Analytical Geometry: Lines & Inclination (Grade 11) — Worksheet & Memo

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Grade 11 · Analytical Geometry

Analytical Geometry: Lines & Inclination (Grade 11)

MARKING GUIDELINE

Marks

Duration

40 minutes

Questions

Name:

Class:

Date:

Mark

/ 25

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.

1.1
The gradient of a line with an angle of inclination of 45° is …
(1)
A)1✓
B)45
C)0
D)√3
Answer: A — m = tan θ, and tan 45° = 1.
- B — quoted the ANGLE instead of its tangent
- C — that is the gradient of a HORIZONTAL line
- D — that is tan 60°
1.2
A line has gradient √3. Its angle of inclination is …
(1)
A)60°✓
B)30°
C)45°
D)120°
Answer: A — tan θ = √3, so θ = 60°.
- B — tan 30° = 1/√3, the reciprocal
- C — tan 45° = 1
- D — tan 120° = −√3, which is negative
1.3
A line has gradient −1. Its angle of inclination is …
(1)
A)135°✓
B)45°
C)−45°
D)225°
Answer: A — An angle of inclination is taken between 0° and 180°, and tan 135° = −1.
- B — tan 45° = +1, not −1
- C — an angle of inclination is never quoted as negative
- D — 225° lies outside the 0° to 180° range
1.4
The line through (2 ; 3) with gradient 4 has equation …
(1)
A)y = 4x − 5✓
B)y = 4x + 5
C)y = 4x + 3
D)y = 2x + 3
Answer: A — y − 3 = 4(x − 2) gives y = 4x − 8 + 3 = 4x − 5.
- B — made a sign slip on the −8
- C — used the point's y-value as the y-intercept
- D — used the point's x-value as the gradient
1.5
The line through A(1 ; 2) and B(5 ; 10) has gradient …
(1)
A)2✓
B)12
C)4
D)8
Answer: A — m = (10 − 2) ÷ (5 − 1) = 8 ÷ 4 = 2.
- B — inverted the formula
- C — that is the change in x
- D — that is the change in y
1.6
A line perpendicular to y = 3x − 1 has gradient …
(1)
A)−13✓
B)13
C)−3
D)3
Answer: A — The perpendicular gradient is the negative reciprocal of 3.
- B — took the reciprocal but left off the minus
- C — changed the sign but never took the reciprocal
- D — that is the PARALLEL gradient
1.7
The midpoint of A(−1 ; 4) and B(5 ; −2) is …
(1)
A)(2 ; 1)✓
B)(4 ; 2)
C)(3 ; 3)
D)(6 ; −6)
Answer: A — ((−1 + 5) ÷ 2 ; (4 + (−2)) ÷ 2) = (2 ; 1).
- B — added the coordinates without halving them
- C — averaged the wrong pairs of values
- D — SUBTRACTED the coordinates instead of averaging them
1.8
Two lines are PARALLEL when their angles of inclination are …
(1)
A)equal✓
B)supplementary
C)complementary
D)differing by 90°
Answer: A — Equal inclinations give equal gradients.
- B — supplementary inclinations give gradients that are negatives of each other
- C — complementary inclinations give reciprocal gradients
- D — that describes PERPENDICULAR lines
1.9
The angle of inclination of the line 2y = 6x + 4 is …
(1)
A)71,57°✓
B)60°
C)18,43°
D)80,54°
Answer: A — Divide by 2 to get y = 3x + 2, so m = 3 and θ = tan⁻¹(3) = 71,57°.
- B — tan 60° = √3, not 3
- C — that is tan⁻¹(1/3), the PERPENDICULAR line's inclination
- D — used the gradient 6 without first dividing the equation by 2
1.10
The line joining (0 ; 0) and (3 ; 3) makes an angle with the x-axis of …
(1)
A)45°✓
B)30°
C)60°
D)90°
Answer: A — The gradient is 3 ÷ 3 = 1, and tan⁻¹(1) = 45°.
- B — tan 30° = 1/√3, not 1
- C — tan 60° = √3, not 1
- D — a gradient of 1 is not vertical

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

[15 MARKS]

In the diagram below, A(−2 ; 5), B(4 ; −3) and C(6 ; 1) are the vertices of △ABC drawn in the Cartesian plane.

2.1
Calculate the length of AB.
(3)
AB = (4 − (−2))² + (−3 − 5)² (1)
= 36 + 64 (1) = 10 units (1)
2.2
Calculate the gradient of AB, and hence the angle of inclination of AB, correct to TWO decimal places.
(5)
m_AB = −3 − 54 + 2 = −43 (2)
tan θ = −43 (1)
The gradient is negative, so θ is obtuse (1)
θ = 126,87° (1)
2.3
Determine the equation of the line through C parallel to AB.
(4)
Parallel lines have equal gradients: m = −43 (1)
1 = −43(6) + c (2)
c = 9, so y = −43x + 9 (1)
2.4
Determine whether AB ⊥ BC. Show ALL your working.
(3)
m_BC = 1 − (−3)6 − 4 = 2 (1)
m_AB × m_BC = −43 × 2 = −83 (1)
The product is not −1, so AB is NOT perpendicular to BC (1)

TOTAL: 25 marks

This question paper consists of 2 questions.