Coordinate Geometry: Additional Mathematics Tingkatan 4 questions

Section A: objective questions

12 marks

Choose the correct answer. Each question carries 1 mark.

A(−3, 1) and B(7, −4). Point P divides AB in the ratio AP : PB = 2 : 3. The coordinates of P are

(1, −1)
(3, −2)
(2, −2)
(0, 1)

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x = (3(−3) + 2(7))/5 = 1 and y = (3(1) + 2(−4))/5 = −1. (3, −2) comes from swapping m and n.

For the teacher: Applying, low difficulty, standard 7.1.2

Textbook p. 178

The midpoint of the line joining A(h, 1) and B(7, k) is M(2, 3). Find h + k.

8
−2
2
12

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(h + 7)/2 = 2 gives h = −3. (1 + k)/2 = 3 gives k = 5. h + k = 2.

For the teacher: Applying, low difficulty, standard 7.1.2

Textbook p. 179

Point P(3, −1) lies on the line joining A(−3, 2) and B(13, −6). The ratio AP : PB is

5 : 3
3 : 5
3 : 8
5 : 8

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Let AP : PB = m : n. (n(−3) + m(13))/(m + n) = 3 gives 10m = 6n, so m : n = 3 : 5. Check with y: (2(5) + (−6)(3))/8 = −1. 5 : 3 swaps the ratio; 3 : 8 is AP : AB.

For the teacher: Applying, medium difficulty, standard 7.1.2

Textbook p. 180

The straight line 2x + 5y = 10 is perpendicular to the line y = kx + 3. Find the value of k.

−5/2
2/5
−2/5
5/2

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2x + 5y = 10 gives y = −(2/5)x + 2, so m = −2/5. Perpendicular means k × (−2/5) = −1, so k = 5/2. −5/2 only changes the sign.

For the teacher: Understanding, low difficulty, standard 7.2.1

Textbook p. 185

The line 4x − 2y = 9 is parallel to y = (p + 1)x − 3. Find the value of p.

1
2
3
−1

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4x − 2y = 9 gives y = 2x − 9/2, so m = 2. Parallel means p + 1 = 2, so p = 1. p = 2 forgets the +1.

For the teacher: Applying, low difficulty, standard 7.2.1

Textbook p. 186

The vertices of triangle ABC are A(0, 1), B(4, 3) and C(6, c). Given ∠ABC = 90°, find c.

4
7
−1
5

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m_AB = 2/4 = 1/2, so m_BC = −2. (c − 3)/(6 − 4) = −2, so c − 3 = −4 and c = −1. c = 4 uses m_BC = 1/2 (parallel instead of perpendicular); c = 7 drops the minus sign.

For the teacher: Applying, medium difficulty, standard 7.2.1

Textbook p. 187

Find the area of triangle ABC with vertices A(−1, 2), B(4, −1) and C(2, 5).

24 units²
12 units²
13 units²
26 units²

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First sum = 1 + 20 + 4 = 25. Second sum = 8 − 2 − 5 = 1. Area = ½|25 − 1| = 12. 24 forgets ½.

For the teacher: Applying, low difficulty, standard 7.3.2

Textbook p. 193

The points (2, −1), (4, p) and (8, 11) lie on a straight line. Find p.

5
7
−3
3

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The gradient from (2, −1) to (8, 11) is 12/6 = 2. From (2, −1) to (4, p): (p + 1)/2 = 2, so p = 3. Setting the area of the triangle to 0 gives the same answer.

For the teacher: Analysing, medium difficulty, standard 7.3.2

Textbook p. 194

The distance of a point P(x, y) from Q(−2, 1) is always 5 units. The equation of the locus of P is

x² + y² + 4x − 2y − 20 = 0
x² + y² − 4x + 2y − 20 = 0
x² + y² + 4x − 2y = 0
x² + y² + 4x − 2y + 30 = 0

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(x + 2)² + (y − 1)² = 25 gives x² + y² + 4x − 2y + 5 = 25, so the constant is −20. The second option has the signs of the centre reversed; the third uses r instead of r².

For the teacher: Applying, low difficulty, standard 7.4.1

Textbook p. 201

Point P moves so that it is always the same distance from A(0, 2) and B(4, 6). The equation of the locus of P is

y = x + 2
x + y = 2
x + y = 6
x − y = 6

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x² + (y − 2)² = (x − 4)² + (y − 6)² gives 8x + 8y = 48, so x + y = 6. y = x + 2 is the line AB itself. The locus is perpendicular to AB through the midpoint (2, 4).

For the teacher: Applying, medium difficulty, standard 7.4.1

Textbook p. 202

Point P(x, y) moves so that its distance from A(1, 0) is twice its distance from the origin O. The equation of the locus is

x² + y² + 2x − 1 = 0
3x² + 3y² + 2x − 1 = 0
3x² + 3y² − 8x + 4 = 0
3x² + 3y² − 2x + 1 = 0

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PA = 2PO, so PA² = 4PO². (x − 1)² + y² = 4x² + 4y² gives 3x² + 3y² + 2x − 1 = 0. The first option wrote 2PO² instead of 4PO².

For the teacher: Analysing, high difficulty, standard 7.4.1

Textbook p. 202

Point P is on the line segment AB and AP : PB = 1 : 1. Point P is

closer to A than to B
closer to B than to A
the point where AB meets the x-axis
the midpoint of AB

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m = n means the two parts are equal, so P is the midpoint. Its coordinates are the averages of the coordinates of A and B.

For the teacher: Remembering, low difficulty, standard 7.1.1

Textbook p. 176

Section B: structured questions

32 marks

Answer all questions. Show your working.

A robot moves in a straight line from E(−2, 1) to F(18, 11) at a steady speed. The journey takes 50 seconds.

[3 marks] (a)Find the coordinates of the robot after 20 seconds.

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EP : PF = 20 : 30 = 2 : 3 [1]1
x = (3(−2) + 2(18))/5 = 6 [1]1
y = (3(1) + 2(11))/5 = 5, so (6, 5) [1]1

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[3 marks] (b)The robot is at Q(14, 9). Find the ratio EQ : QF and the time taken to reach Q.

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(n(−2) + 18m)/(m + n) = 14, so 4m = 16n [1]1
EQ : QF = 4 : 1 [1]1
Time = 4/5 × 50 = 40 seconds [1]1

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For the teacher: Applying, low difficulty, standard 7.1.3

Textbook p. 181

Diagram 1 shows points A(2, 1), B(8, 5) and C(7, 0). The line L passes through C and is perpendicular to AB. L meets AB at F.

Cartesian plane with line AB through A(2, 1) and B(8, 5), and a line L through C(7, 0) crossing AB at F — Diagram 1

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

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(5 − 1)/(8 − 2) = 2/3 [1]1

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[3 marks] (b)Find the equation of L.

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Gradient of L = −3/2 [1]1
y − 0 = −3/2 (x − 7) [1]1
3x + 2y = 21 [1]1

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[3 marks] (c)Find the coordinates of F.

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AB: 2x − 3y = 1 [1]1
Solve with 3x + 2y = 21: 13x = 65 [1]1
x = 5, y = 3, so F = (5, 3) [1]1

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For the teacher: Applying, low difficulty, standard 7.2.2

Textbook p. 188

The vertices of triangle ABC are A(−2, −3), B(6, −1) and C(2, 5).

[3 marks] (a)Find the area of triangle ABC.

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First sum: 2 + 30 − 6 = 26 [1]1
Second sum: −18 − 2 − 10 = −30 [1]1
Area = ½|26 + 30| = 28 units² [1]1

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[2 marks] (b)E(10, k) lies on the straight line AB. Find k.

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Gradient AB = 2/8 = 1/4 [1]1
(k + 3)/12 = 1/4, so k = 0 [1]1

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For the teacher: Applying, medium difficulty, standard 7.3.2

Textbook p. 194

Diagram 2 shows quadrilateral PQRS with P(1, 1), Q(7, 2), R(6, 6) and S(0, 5).

Quadrilateral with vertices P(1, 1), Q(7, 2), R(6, 6) and S(0, 5) — Diagram 2

[3 marks] (a)Find the area of PQRS.

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First sum: 2 + 42 + 30 + 0 = 74 [1]1
Second sum: 7 + 12 + 0 + 5 = 24 [1]1
Area = ½ × 50 = 25 units² [1]1

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[2 marks] (b)Show that PQRS is a parallelogram.

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Gradient PQ = 1/6 = gradient SR [1]1
Gradient QR = −4 = gradient PS, so both pairs are parallel [1]1

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[1 mark] (c)Hence, state the area of triangle PQR.

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½ × 25 = 12.5 units² [1]1

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For the teacher: Applying, low difficulty, standard 7.3.3

Textbook p. 195

Point P(x, y) moves so that PA : PB = 1 : 2, where A is (−1, 0) and B is (5, 0).

[4 marks] (a)Show that the equation of the locus of P is x² + y² + 6x − 7 = 0.

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4PA² = PB² [1]1
4[(x + 1)² + y²] = (x − 5)² + y² [1]1
4x² + 8x + 4 + 4y² = x² − 10x + 25 + y² [1]1
3x² + 3y² + 18x − 21 = 0, so x² + y² + 6x − 7 = 0 [1]1

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[2 marks] (b)Write the equation as (x − a)² + (y − b)² = r² and state the centre and radius.

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(x + 3)² + y² = 16 [1]1
Centre (−3, 0), radius 4 [1]1

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[2 marks] (c)Amir says (1, 0) is on the locus. Is he correct? Check with the ratio PA : PB.

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1 + 0 + 6 − 7 = 0, so it is on the locus [1]1
PA = 2, PB = 4, ratio 1 : 2, so he is correct [1]1

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For the teacher: Evaluating, high difficulty, standard 7.4.1

Textbook p. 202

Section A: objective questions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Section B: structured questions

Question 13

Question 14

Question 15

Question 16

Question 17