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Grade 10 · Analytical Geometry
Analytical Geometry: All the Exam Question Types
MARKING GUIDELINE
Marks
33
Duration
50 minutes
Questions
3
Name:
Class:
Date:
Mark
/ 33
Instructions and Information
- Answer ALL the questions in this question paper.
- Answer QUESTION 1 by circling the letter (A–D) in the answer grid at the end of that section.
- Show ALL calculations clearly.
- Show all units where applicable.
- Number the answers correctly according to the numbering system used in this question paper.
- A non-programmable calculator may be used, unless stated otherwise.
- Write neatly and legibly.
Question 1
[10 MARKS]Four options are given as possible answers to the following questions. Choose the correct answer and circle the letter (A–D) in the grid at the end of this section. If you want to change your choice, put a cross through the wrong letter and circle your new choice.
- 1.1The distance between A(1 ; 2) and B(4 ; 6) is …(1)A)5✓B)7C)25D)2,65Answer: A — d = √((4 − 1)² + (6 − 2)²) = √25 = 5.
- B — ADDED the two differences instead of squaring and rooting
- C — forgot to take the square root
- D — SUBTRACTED the squares instead of adding them
- 1.2The midpoint of P(−2 ; 3) and Q(6 ; −1) is …(1)A)(2 ; 1)✓B)(4 ; 2)C)(8 ; −4)D)(−4 ; 2)Answer: A — Average each coordinate: (−2 + 6) ÷ 2 = 2 and (3 + (−1)) ÷ 2 = 1.
- B — added the coordinates but never halved them
- C — SUBTRACTED the coordinates instead of averaging them
- D — subtracted the x-values but added the y-values before halving
- 1.3The gradient of the line through (2 ; 5) and (6 ; 13) is …(1)A)2✓B)12C)−2D)4Answer: A — m = (13 − 5) ÷ (6 − 2) = 8 ÷ 4 = 2.
- B — inverted the formula — it is the change in y OVER the change in x
- C — subtracted in opposite orders on the top and the bottom
- D — that is the change in x, not the gradient
- 1.4Two lines are PERPENDICULAR when …(1)A)m1 × m2 = −1✓B)m1 = m2C)m1 + m2 = 0D)m1 × m2 = 1Answer: A — Perpendicular gradients are negative reciprocals of each other.
- B — equal gradients means PARALLEL
- C — that makes them negatives of each other, not negative RECIPROCALS
- D — the product must be −1, not +1
- 1.5A line has gradient 3. A line perpendicular to it has gradient …(1)A)−13✓B)13C)−3D)3Answer: A — The perpendicular gradient is the negative reciprocal, −1/3.
- B — took the reciprocal but left off the minus
- C — changed the sign but never took the reciprocal
- D — that is the gradient of a PARALLEL line
- 1.6A(1 ; 2), B(3 ; 6) and C(5 ; 10) are collinear because …(1)A)mAB = mBC = 2✓B)AB = BCC)they form a triangleD)AC is the longest of the three lengthsAnswer: A — Collinear points lie on ONE straight line, so every pair of them has the same gradient.
- B — equal lengths do not put three points on one line
- C — collinear points do NOT form a triangle
- D — a longest side describes a triangle, not a straight line
- 1.7The line through (0 ; −3) with gradient 2 has equation …(1)A)y = 2x − 3✓B)y = 2x + 3C)y = −3x + 2D)y = 3x − 2Answer: A — y = mx + c with m = 2 and c = −3.
- B — the y-intercept is −3, not +3
- C — swapped the gradient and the y-intercept
- D — swapped them and changed the sign as well
- 1.8P(x ; 4) is 5 units from Q(1 ; 0). Then x = …(1)A)4 or −2✓B)4 onlyC)6 or −4D)3 or −3Answer: A — (x − 1)² + 4² = 5² gives (x − 1)² = 9, so x − 1 = ±3.
- B — a square root has TWO values, so there is a second answer
- C — solved (x − 1)² = 25 by forgetting to subtract the 16
- D — solved x² = 9 and never moved the −1 across
- 1.9The distance from the origin to (−6 ; 8) is …(1)A)10✓B)14C)2D)100Answer: A — √((−6)² + 8²) = √100 = 10.
- B — added the two coordinates' sizes instead of squaring and rooting
- C — added −6 and 8 as they stand
- D — forgot to take the square root
- 1.10Line AB has gradient 4. A line PARALLEL to AB has gradient …(1)A)4✓B)−14C)−4D)14Answer: A — Parallel lines have EQUAL gradients.
- B — that is the PERPENDICULAR gradient
- C — changing the sign does not make a line parallel
- D — taking the reciprocal does not make a line parallel
Answer grid — marking guideline
| 1.1 | A | B | C | D |
| 1.2 | A | B | C | D |
| 1.3 | A | B | C | D |
| 1.4 | A | B | C | D |
| 1.5 | A | B | C | D |
| 1.6 | A | B | C | D |
| 1.7 | A | B | C | D |
| 1.8 | A | B | C | D |
| 1.9 | A | B | C | D |
| 1.10 | A | B | C | D |
Question 2
[11 MARKS]A(2 ; 10) and B(6 ; −2) are two points on the Cartesian plane.
- 2.1Calculate the length of AB. Leave your answer in simplest surd form.(4)AB = (6 − 2)2 + (−2 − 10)2 (2)
= 160 (1) = 410 (1) - 2.2Calculate the coordinates of M, the midpoint of AB.(3)M = (2 + 62 ; 10 + (−2)2) (2) = (4 ; 4) (1)
- 2.3Calculate the gradient of AB, and hence write down the gradient of any line perpendicular to AB.(4)mAB = −2 − 106 − 2 = −3 (2)
Perpendicular: m = 13 (1) — the product must be −1 (reason 1)
Question 3
[12 MARKS]Answer the questions below.
- 3.1Determine the equation of the line through A(2 ; 10) and B(6 ; −2) in the form y = mx + c.(4)m = −3 (1)
10 = −3(2) + c (2)
c = 16, so y = −3x + 16 (1) - 3.2Determine whether the point C(4 ; 4) lies on the line AB. Show your working.(3)−3(4) + 16 = 4 (2)
The y-value matches, so C lies on AB (1) - 3.3Points P(−8 ; 0), Q(−5 ; −8) and R(x ; −14) are collinear. Calculate the value of x.(5)Collinear means mPQ = mQR (1)
mPQ = −8 − 0−5 + 8 = −83 (1)
−14 + 8x + 5 = −83 (1)
−18 = −8(x + 5) (1)
x = −2,75 (1)
TOTAL: 33 marks
This question paper consists of 3 questions.