10 questions · Form 4 Additional Mathematics Bab 7: Coordinate Geometry
A point P(x, y) moves such that its distance from a fixed point A(2, -3) is always 5 units. Find the equation of the locus of P.
Prefer reading to quizzing? All 10 questions are listed below with the answer and explanation under each one.
1. A point P(x, y) moves such that its distance from a fixed point A(2, -3) is always 5 units. Find the equation of the locus of P.
Answer: A
PA² = 5² => (x - 2)² + (y - (-3))² = 25 => (x - 2)² + (y + 3)² = 25 => x² - 4x + 4 + y² + 6y + 9 = 25 => x² + y² - 4x + 6y - 12 = 0.
2. Find the coordinates of the midpoint of the line segment joining (-3, 8) and (5, -2).
Answer: A
M = (-3 + 52, 8 + (-2)2) = (22, 62) = (1, 3).
3. The points A(1, 3), B(4, k), and C(7, 11) are collinear. Find the value of k.
Answer: A
Gradient AB = Gradient AC => k - 34 - 1 = 11 - 37 - 1 => k - 33 = 86 = 43 => k - 3 = 4 => k = 7.
4. Line L passes through (2, 5) and (4, 9). Line N passes through (1, 1) and (3, k). If L is parallel to N, find k.
Answer: A
Gradient of L: m_L = 9 - 54 - 2 = 42 = 2. Since L ∥ N, m_N = 2 => k - 13 - 1 = 2 => k - 12 = 2 => k - 1 = 4 => k = 5.
5. Calculate the area of the triangle with vertices A(1, 2), B(5, 3), and C(3, 7).
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.
6. 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.
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).
7. If line L₁ has equation 2x - 3y + 6 = 0 and line L₂ is perpendicular to L₁, find the gradient of L₂.
Answer: A
Rearranging L₁: 3y = 2x + 6 => y = (23)x + 2. So gradient m₁ = 23. For L₂ ⊥ L₁: m₂ = -1m₁ = -1 / (23) = -32.
8. 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?
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.
9. Three vertices of a parallelogram ABCD are A(1, 1), B(4, 2), and C(5, 6). Find the coordinates of vertex D.
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).
10. 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).
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).