$, $f = 2$: equation $x^2 + y^2 - 4x + 4y = 0$ [0.5]

Answer: $x^2 + y^2 + 4x - 4y = 0$ and $x^2 + y^2 - 4x + 4y = 0$ [4]

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Section D: Coordinate Geometry Applications (20 marks)

14. $y = ax^n$

(a) Take logarithms (any base): $\log y = \log a + n \log x$ [1] Plot $\log y$ against $\log x$. [0.5] The gradient is $n$ and the vertical intercept is $\log a$. [1.5]

Answer: Plot $\log y$ vs $\log x$; gradient = $n$, intercept = $\log a$ [3]

(b) Compute $\log x$ and $\log y$ (using base 10 or natural log):

$x$	$y$	$\log_{10} x$	$\log_{10} y$
2	3.2	0.301	0.505
4	12.8	0.602	1.107
6	28.8	0.778	1.459
8	51.2	0.903	1.709
10	80.0	1.000	1.903

[1 for correct computation]

Plotting $\log y$ vs $\log x$ gives a straight line. [0.5] Gradient $n$: Using points (0.301, 0.505) and (1.000, 1.903): $n = \frac{1.903 - 0.505}{1.000 - 0.301} = \frac{1.398}{0.699} \approx 2.00$ [1] Intercept $\log a$: Using $\log y = n\log x + \log a$ with point (0.301, 0.505): $0.505 = 2(0.301) + \log a \implies \log a = 0.505 - 0.602 = -0.097$ $a = 10^{-0.097} \approx 0.80$ [1.5]

Answer: $n \approx 2$, $a \approx 0.80$ [4]

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15. $y = \frac{4}{x - 2}$, $x > 2$

(a) Vertical asymptote: denominator = 0 $\implies x = 2$ [1] Answer: $x = 2$ [1]

(b) As $x \to \infty$, $y \to 0$. Horizontal asymptote: $y = 0$ [1] Answer: $y = 0$ [1]

(c) Intersection: $\frac{4}{x - 2} = x + k$ $4 = (x + k)(x - 2) = x^2 - 2x + kx - 2k$ $x^2 + (k - 2)x - 2k - 4 = 0$ [1] For two distinct points, discriminant > 0: $(k - 2)^2 - 4(1)(-2k - 4) > 0$ $k^2 - 4k + 4 + 8k + 16 > 0$ $k^2 + 4k + 20 > 0$ [1] Discriminant of this quadratic: $4^2 - 4(1)(20) = 16 - 80 = -64 < 0$ [0.5] Since coefficient of $k^2$ is positive and discriminant < 0, $k^2 + 4k + 20 > 0$ for all real $k$. [1] But also need $x > 2$ for domain. Check: solutions of quadratic are $x = \frac{-(k-2) \pm \sqrt{k^2 + 4k + 20}}{2}$. For $x > 2$, we need the larger root > 2. Since the quadratic in $k$ is always positive, both roots are real. The larger root is $\frac{2 - k + \sqrt{k^2 + 4k + 20}}{2} > 2$ for all $k$ (can be verified). So $k$ can be any real number. [0.5]

Answer: All real values of $k$ [4]

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16. P(–3, 1), Q(1, 5), R(5, 1)

(a) PQ = $\sqrt{(1 - (-3))^2 + (5 - 1)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$ [1] QR = $\sqrt{(5 - 1)^2 + (1 - 5)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$ [0.5] Thus PQ = QR. [0.5]

Answer: Shown [2]

(b) Area of triangle PQR: Using coordinates: $\frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$ $= \frac{1}{2}|(-3)(5 - 1) + 1(1 - 1) + 5(1 - 5)|$ $= \frac{1}{2}|(-3)(4) + 0 + 5(-4)| = \frac{1}{2}|-12 - 20| = \frac{1}{2}(32) = 16$ [2]

Answer: 16 square units [2]

(c) For rhombus PQRS, diagonals bisect each other. Midpoint of PR: $\left(\frac{-3+5}{2}, \frac{1+1}{2}\right) = (1, 1)$ [0.5] Midpoint of QS must also be (1, 1). Let S = $(x, y)$: $\frac{1 + x}{2} = 1 \implies x = 1$; $\frac{5 + y}{2} = 1 \implies y = -3$ [1] So S = $(1, -3)$. [0.5]

Answer: $(1, -3)$ [2]

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17. $L: 3x + 4y = 24$

(a) Meets $x$-axis (A): $y = 0 \implies 3x = 24 \implies x = 8$. A = $(8, 0)$ [0.5] Meets $y$-axis (B): $x = 0 \implies 4y = 24 \implies y = 6$. B = $(0, 6)$ [0.5] Answer: A$(8, 0)$, B$(0, 6)$ [2]

(b) Area of triangle OAB = $\frac{1}{2} \times 8 \times 6 = 24$ [1] Answer: 24 square units [1]

(c) AB is diameter. Centre = midpoint of AB = $\left(\frac{8+0}{2}, \frac{0+6}{2}\right) = (4, 3)$ [1] Radius = half of AB = $\frac{1}{2}\sqrt{(8-0)^2 + (0-6)^2} = \frac{1}{2}\sqrt{64 + 36} = \frac{1}{2}\sqrt{100} = 5$ [1] Equation: $(x - 4)^2 + (y - 3)^2 = 25$ [1]

Answer: $(x - 4)^2 + (y - 3)^2 = 25$ [3]

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18. $y = x^2 - 2x - 3$, $y = x + 1$

(a) Intersection: $x^2 - 2x - 3 = x + 1$ $x^2 - 3x - 4 = 0$ $(x - 4)(x + 1) = 0$ [1] $x = 4$ or $x = -1$ [0.5] When $x = 4$: $y = 4 + 1 = 5$. P or Q = $(4, 5)$ [0.5] When $x = -1$: $y = -1 + 1 = 0$. Q or P = $(-1, 0)$ [0.5] Answer: $(-1, 0)$ and $(4, 5)$ [3]

(b) Length PQ = $\sqrt{(4 - (-1))^2 + (5 - 0)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}$ [2] Answer: $5\sqrt{2}$ units [2]

(c) Midpoint of PQ: $\left(\frac{-1+4}{2}, \frac{0+5}{2}\right) = (1.5, 2.5)$ [0.5] Gradient of PQ: $\frac{5 - 0}{4 - (-1)} = \frac{5}{5} = 1$ [0.5] Gradient of perpendicular bisector: $-1$ [0.5] Equation: $y - 2.5 = -1(x - 1.5)$ $y - 2.5 = -x + 1.5$ $y = -x + 4$ [0.5]

Answer: $y = -x + 4$ [2]

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19. $y = \frac{2x + 1}{x - 1}$

(a) $\frac{2x + 1}{x - 1} = \frac{2(x - 1) + 3}{x - 1} = 2 + \frac{3}{x - 1}$ [2] Answer: $2 + \frac{3}{x - 1}$ [2]

(b) Vertical asymptote: $x = 1$ (denominator = 0) [1] Horizontal asymptote: as $x \to \pm\infty$, $\frac{3}{x-1} \to 0$, so $y \to 2$. Asymptote: $y = 2$ [1] Answer: $x = 1$, $y = 2$ [2]

(c) Meets $y$-axis ($x = 0$): $y = \frac{2(0) + 1}{0 - 1} = \frac{1}{-1} = -1$. Point: $(0, -1)$ [1] Meets $x$-axis ($y = 0$): $0 = \frac{2x + 1}{x - 1} \implies 2x + 1 = 0 \implies x = -\frac{1}{2}$. Point: $\left(-\frac{1}{2}, 0\right)$ [1] Answer: $(0, -1)$ and $\left(-\frac{1}{2}, 0\right)$ [2]

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20. A(–2, 1), B(4, 3), C(6, –1), D(0, –3)

(a) For parallelogram, opposite sides are parallel (equal gradients): Gradient of AB: $\frac{3 - 1}{4 - (-2)} = \frac{2}{6} = \frac{1}{3}$ [0.5] Gradient of DC: $\frac{-3 - (-1)}{0 - 6} = \frac{-2}{-6} = \frac{1}{3}$ [0.5] So AB $\parallel$ DC. Gradient of BC: $\frac{-1 - 3}{6 - 4} = \frac{-4}{2} = -2$ [0.5] Gradient of AD: $\frac{-3 - 1}{0 - (-2)} = \frac{-4}{2} = -2$ [0.5] So BC $\parallel$ AD. Thus ABCD is a parallelogram.

Answer: Shown [2]

(b) Intersection of diagonals is midpoint of both AC and BD. Midpoint of AC: $\left(\frac{-2+6}{2}, \frac{1+(-1)}{2}\right) = (2, 0)$ [1] (Check with BD: $\left(\frac{4+0}{2}, \frac{3+(-3)}{2}\right) = (2, 0)$) [1]

Answer: $(2, 0)$ [2]

(c) For rhombus, all sides equal (or diagonals perpendicular). AB = $\sqrt{(4 - (-2))^2 + (3 - 1)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}$ [0.5] BC = $\sqrt{(6 - 4)^2 + (-1 - 3)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$ [0.5] Since AB $\neq$ BC, ABCD is not a rhombus. [1]

Answer: Not a rhombus; adjacent sides have different lengths ($2\sqrt{10} \neq 2\sqrt{5}$) [2]

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END OF MARKING SCHEME