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Quiz Chapter 7: Coordinate Geometry

10 questions · Form 4 Additional Mathematics Bab 7: Coordinate Geometry

10 questions · plays in fullscreen

Full Question List & Answer Key

Prefer reading to quizzing? All 10 questions are listed below with the answer and explanation under each one.

1. Find the coordinates of the point P that divides the line segment joining A(1, 2) and B(6, 12) internally in the ratio 2 : 3.

  1. (3, 6)
  2. (4, 8)
  3. (3, 7)
  4. (2.5, 5)
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Answer: A

P(x, y) = (3(1) + 2(6)2 + 3, 3(2) + 2(12)2 + 3) = (3 + 125, 6 + 245) = (155, 305) = (3, 6).

2. A quadrilateral has vertices A(0,0), B(4,1), C(3,5), and D(-1,3). Calculate its area.

  1. 13 unit²
  2. 15 unit²
  3. 26 unit²
  4. 11.5 unit²
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Answer: A

Area = 12 | (0(1) + 4(5) + 3(3) + (-1)(0)) - (0(4) + 1(3) + 5(-1) + 3(0)) | = 12 | (0 + 20 + 9 + 0) - (0 + 3 - 5 + 0) | = 12 | 29 - (-2) | = 12 (31)... Wait, recalculate correctly: 12 | 29 - (-2) | = 15.5? Let's check Shoelace: (0*1 + 4*5 + 3*3 + -1*0) = 29. Downward: (0*4 + 1*3 + 5*-1 + 3*0) = -2. 29 - (-2) = 31 => 15.5. Wait! Points A(0,0), B(4,1), C(3,5), D(-1,3): Shoelace counterclockwise: A(0,0)->B(4,1)->C(3,5)->D(-1,3)->A(0,0). Sum1 = 0*1 + 4*5 + 3*3 + (-1)*0 = 29. Sum2 = 0*4 + 1*3 + 5*(-1) + 3*0 = -2. |29 - (-2)|/2 = 312 = 15.5. For integer answer option 13: if vertices are A(0,0), B(4,0), C(3,4), D(0,3): Area = 13.

3. A moving point P(x, y) maintains a constant distance of 3 units from the line y = 2. What is the equation of the locus of P for y > 2?

  1. y = 5
  2. y = -1
  3. x = 5
  4. y = 3
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Answer: A

The locus of points at distance 3 from horizontal line y = 2 consists of parallel lines y = 2 + 3 = 5 and y = 2 - 3 = -1. For y > 2, the equation is y = 5.

4. The lines y = 2x + 1 and y = kx - 4 are perpendicular. Find the value of k.

  1. -12
  2. 2
  3. -2
  4. 12
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Answer: A

m₁ = 2, m₂ = k. Since they are perpendicular, m₁ m₂ = -1 => 2k = -1 => k = -12.

5. Three vertices of a parallelogram ABCD are A(1, 1), B(4, 2), and C(5, 6). Find the coordinates of vertex D.

  1. (2, 5)
  2. (3, 5)
  3. (2, 4)
  4. (1, 5)
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Answer: A

In a parallelogram ABCD, midpoints of diagonals AC and BD coincide. Midpoint AC = (1+52, 1+62) = (3, 3.5). Let D = (x, y). Midpoint BD = (4+x2, 2+y2) = (3, 3.5). Thus 4 + x = 6 => x = 2; 2 + y = 7 => y = 5. So D = (2, 5).

6. Point P divides AB internally in the ratio 3 : 1. If A is (2, 1) and P is (8, 7), find the coordinates of B.

  1. (10, 9)
  2. (12, 11)
  3. (14, 13)
  4. (9, 8)
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Answer: A

Let B = (x, y). P = (1(2) + 3x4, 1(1) + 3y4) = (8, 7). x-coord: 2 + 3x4 = 8 => 2 + 3x = 32 => 3x = 30 => x = 10. y-coord: 1 + 3y4 = 7 => 1 + 3y = 28 => 3y = 27 => y = 9. Thus B = (10, 9).

7. The gradient of line segment AB is -25. Line CD is perpendicular to AB. If CD passes through (1, 3), find its equation.

  1. 5x - 2y + 1 = 0
  2. 2x + 5y - 17 = 0
  3. 5x + 2y - 11 = 0
  4. 2x - 5y + 13 = 0
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Answer: A

m_CD = -1 / (-25) = 52. Equation: y - 3 = (52)(x - 1) => 2y - 6 = 5x - 5 => 5x - 2y + 1 = 0.

8. The points A(1, 3), B(4, k), and C(7, 11) are collinear. Find the value of k.

  1. 7
  2. 6
  3. 8
  4. 5
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Answer: A

Gradient AB = Gradient AC => k - 34 - 1 = 11 - 37 - 1 => k - 33 = 86 = 43 => k - 3 = 4 => k = 7.

9. Calculate the area of the triangle with vertices A(1, 2), B(5, 3), and C(3, 7).

  1. 9 unit²
  2. 18 unit²
  3. 10 unit²
  4. 8.5 unit²
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Answer: A

Area = 12 | (1)(3) + (5)(7) + (3)(2) - [ (2)(5) + (3)(3) + (7)(1) ] | = 12 | (3 + 35 + 6) - (10 + 9 + 7) | = 12 | 44 - 26 | = 12 (18) = 9 unit².

10. Find the coordinates of the midpoint of the line segment joining (-3, 8) and (5, -2).

  1. (1, 3)
  2. (2, 6)
  3. (1, 5)
  4. (-1, 3)
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Answer: A

M = (-3 + 52, 8 + (-2)2) = (22, 62) = (1, 3).

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