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Quiz Chapter 7: Coordinate Geometry

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

Question 1 of 10Score: 0

The point P(k, 5) divides the line segment joining A(-1, 2) and B(7, 8) internally in the ratio m : n. Find the ratio m : n.

Full Question List & Answer Key

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

1. The point P(k, 5) divides the line segment joining A(-1, 2) and B(7, 8) internally in the ratio m : n. Find the ratio m : n.

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

Using y-coordinate: y = n y₁ + m y₂m + n => 5 = 2n + 8mm + n => 5m + 5n = 8m + 2n => 3n = 3m => mn = 11 => 1 : 1.

2. Find the equation of the locus of a moving point P(x, y) such that its distance from A(0, 3) is twice its distance from B(3, 0).

  1. 3x² + 3y² - 24x + 6y + 27 = 0
  2. x² + y² - 8x + 2y + 9 = 0
  3. 3x² + 3y² + 24x - 6y - 27 = 0
  4. x² + y² - 6x + 6y + 18 = 0
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Answer: A

PA = 2 PB => PA² = 4 PB² => x² + (y - 3)² = 4[(x - 3)² + y²] => x² + y² - 6y + 9 = 4[x² - 6x + 9 + y²] => x² + y² - 6y + 9 = 4x² - 24x + 36 + 4y² => 3x² + 3y² - 24x + 6y + 27 = 0 (which simplifies to x² + y² - 8x + 2y + 9 = 0).

3. 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).

4. 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.

5. 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.

6. Find the y-intercept of the line that is perpendicular to 4x - 2y + 5 = 0 and passes through (2, 1).

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

4x - 2y + 5 = 0 => 2y = 4x + 5 => y = 2x + 2.5, so m₁ = 2. Perpendicular gradient m₂ = -12. Line: y - 1 = -12(x - 2) => y = -12 x + 1 + 1 => y = -12 x + 2. The y-intercept is 2.

7. 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.

8. Find the perpendicular distance from the origin (0,0) to the midpoint of the line joining A(2, 6) and B(6, 2).

  1. 4√2 units
  2. 4 units
  3. 2√2 units
  4. 8 units
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Answer: A

Midpoint M = (2+62, 6+22) = (4, 4). Distance OM = √(4² + 4²) = √32 = 4√2 units.

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. 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).

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