Perfect — let’s carefully reformat and transcribe Exercise 8.

1 (pages 1–6) while keeping

every calculation exact, writing each equality on a new line, and polishing grammar/clarity.

Exercise 8.1 — Geometry of Straight Lines

Q1.
Find the gradient (slope) of line whose inclination is:

(i) 0∘
Solution:
m=tan θ
∘
m=tan 0
m=0
So, the slope is 0.

(ii) 30∘
Solution:
m=tan θ
∘
m=tan 30
m ≈ 0.577

(iii) 60∘
Solution:
m=tan θ
∘
m=tan 60
m ≈1.732
(iv) 90∘
Solution:
m=tan θ
∘
m=tan 90 =∞
So, m is undefined.

(v) 120∘
Solution:
m=tan θ
∘
m=tan 120
m ≈− 1.732

(vi) 150∘
Solution:
m=tan θ
∘
m=tan 150
m ≈− 0.577

(vii) 170∘
Solution:
m=tan θ
∘
m=tan 170
m ≈− 0.176

(viii) 45 ∘ 5′
Solution:
m=tan θ
∘ ′
m=tan 45 5
 m ≈1.018

Q2.
Find inclination of the line whose slope is:

(i) m=0
tanθ=0
−1
θ=tan (0)
∘
θ=0
So, inclination = 0∘.

(ii) m=0.577
tanθ=0.577
−1
θ=tan (0.577)
∘
θ ≈ 30

(iii) m=− 1.732

tanθ=−1.732
−1
θ=tan (−1.732)
∘ ∘
θ ≈ −60 or 120
Since inclination is measured between 0∘ and 180∘:
∘
θ=120

(iv) m=− 0.364

tanθ=− 0.364
−1
θ=tan (−0.364 )
∘
θ ≈ −20
Add 180∘:
∘
θ=160
But in the image they approximated:
∘
θ ≈ 20
So, inclination = 20∘.

Q3.
Find gradient and inclination of lines joining:

(i) A(2, 6), B(5 ,8)

y2 − y1
m=
x2− x1
8 −6
m=
5 −2
2
m=
3
Inclination:
−1
θ=tan (m)

θ=tan− 1 ( 32 )
∘
θ ≈ 33.69

(ii) C (−2 , 4), D(1 ,− 3)

y2 − y1
m=
x2− x1
−3 − 4
m=
1−(− 2)
−7
m=
3
Inclination:
 θ=tan − 1 ( −37 )
∘
θ ≈ −66.8
Add 180∘:
∘
θ=113.2
So, gradient = −7 /3, inclination = 113.2∘ .

(iii) E(5 ,− 2), F (− 2 ,− 3)

−3 −(−2)
m=
−2 −5
−1
m=
−7
1
m=
7
Inclination:

θ=tan
−1
( 71 )
∘
θ ≈ 8.13

Q4.
If A(− 2 ,6) and B(7 , −3), find the slope of line:

(i) Parallel to AB
− 3− 6
m A B=
7 −(−2)
−9
m A B=
9
m A B=−1

So, parallel slope = -1.

(ii) Perpendicular to AB
1
m ⟂ =−
m
1
m⟂ =−
−1
m⟂ =1

So, perpendicular slope = 1.

Q5.
Find x if the slope of line passing through A(3 , x ) and B(5 , 8) is 4.
Solution:
y2 − y1
m=
x2− x1
8−x
4=
5−3
8−x
4=
2
8=8 − x
0=− x
x=0

Q6.
Find k if lines passing through A(k ,2), B(3 ,5) and C (5 ,− 1), D(8 , 7) are parallel.
Solution:
Slope of AB:
5 −2
m A B=
3 −k
3
m A B=
3 −k
Slope of CD:
 7 −(− 1)
mC D =
8−5
8
mC D =
3
For parallel lines:
m A B=mC D

3 8
=
3− k 3
Cross multiply:
3 ×3=8(3 − k )
9=24 − 8 k
9 −24=− 8 k
−15=−8 k
15
k=
8

Q7.
Find k if lines passing through P(− 1, 2),Q(4 , 7) and R(2 , k ), S (7 , 10) are perpendicular.
Solution:
Slope of PQ:
7 −2
mP Q =
4 −(− 1)
5
mP Q =
5
mP Q =1

Slope of RS:
10 −k
m R S=
7−2
10 −k
m R S=
5
For perpendicularity:
 mP Q × mR S=−1

10− k
1× =−1
5
10− k
=− 1
5
Multiply through:
10 − k=−5
− k=− 5− 10
− k=− 15
k =15

✅ Reformatted solutions for Q1–Q7 completed (pages 1–6).

Would you like me to continue transcribing pages 7–12 (Q8 onward) in the same cleaned
format?

8 Related Questions You Might Ask Next

1. Can you finish Exercise 8.1 (pages 7–12)?
2. Can you create a summary sheet of slope & inclination rules?
3. Can you provide graphs for some examples?
4. Can you make a LaTeX version for neat typesetting?
5. Can you create new practice problems (parallel & perpendicular lines)?
6. Can you explain why perpendicular slopes multiply to -1?
7. Can you prepare a revision test from this unit?
8. Can you cross-check numerical accuracy (esp. decimals)?
💡 Tip: Always remember: slope of a line at inclination θ is m=tan θ. For negative values, check
if you need to add 180∘ to keep inclination between 0∘ and 180∘.

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Shall I proceed with pages 7–12 (remaining of Exercise 8.1)?