Chapter 20 Exercise 20 - Biology Leaving Cert.shevlinbiology.webs.com/chapter 20 solutions.pdf ·...

Active Maths 2 (Strands 1–5): Ch 20 Solutions

Chapter 20 Exercise 20.1

1

Q. 1. (i)

x

y

A = (2,2)

B = (–4,6)

–6

–4

–2

2

4

6

–2 2 4 6 8 10 12 1400

–4–6

(ii) C ( 2 − 4 ______ 2 , 2 + 6 ______ 2 ) = (−1,4)

(iii) |AC| = √___________________

(−1 − 2)2 + (4 − 2)2

= √_____________

(−3)2 + (−2)2

= √______

9 + 4

= √___

13

|CB| = √___________________

(−1 + 4)2 + (4 − 6)2

= √___________

(3)2 + (−2)2

= √______

9 + 4

= √___

13

∴ |AC| = |CB|

(iv) Slope of AB = 6 − 2 _______ −4 − 2 = 4 ___ −6 = –2 ___ 3

y − 2 = – 2 __ 3 (x − 2)

3y − 6 = –2(x − 2) 3y – 6 = –2x + 4 2x + 3y – 10 = 0

(v) AB cuts the y-axis at x = 0. 0 + 3y – 10 = 0 3y = 10

y = 10 ___ 3

|AB| cuts the y-axis at ( 0, 10 ___ 3 ) .

(vi) AB cuts the x-axis at y = 0

2x + 3(0) – 10 = 0

2x = 10

x = 5

AB cuts the x-axis at (5,0).

Q. 2. (i)

x

y

A = (2,4)

C = (–2,0) B = (6,0)

–6

–4

–2

2

4

6

–2 2 4 6 8 10 12 1400

–4–6

(ii) M ( 6 − 2 ______ 2 , 0 + 0 ______ 2 ) = M(2,0)

(iii) AB: slope = 0 − 4 ______ 6 − 2 = −4 ___ 4 = −1

(iv) y − y1 = m(x − x1)

y − 4 = −1(x − 2)

y − 4 = −x + 2

x + y − 6 = 0

(v) x + y = 6 x – y = –2

2x = 4

x = 2

2 + y = 6

y = 4

(2,4)

(vi) h = 4 base = 8

Area = 1 __ 2 × b × ⊥h

= 1 __ 2 × 8 × 4

= 16 units2x

y

–4–3

–2–1

123456

0–1 1 2 3 4 5 6 7 8 9 10 11 120–2–3–4–5–6–7

C = (–2,0)

A = (2,4)

B = (6,0)

2 Active Maths 2 (Strands 1–5): Ch 20 Solutions

Q. 3. (i) m: x + 4y − 8 = 0

4y = −x + 8

y = −1 ___ 4 x + 2

Slope = −1 ___ 4 , C = 2

x

y

–5–4–3–2–1

123456

00–1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17–2–3–4

n

m

n: x − y – 3 = 0

−y = − x + 3

y = x − 3

Slope = 1, C = −3

(ii)

x

y

–5–4–3–2–1

123456

00–1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17–2–3–4

n

m(4,1)

(4,1)

(iii) x + 4y – 8 = 0 – x –+ y –+ 3 = 0

5y − 5 = 0

5y = 5

y = 1

x − 1 − 3 = 0

x − 4 = 0

x = 4

(4,1)

(iv) At y-axis, x = 0

0 + 4y − 8 = 0

4y = 8

y = 2

(0,2)

3Active Maths 2 (Strands 1–5): Ch 20 Solutions

Q. 4. (i) 2x − y + 3 = 0

−y = −2x − 3

y = 2x + 3

Slope = 2, C = 3

5x + y − 10 = 0

y = −5x + 10

Slope = −5, C = 10

x0

0–1–1

123456789

10

1 2 3 4 5 6 7 8 9 10 11 12–2–3–4–5–6–7

s ry

(ii)

x0

0–1–1

123456789

10

1 2 3 4 5 6 7 8 9 10 11 12–2–3–4–5–6–7

s ry

(1,5)

(1,5)

(iii) 5x + y – 10 = 0 2x – y + 3 = 0

7x − 7 = 0

7x = 7

x = 1

5(1) + y − 10 = 0

y − 5 = 0

y = 5

(1,5)

(iv) At x-axis, y = 0

2x − 0 + 3 = 0

2x = − 3

x = −3 ___ 2

( −3 ___ 2 ,0 )


Q. 5. (i) 3x + 2y = 18

2y = − 3x + 18

y = −3 ___ 2 x + 9

Slope = −3 ___ 2 , c = 9

y = 3x

Slope = 3, c = 0

–6 –4 –2

2

46

810

0 2 4 6 8 10 12 14 16 18x

yp q

(ii)

–6 –4 –2

2

46

810

0 2 4 6 8 10 12 14 16 18x

yp q

(2,6)

(2,6)

(iii) 3x + 2y = 18 3x + 2(3x) = 18 3x + 6x = 18 9x = 18 x = 2

y = 3x

y = 3(2) y = 6

(2,6)

Q. 6. (i) d = √_________________

(5 + 3)2 + (4 + 2)2

= √_______

82 + 62

= √____

100

= 10 units

(ii) 10 × 1 ___ 10 = 1 km

(iii) T = D __ S = 1 ___ 2.5 = 2 __ 5 hr = 24 minutes

Q. 7. (i) h: 3x + y = 5

(1,2) ⇒ 3(1) + 2 = 5?

3 + 2 = 5?

5 = 5

∴ (1,2) ∈ h

(ii) (k,0) ⇒ 3(k) + 0 = 5

3k = 5

k = 5 __ 3

(iii) Let y = 3

⇒ 3x + 3 = 5

3x = 2

x = 2 __ 3

( 2 __ 3 ,3 ) Let y = −1

⇒ 3x − 1 = 5

3x = 6

x = 2

(2,−1)

Let y = 5

⇒ 3x + 5 = 5

3x = 0

x = 0

(0,5)

(iv) Slope = 5 + 1 ______ 0 − 2 = 6 ___ −2 = −3

(v) 3x + y = 5

y = −3x + 5

m = −3

Q. 8. (i) y = 4x − 3

(ii) y = −3x + 5

Q. 9. (2,4) → (3,−1) → (4,−6)

a = 4

b = −6

Q. 10. (i) m = 2 − 6 _______ 12 − 4 = −4 ___ 8 = −1 ___ 2

y − y1 = m(x − x1)

y − 6 = −1 ___ 2 (x − 4)

y − 6 = −1 ___ 2 x + 2

2y − 12 = − x + 4

l1: x + 2y = 16


(ii) l2: y = −4x + 0

4x + y = 0

(iii) x + 2y = 16

x + 2(−4x) = 16

x − 8x = 16

−7x = 16

x = −16 ____ 7

y = −4 ( −16 ____ 7 ) y = 64 ___ 7

( −16 ____ 7 , 64 ___ 7 ) Midpoint ( 4 − 16 __ 7

______ 2 , 6 + 64 __ 7

______ 2 ) = ( 12 ___ 14 , 108 ____ 14 ) = ( 6 __ 7 , 54 ___ 7 ) Q. 11. (a) A(1,1)

B(11,1)

(b)

20181614121086420–2

2

46

8

–2

0–4

x

y D = (6,8)

B = (11,1)A = (1,1)

(c)

20181614121086420–2

2

46

8

–2

0–4

x

y D = (6,8)

B = (11,1)A = (1,1) C = (6,1)

(6,1)

(d)

20181614121086420–2

2

46

8

–2

0–4

x

y D = (6,8)

B = (11,1)A = (1,1) C = (6,1)

(6,1)

(e) |AD| = √_________________

(6 − 1)2 + (8 − 1)2

= √_________

(5)2 + (7)2

= √________

25 + 49

= √___

74

|BD| = √__________________

(6 − 11)2 + (8 − 1)2

= √___________

(−5)2 + (7)2

= √________

25 + 49

= √___

74

(f) ABD is an isosceles Δ as two sides are equal in measure.

(g) |AD| = |BD| shown in (e)

|AC| = |BC| as C is the midpoint

CD| = |CD| as it is the common side

∴ Δ ACD ≡ Δ BCD by SSS.

Q. 12. (i)

D = (4,8)

A = (2,3)

B = (10,4)

x

y

C = (12,9)


(ii) |AD| = √_________________

(4 − 2)2 + (8 − 3)2

= √_________

(2)2 + (5)2

= √_______

4 + 25

= √___

29

|BC| = √___________________

(12 − 10)2 + (9 − 4)2

= √_________

(2)2 + (5)2

= √_______

4 + 25

= √___

29

∴ |AD| = |BC|

(iii) Midpoint of [AC]:

E ( 2 + 12 _______ 2 , 3 + 9 ______ 2 ) = ( 14 ___ 2 , 12 ___ 2 ) = (7,6)

Midpoint of [BD]

F ( 10 + 4 _______ 2 , 4 + 8 ______ 2 ) = ( 14 ___ 2 , 12 ___ 2 ) = (7,6)

E = F

∴ [AC] and [BD] bisect each other.

(iv) Yes, I can now conclude that ABCD is a parallelogram as opposite sides are equal in length and the diagonals bisect each other.

Exercise 20.2

Q. 1. Slope Slope of parallel line

Slope of perpendicular

line– 1 _ 2 – 1 _ 2 2

3 _ 4 3 _ 4 – 4 _ 3

– 1 _ 3 – 1 _ 3 3

– 8 __ 11 – 8 __ 11 11 __ 8

13 13 – 1 __ 13

Q. 2. 5 __ 4 Q. 3. 4 __ 3 Q. 4. − 1 __ 2

Q. 5. −4 Q. 6. 1 __ 3

Q. 7. (i) 2x + y − 8 = 0

y = −2x + 8

m = −2

(ii) 3x − y + 6 = 0

− y = −3x − 6

y = 3x + 6

m = 3

(iii) 2x − 2y = 9

−2y = −2x + 9

y = x − 9 __ 2

m = 1

(iv) 3(x − y) = 4

x − y = 4 __ 3

−y = −x + 4 __ 3

y = x − 4 __ 3

m = 1

(v) 2x − 3(y − 2) = 0

2x − 3y + 6 = 0

−3y = −2x − 6

y = 2 __ 3 x + 2

m = 2 __ 3

(vi) 4x = 6 − 5y

5y = −4x + 6

y = −4 ___ 5 x + 6 __ 5

m = − 4 __ 5

Q. 8. 3x + y = 6

y = − 3x + 6 m = −3 c = 6 Line ⊥ 3x + y = 6 has m = 1 __ 3

∴ Equation is y = 1 __ 3 x + 6

3y = x + 18

Q. 9. 2x − y = 3 −y = −2x + 3 y = 2x −3 m = 2, c = −3 Line ⊥ to 2x − y = 3 has slope − 1 __ 2

∴ y = −1 ___ 2 x − 3

2y = −x −6

x + 2y + 6 = 0


Q. 10. x + 2y = 4 2y = −x + 4

y = −1 ___ 2 x + 2

m = −1 ___ 2 , c = 2

Slope of line ⊥ to x + 2y = 4 is 2

∴ y = 2x + 2

Q. 11. x − 3y + 6 = 0

−3y = −x −6

y = 1 __ 3 x + 2

m = 1 __ 3 , c = 2

Line || to x − 3y + 6 = 0 has slope 1 __ 3

y − y1 = m(x − x1)

y − 4 = 1 __ 3 (x − 2)

3y − 12 = x − 2

x − 3y + 10 = 0

Q. 12. 2x + 3y + 6 = 0

3y = −2x − 6

y = −2 ___ 3 x − 2

m = −2 ___ 3 , c = −2

Line ⊥ to 2x + 3y + 6 = 0 has slope 3 __ 2

y − y1 = m(x − x1)

y + 6 = 3 __ 2 (x − 1)

2y + 12 = 3x − 3

3x − 2y − 15 = 0

Q. 13. (i) a: 2x + 3y − 8 = 0

3y = −2x + 8

y = −2 ___ 3 x + 8 __ 3

b: 2x − 3y + 6 = 0

−3y = −2x − 6

y = 2 __ 3 x + 2

m1m2 = ( −2 ___ 3 ) ( 2 __ 3 ) ≠ 1

∴ a ⊥ b

(ii) a: −2x + 3y − 8 = 0

3y = 2x + 8

y = 2 __ 3 x + 8 __ 3

b: 2x − 3y + 14 = 0

−3y = −2x − 14

y = 2 __ 3 x + 14 ___ 3

m1 = m2

∴ a ⊥ b

(iii) a: 2x − 6y – 8 = 0

− 6y = −2x + 8

y = 1 __ 3 x − 8 __ 6

b: x − 3y + 1 = 0

−3y = −x −1

y = 1 __ 3 x + 1 __ 3

m1 = m2

∴ a ⊥ b

(iv) a: 2x − 4y − 3 = 0

−4y = −2x + 3

y = 1 __ 2 x − 3 __ 4

b: y = −2x + 10

m1m2 = ( 1 __ 2 ) (−2) = −1

∴ a ⊥ b

(v) a: −2x + 3y − 3 = 0

3y = 2x + 3

y = 2 __ 3 x + 1

b: 3y = −2x + 10

y = −2 ___ 3 x + 10 ___ 3

m1m2 = ( 2 __ 3 ) ( −2 ___ 3 ) ≠ −1

∴ a ⊥ b

(vi) a: y = 2x − 7

b: x +2y = 3

2y = −x + 3

y = −1 ___ 2 x + 3 __ 2

m1m2 = (2) ( −1 ___ 2 ) = −1

∴ a ⊥ b

Q. 14. (i) (1,2)

y = 3 is horizontal,

∴ Line ⊥ to y = 3 must be vertical.

Vertical line going through (1,2) has equation x = 1.


(ii) It is not possible to verify as slope of y = 3 is 0 and slope of x = 1 is undefined.

Q. 15. l: 5x − y + 4 = 0

−y = −5x − 4

y = 5x + 4

m = 5

Slope of k = 5

y − y1 = m(x − x1)

y + 3 = 5(x − 4)

y + 3 = 5x − 20

5x − y − 23 = 0

Q. 16. m: 3x − 6y + 11 = 0

−6y = − 3x − 11

y = 1 __ 2 x + 11 ___ 6

n has slope −2

y − y1 = m(x − x1)

y − 2 = −2(x + 4)

y − 2 = −2x − 8

2x + y + 6 = 0

Q. 17. p: 7x − y + 14 = 0

−y = −7x − 14

y = 7x + 14

q has slope 7

y − y1 = m(x − x1)

y −3 = 7(x + 1)

y − 3 = 7x + 7

7x − y + 10 = 0

Q. 18. l: x − y + 4 = 0

−y = − x − 4

y = x + 4

k has slope −1

y − y1 = m(x − x1)

y − 2 = −1(x − 2)

y − 2 = −x + 2

x + y − 4 = 0

Revision Exercises

Q. 1. (i) m = rise ___ run = 6 __ 5

(ii) C ( 1 + 6 ______ 2 , − 2 + 4 ________ 2 ) = ( 3 1 _ 2 ,1 ) (iii) |AB| = √

___________________ (1 − 6)2 + (− 2 − 4)2

= √______________

(− 5)2 + (− 6)2

= √________

25 + 36

= √___

61

(iv) Slope of k = −5 ___ 6

C ( 3 1 _ 2 ,1 ) y − y1 = m(x − x1)

y − 1 = −5 ___ 6 ( x − 7 __ 2 ) 6y − 6 = − 5x + 35 ___ 2

12y − 12 = –10x + 35

k: 10x + 12y − 47 = 0

(v) At x-axis, y = 0 10x − 47 = 0 10x = 47 x = 4.7 (4.7,0)

At y-axis, x = 0 12y − 47 = 0

12y = 47

y = 47 ___ 12 ( 0, 47 ___ 12 ) Q. 2. (i) l: 3x − 4y + 7 = 0

P(−1,h): 3(−1) − 4(h) + 7 = 0

−3 −4h + 7 = 0

− 4h + 4 = 0

− 4h = − 4

h = 1

m: 4x + 3y − 24 = 0

Q(k,0): 4(k) + 3(0) − 24 = 0

4k − 24 = 0

4k = 24

k = 6


(ii) 3x − 4y + 7 = 0 × 3 4x + 3y − 24 = 0 × 4

9x − 12y + 21 = 0 16x + 12y − 96 = 0

25x − 75 = 0

25x = 75

x = 3

3x − 4y + 7 = 0 3(3) − 4y + 7 = 0 9 − 4y + 7 = 0 −4y = −16 y = 4

(3,4)

(iii)

x

m

l

R = (3,4)

P = (–1,1) Q = (6,0)–6 –4 –2

–2

2

468

–4

00

2 4 6 8 10 12 14 16 18

(iv) ∠PRQ is a right angle

⇒ PR ⊥ RQ

⇒ l ⊥ m

l: 3x − 4y + 7 = 0

−4y = −3x − 7

y = 3 __ 4 x + 7 __ 4

m: 4x + 3y − 24 = 0

3y = −4x + 24

y = −4 ___ 3 x + 8

m1m2 = ( 3 __ 4 ) ( −4 ___ 3 ) = −1

∴ l ⊥ m

∴ ∠PRQ is a right angle.

Q. 3. (i) P(−1,2) R(3,4)

M ( −1 + 3 _______ 2 , 2 + 4 ______ 2 ) = M ( 2 __ 2 , 6 __ 2 ) = M(1,3)

(ii) m = 4 − 2 ________ 3 − (−1) = 2 __ 4 = 1 __ 2

(iii) Line ⊥ to [PR] has slope −2 y − y1 = m(x − x1) y − 3 = − 2(x − 1) y − 3 = −2x + 2 2x + y − 5 = 0 (iv) k: x − 2y = 0 l: 2x + y = 5 × 2 x − 2y = 0 4x + 2y = 10 5x = 10 x = 2

x − 2y = 0 2 − 2y = 0 − 2y = − 2 y = 1 N(2,1)

Q. 4. l: 2x + 3y = − 1

(i) 3y = −2x −1

y = −2 ___ 3 x − 1 __ 3

Slope of k = 3 __ 2

y − y1 = m(x − x1)

y − 2 = 3 __ 2 (x − 3)

2y − 4 = 3x − 9

k: 3x − 2y −5 = 0

(ii) k and l are ⊥

2x + 3y = −1 ×2

3x − 2y = 5 ×3

4x + 6y = −2 9x − 6y = 15

13x = 13 x = 1

2x + 3y = −1 2(1) + 3y = −1 3y = −3 y = −1

Point of intersection of k and l is (1,−1)

P (3,2) → (1,−1) → P‘ (−1,−4)


Q. 5. (i) AD: m1 = −3 − (−1)

__________ 1 − 2 = −3 + 1 _______ −1

= −2 ___ −1 = 2

BC: m2 = 1 − (−2)

________ −2 − 4 = 1 + 2 ______ −6

= 3 ___ −6 = −1 ___ 2

m1m2 = 2 ( −1 ___ 2 ) = −1

∴ AD ⊥ BC

(ii) |BC| = √_____________________

(−2 − 4)2 + (1 − (−2))2

= √___________

(−6)2 + (3)2

= √_______

36 + 9

= √___

45

(iii) Area of ΔABC = 1 __ 2 × b × ⊥h

= 1 __ 2 × √___

45 × √__

5

= 1 __ 2 √___

90

= √__

9 √___

10 _______ 2 = 3 __ 2 √___

10 units2

Q. 6. (i) l: 3x −5y + 15 = 0

At x-axis, y = 0

3x + 15 = 0

3x = −15

x = −5

(−5,0) = C

m: 3x + 4y − 12 = 0

3x −12 = 0

3x = 12

x = 4

D = (4,0)

(ii) 3x − 5y + 15 = 0 – 3x

+– 4y –+ 12 = 0

−9y + 27 = 0

−9y = −27

y = 3

3x − 5(3) + 15 = 0

3x − 15 + 15 = 0

3x = 0

x = 0

E = (0,3)

(iii)

x

y

–210–8 –6 –4 –2–2

246

–4

00

2 4 6 8 10 12 14

E = (0,3)C = (–5,0) D = (4,0)

(iv) Area ΔCDE = 1 __ 2 × b × ⊥h

= 1 __ 2 × 9 × 3

= 27 ___ 2

= 13.5 units2

Q. 7. (i) R ( 2 − 2 ______ 2 , −3 + 1 _______ 2 ) = R ( 0 __ 2 , −2 ___ 2 ) = R(0,−1)

(ii) Slope PQ = 1 − (−3)

________ −2 − 2 = 1 + 3 ______ −4

= 4 ___ −4 = −1

∴ Slope of k = 1

k: y − y1 = m(x − x1)

y + 1 = 1 (x − 0)

y + 1 = x

k: y = x − 1

(iii) k: y = x − 1

S(3,2) ⇒ 2 = 3 − 1?

2 = 2

∴ S ∈ k

(iv) If ΔPQS is isosceles then |PS| = |QS|

|PS| = √___________________

(2 − 3)2 + (− 3 −2)2

= √_____________

(−1)2 + (−5)2

= √_______

1 + 25

= √___

26

|QS| = √__________________

(−2 −3)2 + (1 − 2)2

= √_____________

(−5)2 + (−1)2

= √_______

25 + 1

= √___

26

∴ ΔPQS is isosceles


Q. 8. (a) Slope of l1 = rise ___ run = 275 ____ 25 = 11 ___ 1 = 11

Slope of l2 = rise ___ run = 340 ____ 40 = 17 ___ 2 = 8.5

(b) At a steady speed of 60 km/hr John’s car can travel 11 km per litre of fuel consumed and at a steady speed of 100 km/hr John’s car travels 8.5 km per litre of fuel consumed.

(c) At 60 km/hr: 200 ____ 11 × 149.9

= 2,725 5 __ 11 c

= €27.25

At 100 km/hr: 200 ____ 8.5 × 149.9

= 3,527 1 __ 17 c

= €35.27

€35.27 − €27.25 = €8.02 cheaper

Q. 9. (a) Line 4 has greatest slope as slope = 5

(b) Lines 1 and 2 are parallel as they both have slope 3.

(c)

x

y

–4 –3 –2 –1–1

123

–2–3–4–5–6–7–8

0 1 2 3 4 5 6 7 8 9 10 11

(d) It represents line 5.

(e) Line 6

(f) y = x − 7 – y = – 4x –+ 16

0 = −3x + 9

−9 = −3x

3 = x

y = 3 − 7

y = −4

(3,−4)

(g) Line 4: If x = 3, y = 3 − 7 = –4

Line 6: If x = 3 , y = 4(3) −16

= 12 −16

= −4


Q. 10. (a) A(1,4)

(b)

x

D = (0,4)A = (1,4)

B = (2,0)E = (–6,0)

F = (4,–4)C = (–4,4)

–6–8–10 –4 –2–2

2

46

–4

00

2 4 6 8 10 12 14

(c) Midpoint [DF]

= ( 0 + 4 ______ 2 , 4 − 4 ______ 2 ) = ( 4 __ 2 , 0 __ 2 ) = (2,0)

(d) Slope BF = rise ___ run = −2 ___ 1 = −2

y intercept BF = D = 4

∴ BF: y = −2x + 4

(e) Slope CE = rise ___ run = −2 ___ 1 = −2

(f) y − y1 = m(x − x1)

y − 0 = −2(x + 6)

y = −2x − 12

2x + y + 12 = 0

(g) Area ΔBCE : Area ΔBCF

1 : 1

(h) Yes, ΔBCE and ΔBCF are congruent because they have equal area, |BE| = |CF| and |EC| = |BF|

Q. 11. (i) k: 4x−3y + 12 = 0

At x-axis, y = 0

4x + 12 = 0

4x = −12

x = −3

A (−3,0)

At y-axis, x = 0

−3y + 12 = 0

−3y = −12

y = 4

B (0,4)

(ii)

–10 –8 –6 –4 –2–2

246

m y

x

–4

00

2 4 6 8 10 12 14

B = (0,4)

A = (–3,0)

k

(iii) k: 4x − 3y + 12 = 0

−3y = −4x − 12

y = 4 __ 3 x + 4

Slope of m = −3 ___ 4

m: y − y1 = m(x − x1)

y − 4 = −3 ___ 4 (x − 0)

4y − 16 = −3x

m: 3x + 4y − 16 = 0

Q. 12. (i) x +2y −10 = 0 – x –+ y

+– 2 = 0

3y − 12 = 0 3y = 12 y = 4 x − y + 2 = 0 x − 4 + 2 = 0 x − 2 = 0 x = 2 R (2,4)

(ii) k: x − y + 2 = 0 Q (3,5): 3 − 5 + 2 = 0? 0 = 0 ∴ Q ∈ k

(iii) |QR| = √_________________

(3 − 2)2 + (5 − 4)2

= √_________

(1)2 + (1)2

= √__

2

Q. 13. (i) x + y − 8 = 0 – x –+ 2y

+– 10 = 0

3y − 18 = 0 3y = 18 y = 6 x + y −8 = 0 x + 6 − 8 = 0 x − 2 = 0 x = 2 R (2,6)


(ii) a: x + y − 8 = 0 At x-axis, y = 0

x − 8 = 0

x = 8 Q (8,0)

(iii) b: x −2y + 10 = 0 At y-axis, x = 0

−2y + 10 = 0

−2y = −10

y = 5 P (0,5)

(iv)

–11–10–9 –8 –7 –6 –5 –4 –3 –2 –1

–2–1

10

23456789

0 1 2 3 4 5 6 7 8 9 10 11 12

R = (2,6)

Q = (8,0)

P = (0,5)

b

a

y

x

(v) T ( 0 + 8 ______ 2 , 5 + 0 ______ 2 ) = T ( 4,2 1 _ 2 ) (vi) Slope PT = slope PQ = −5 ___ 8

y-intercept = 5

∴ y = mx + c

y = −5 ___ 8 x + 5

8y = −5x + 40

5x + 8y = 40

Q. 14.

Line Slopep positiveq 0r undefi neds negativet 0

Q. 15. (i) C ( −1 + 5 _______ 2 , −3 −7 _______ 2 ) = C ( 4 __ 2 , −10 ____ 2 ) = C(2,−5)

(ii) Slope AB = −7 + 3 _______ 5 + 1 = −4 ___ 6 = −2 ___ 3

(iii) Slope p = 3 __ 2 c(2,−5)

p: y − y1 = m(x − x1)

y + 5 = 3 __ 2 (x − 2)

2y + 10 = 3x − 6 3x − 2y − 16 = 0

(iv) p: 3x − 2y − 16 = 0 −2y = −3x + 16

y = 3 __ 2 x − 8

(v) y-intercept (0,−8)

Q. 16. (a) Slope = rise ___ run = 3 __ 1 = 3

y-intercept = 2 ∴ y = 3x + 2

(b)

–10 –8 –6 –4 –2

2

4

6

8

10l2y

x

l1

00

2 4 6 8 10 12 14

l3


(c)

–10 –8 –6 –4 –2

2

4

6

8

10l2y

x

l1

00

2 4 6 8 10 12 14

l3

l4(0,4)

(d) l4: y = ( −1 ___ 3 ) x + 4

(27,−4): −4 = ( −1 ___ 3 ) 27 + 4?

−4 = −9 + 4? −4 = −5 × ∴ (27,−4) ∉ l4

Q. 17. (i) Slope of AB = 4 − 0 ______ 0 − 3 = −4 ___ 3

∴ Slope of BC = 3 __ 4

(ii) B(0, 4) C(8, k)

|BC| = √___________________

(x2 − x1)2 + (y2 − y1)

2

= √_________________

(8 − 0)2 + (k − 4)2

= √_________________

64 + k2 − 8k + 16

= √____________

k2 − 8k + 80

(iii) |BC| = 10

√____________

k2 − 8k + 80 = 10

k2 − 8k + 80 = 100

k2 − 8k − 20 = 0

(k + 2)(k − 10) = 0

k = −2 OR k = 10

# as k is positive

∴ k = 10

(iv) |AB| = √_________________

(0 − 3)2 + (4 − 0)2

= √_______

9 + 16 = √

___ 25

= 5 ∴ Area ABCD = 5(10) = 50 units2

Q. 18. (i) Slope AB = 4 − 2 ______ 3 − 1 = 2 __ 2 = 1

(ii) Slope BC = −2 − 4 _______ 9 − 3 = −6 ___ 6 = −1

m1m2 = (1)(−1) = −1 ∴ AB ⊥ BC

(iii) y − y1 = m(x − x1) y − 4 = −1(x −3) y − 4 = −x + 3 x + y − 7 = 0

(iv) |AB| = √_________________

(3 − 1)2 + (4 − 2)2 = √

_______ 22 + 22

= √__

8 = √

__ 4 √

__ 2

= 2 √__

2 ∴ p = 2

(v) |BC| = √__________________

(9 − 3)2 + (−2 −4)2 = √

__________ 62 + (−6)2

= √___

72 = √

_______ 36 × 2

= 6 √__

2 ∴ Area ΔABC = 1 __ 2 ( 2 √

__ 2 ) ( 6 √

__ 2 )

= 6(2) = 12 units2

Q. 19. (a) |BC| = √___________________

(10 − 3)2 + (9 − 10)2

= √_______

49 + 1

= √___

50

= 7.071 km

(b) |BE| = √__________________

(6 − 3)2 + (6 − 10)2

= √_______

9 + 16

= √___

25

= 5

|BE| = |CE| as E is centre of a circle

∴ distance from B to E to C = 10 km

10 km − 7.071 km = 2.929 km

≈ 3 km

(c) Slope AB = 10 − 8.5 ________ 3 − 1 = 1. 5 ____ 2 = 3 __ 4

y − y1 = m(x − x1) y − 10 = 3 __ 4 (x − 3) 4y − 40 = 3x − 9 3x − 4y + 31 = 0 is equation of

Tangent Street

(d) Slope = −4 ___ 3

y − y1 = m(x − x1)

y − 8 = −4 ___ 3 (x − 17)

3y − 24 = −4x + 68

4x + 3y − 92 = 0 is equation of Perpendicular Avenue


(e) 3x − 4y + 31 = 0 × 3 4x + 3y − 92 = 0 × 4 9x − 12y + 93 = 0 16x + 12y − 368 = 0 25x − 275 = 0 25x = 275 x = 11 3(11) − 4y + 31 = 0 33 − 4y + 31 = 0 −4y + 64 = 0 −4y = −64 y = 16

(11,16) co-ordinates of museum

(f) From Town Hall (0,0) to intersection of Straight Road and East Drive (1,0)

1 km Then North on Straight Road to

A(1,8.5) 8.5 km Then travel from A on Tangent

Street to Museum (11,16)

√_____________________

(11 − 1)2 + (16 − 8.5)2

√___________

102 + (7.5)2

√____________

100 + 56.25 = √_______

156.25

= 12.5 km

Total distance travelled is 1 + 8.5 + 12.5 = 22 km

Q. 20. (a) A(6,1) B(2, −1)

Slope of AB = −1 − 1 _______ 2 − 6 = −2 ___ −4 = 1 __ 2

y − y1 = m (x − x1)

y + 1 = 1 __ 2 (x − 2)

2y + 2 = x − 2

x − 2y − 4 = 0

(b) At y-axis x = 0

⇒ −2y −4 = 0

−2y = 4

y = −2 C(0,−2)

(c) |AB| = √___________________

(2 − 6)2 + (−1 − 1)2

= √_____________

(−4)2 + (−2)2

= √___

20

= √__

4 √__

5

= 2 √__

5

|AC| = √___________________

(0 − 6)2 + (−2 − 1)2

= √_______

36 + 9

= √___

45

= √__

9 √__

5

= 3 √__

5

∴ |AB|

_____ |AC|

= 2 √__

5 ____ 3 √

__ 5 = 2 __ 3

Q. 21.

Slope Line

2 k

1 __ 8 m

0 l

– 1 __ 4 j

–1 n

Chapter 20 Exercise 20 - Biology Leaving Cert.shevlinbiology.webs.com/chapter 20 solutions.pdf ·...

Documents

Transcript of Chapter 20 Exercise 20 - Biology Leaving Cert.shevlinbiology.webs.com/chapter 20 solutions.pdf ·...