Maths ch7 coordinate geometry
37 marks. Read the notes, then tap Start test. Answers and your score appear only after you submit.
Quick notes
Distance formula: the distance between P(x₁, y₁) and Q(x₂, y₂) is √[(x₂ − x₁)² + (y₂ − y₁)²].
The distance of P(x, y) from the origin is √(x² + y²).
Section formula: the point dividing the join of (x₁, y₁) and (x₂, y₂) internally in the ratio m₁ : m₂ is
((m₁x₂ + m₂x₁)/(m₁ + m₂), (m₁y₂ + m₂y₁)/(m₁ + m₂)).
Midpoint: ((x₁ + x₂)/2, (y₁ + y₂)/2).
A point on the x-axis is (x, 0); a point on the y-axis is (0, y).
Part A: Multiple choice 16 questions
1. The distance between (2, 3) and (4, 1) is
- 2
- 2√2
- 4
- √2
√(2² + 2²) = √8 = 2√2.
2. The distance of (−6, 8) from the origin is
- 10
- 14
- 2
- 100
√(36 + 64) = 10.
3. The midpoint of (2, 3) and (4, 7) is
- (6, 10)
- (1, 2)
- (3, 5)
- (2, 4)
((2 + 4)/2, (3 + 7)/2) = (3, 5).
4. Every point on the x-axis has the form
- (0, y)
- (x, x)
- (y, 0)
- (x, 0)
The y-coordinate is 0.
5. The distance between (0, 0) and (36, 15) is
- 39
- 51
- 21
- 45
√(1296 + 225) = √1521 = 39.
6. The point dividing the join of (−1, 7) and (4, −3) in the ratio 2 : 3 is
- (2, 1)
- (1, 3)
- (3, 1)
- (0, 5)
x = (8 − 3)/5 = 1, y = (−6 + 21)/5 = 3.
7. The distance between (a, b) and (−a, −b) is
- 2√(a² + b²)
- √(a² + b²)
- 2(a + b)
- 0
√[(2a)² + (2b)²] = 2√(a² + b²).
8. If the distance between (2, −3) and (10, y) is 10, then y is
- 3 only
- −9 only
- 3 or −9
- 9 or −3
64 + (y + 3)² = 100, so y + 3 = ±6.
9. The points (1, 5), (2, 3) and (−2, −11) are
- collinear
- vertices of a right triangle
- vertices of an equilateral triangle
- not collinear
√5 + √212 ≠ √265, so the three points do not lie on one line.
10. The y-axis divides the join of (5, −6) and (−1, −4) in the ratio
- 1 : 5
- 5 : 1
- 1 : 1
- 2 : 3
(−k + 5)/(k + 1) = 0 gives k = 5, so 5 : 1.
11. The midpoint of (−3, 10) and (6, −8) is
- (1.5, 1)
- (3, 2)
- (4.5, 9)
- (1, 1.5)
((−3 + 6)/2, (10 − 8)/2) = (1.5, 1).
12. (1, 2), (4, y), (x, 6), (3, 5) are vertices of a parallelogram in order. Then (x, y) is
- (3, 6)
- (6, 3)
- (5, 4)
- (4, 5)
Diagonals bisect each other: (1 + x)/2 = 7/2 and (2 + 6)/2 = (y + 5)/2.
13. The abscissa of a point is its
- distance from origin
- y-coordinate
- x-coordinate
- quadrant
Abscissa = x-coordinate, ordinate = y-coordinate.
14. The point on the x-axis equidistant from (2, −5) and (−2, 9) is
- (7, 0)
- (0, −7)
- (−2, 0)
- (−7, 0)
(x − 2)² + 25 = (x + 2)² + 81 gives x = −7.
15. The distance between (−5, 7) and (−1, 3) is
- 4√2
- 8
- 4
- 2√2
√(16 + 16) = 4√2.
16. The point (−3, −4) lies in quadrant
- I
- II
- III
- IV
Both coordinates are negative.
Part B: Fill in the blanks 8 blanks
Part C: Questions and answers 9 questions
23. Find the point on the x-axis that is equidistant from (2, −5) and (−2, 9).
Point: (−7, 0).
24. Show that (−1, −2), (1, 0), (−1, 2) and (−3, 0) are the vertices of a square.
Diagonals: between (−1, −2) and (−1, 2) is 4; between (1, 0) and (−3, 0) is 4.
Equal sides and equal diagonals, so it is a square.
25. Find the points of trisection of the line segment joining (4, −1) and (−2, −3).
Ratio 2 : 1: ((−4 + 4)/3, (−6 − 1)/3) = (0, −7/3).
26. In what ratio does (−4, 6) divide the join of (−6, 10) and (3, −8)?
Ratio = 2 : 7. (Check y: (−16 + 70)/9 = 6 ✓)
27. Find x if the distance between (x, −1) and (3, 2) is 5.
x = 7 or x = −1.
28. Find a relation between x and y such that (x, y) is equidistant from (3, 6) and (−3, 4).
3x + y − 5 = 0.
29. Show that (1, 7), (4, 2), (−1, −1) and (−4, 4) are the vertices of a square.
Diagonals: √(4 + 64) = √68 and √(64 + 4) = √68.
Equal sides and equal diagonals → square.
30. Find the coordinates of the point dividing the join of (−1, 7) and (4, −3) in the ratio 2 : 3.
Point: (1, 3).
31. Find the distance between (−5, 7) and (−1, 3).