Chapter 11 Exercise 11.1
__
(i) x2 + y2 = 4
Q. 1.
(vii) √3
(ii) x2 + y2 = 9
__
(viii) √ 5
(iii) x2 + y2 = 1
3
(ix) __
2
(iv) x2 + y2 = 144
(v) x2 + y2 = 25
Q. 6.
(vi) x2 + y2 = 2
(vii) x2 + y2 = 5
9
(viii) x2 + y2 = ___
16
1
(ix) x2 + y2 = __
4
72
(x) x2 + y2 = __
3
49
x2 + y2 = ___
9
10
(x) ___
3
7
__
(xi)
5
11
(xii) ___
6
(i)
7
(0,0)
()
Q. 2.
(ii)
__________________
(7 −
+ (24 −
√_________
0)2
0)2
√49 + 576
____
√625
4
(0,0)
25
x2 + y2 = 625
Q. 3.
____________________
|OB| = √ (−6 − 0)2 + (−8 − 0)2
________
= √ 36 + 64
= 10
(iii)
x2 + y2 = 102
3
x2 + y2 = 100
Q. 4.
(0,0)
_________________
|OC| = √ (3 − 0)2 + (4 − 0)2
_______
= √ 9 + 16
___
= √ 25
=5
x2
+
y2
(iv)
= 25
Q. 5. All centres (0,0)
(i) 5
(iv) 9
(ii) 10
(v) 2
(iii) 8
(vi) 7
5
(0,0)
Active Maths 3 Book 2 – Strands 1–5 – Ch 11 Solutions
1
(v)
Q. 9.
(0,0)
(i) (0,0)
(ii) r = 2
1
–
2
(iii)
2
(0,0)
(vi)
1
–
4
(iv) C = 2pr
(0,0)
= 2(p)(2)
Calculator for p
= 12.57
Q. 10.
(i) (0,0)
(ii) r = 6
Q. 7.
(i) A(1,0)
(iii)
B(0,1)
6
C(−1,0)
(0,0)
D(0,−1)
(ii) A(3,0)
B(0,3)
C(−3,0)
D(0,−3)
(iv) C = 2pr
(iii) A(2,0)
= 2p(6)
B(0,2)
= 12p
C(−2,0)
= 37.699
D(0,−2)
(iv) A(6,0)
B(0,6)
Q. 8.
(v)
(36)2 + (15)2
√___________
C(−6,0)
√1296 + 225
_____
√1521
D(0,−6)
= 39
(i) x2 + y2 = 25
(ii) x2 + y2 = 100
(iii) x2 + y2 = 144
(iv) x2 + y2 = 49
2
≅ 37.70
____________
Active Maths 3 Book 2 – Strands 1–5 – Ch 11 Solutions
Yes.
Q. 11.
(i) Centre = C = Midpoint [AB]
3+3 4−4
C = −______, ______
2
2
C = (0,0)
(
(ii)
)
A(–3,4)
C(0,0)
B(3,–4)
(iii) Radius = |BC|
B(3, −4) C (0, 0)
x1 y1
x2 y2
__________________
= √ (0 − 3)2 + (0 + 4)2
_______
= √9 + 16
= 5.
(iv) x2 + y2 = 25
Q. 12. A(−2,1) B(2,−1)
(i) Centre = C = (0,0)
(ii)
A(–2,1)
(0,0)
B(2,–1)
(iii) |AB| = diameter
|AC| = radius
A(−2, 1) C(0, 0)
x2 y2
x1 y1
__________________
|AC| = √ (0 + 2)2 + (0 − 1)2
______
= √4 + 1
__
= √5
(iv) x2 + y2 = r2
x2 + y2 = 5
(v) A = pr2
A = p (5)
= 15.70796327
≈ 15.71 square units
Active Maths 3 Book 2 – Strands 1–5 – Ch 11 Solutions
3
Q. 13.
(i) Centre = (0, 0)
Use midpoint formulas
A(–5,1)
C(0,0)
B(5,–1)
(ii) Radius = |AC|
A(−5, 1)
x1 y1
C(0, 0)
x2 y2
__________________
|AC| = √ (0 + 5)2 + (0 − 1)2
_______
= √ 25 + 1
___
= √ 26
(iii) x2 + y2 = 26
(iv) Area = pr2
(if not given value to take for p, use p symbol on calculator)
___ 2
= p ( √26 )
= 26p
≅ 81.68 square units
(v) No, as area is 81 square units
Circle ≅ 81.68 square units
Q. 3.
Exercise 11.2
Q. 1.
Q. 2.
4
(i) x = 4
(4,3) (4,−3)
(ii) y = 4
(−3,4) (3,4)
(i) (−4,3) (5,0)
(iii) x = −4
(−4,3) (−4,−3)
(ii) (−6,−8) (6,8)
(iv) y = −4
(−3,−4) (3,−4)
(iii) (−2,1) (1,2)
(v) x = 3
(3,4) (3,−4)
(iv) (−5,1) (5,1)
(vi) y = 3
(4,3) (−4,3)
(v) (−2,4)
(vii) x = −3
(−3,4) (−3,−4)
(viii) y = −3
(4,−3) (−4,−3)
(i) x = 2
(2,4) (2,−4)
(ii) y = 2
(−4,2) (4,2)
(iii) x = −2
(−2,4) (−2,−4)
(iv) y = −2
(−4,−2) (4,−2)
Active Maths 3 Book 2 – Strands 1–5 – Ch 11 Solutions
Q. 4.
(i) x2 + y2 = 20 .… A
x = 10 − 3y .…. B
Substitute
B
into
A
(10 − 3y)2 + y2 = 20
100 − 60y + 9y2 + y2 = 20
10y2 − 60y + 80 = 0
y2
÷ 10
− 6y + 8 = 0
(y − 2)(y − 4) = 0
y=2 y=4
y = 2 x = 10 − 3(2)
B
= 10 − 6
=4
x = 4, y = 2
y = 4 x = 10 − 3(4)
x = −2
x = − 2, y = 4
(4,2)
and
(−2,4)
(ii) x2 + y2 = 25 .… A
x − 2y = 5 .… B
x = 5 + 2y
Rewrite B
into
Substitute B
A
(5 + 2y)2 + y2 = 25
25 + 20y + 4y2 + y2 = 25
5y2 + 20y + 25 − 25 = 0
5y2 + 20y = 0
5y(y + 4) = 0
5y = 0 y + 4 = 0
y = 0, y = −4
Find x-values:
y=0 B
x = 5 + 2(0)
y = −4
B
x = 5 + 2(−4)
x=5
x = −3
x = 5, y = 0
x = −3, y = −4
(5,0)
and
(−3,−4)
Active Maths 3 Book 2 – Strands 1–5 – Ch 11 Solutions
5
x2 + y2 = 20 … A
(iii)
y = 2x .… B
x2 + (2x)2 = 20
x2 + 4x2 = 20
5x2 = 20
x2 = 4
__
x = ±√ 4
x = ±2
y = 2(2) = 4
x=2 B
x = 2, y = 4
y = 2(−2) = −4
x=−2 B
x = −2, y = −4
(2,4)
and
(−2,−4)
(iv) x2 + y2 = 40 .… A
y = − 2 .…. B
x2 + (−2)2 = 40
x2 = 40 − 4
x2 = 36
x=±6
x = 6, y = − 2 x = −6, y = − 2
(6,−2)
and
(−6,−2)
Q. 6. x − y = 1 .… A
Q. 5. x + y = 4 .… A
x2
+
y2
x2 + y2 = 13 … B
= 10 .… B
A
x=4−y
B
y)2
(4 −
+
y2
= 10
16 − 8y + y2 + y2 = 10
A
x=1+y
B
(1 + y)2 + y2 = 13
1 + 2y + y2 + y2 = 13
2y2 − 8y + 6 = 0
2y2 + 2y − 12 = 0
y2 − 4y + 3 = 0
y2 + y − 6 = 0
(y − 3)(y − 1)
{ yx == 13 }
(1,3)
or
and
{ yx == 31 }
(3,1)
(y − 2)(y + 3) = 0
y = 2 y = −3
A
x−2=1
x=3
(3,2)
6
Active Maths 3 Book 2 – Strands 1–5 – Ch 11 Solutions
A
x+3=1
x = −2
and
(−2,−3)
(ii) 52 + y2 = 25
l
y = 3 − 2x
s
x2 + (3 − 2x)2 = 26
y2 = 0
x2 + 9 − 12x + 4x2 = 26
y=0
Q. 7.
Only one point of intersection
⇒ tangent.
5x2 − 12x −17 = 0
(5x − 17)(x + 1) = 0
17
x = ___ x = −1
5
17
l y = 3 − 2 ___ l y = 3 − 2(−1)
5
34
___
y=3−
y=5
5
19
y = −___
5
19
17
___
___
and (−1, 5)
,−
5
5
(iii) (2y − 10)2 + y2 = 20
4y2 − 40y + 100 + y2 − 20 = 0
( )
(
5y2 − 40y + 80 = 0
y2 − 8y + 16 = 0
(y − 4)(y − 4) = 0
y=4
Only one point of intersection
⇒ tangent.
)
m x = 4 −7y
Q. 8.
n
(iv) 2(3 − y)2 + 2y2 = 9
(4 − 7y)2 + y2 = 10
2(9 − 6 y + y2) + 2y2 = 9
16 − 56y + 49y2 + y2 = 10
18 − 12y + 2y2 + 2y2 = 9
50y2 − 56y + 6 = 0
4y2 − 12y + 9 = 0
(50y − 6)(y − 1) = 0
6
y = ___ y = 1
50
6
m x = 4 − 7 ___ m x = 4 − 7(1)
50
79
= ___
x = −3
25
6
79 ___
___
and (−3,1)
,
25 50
(2y − 3) (2y − 3)
3
y = __
2
One point of intersection
⇒ tangent.
( )
(
(i) 4 units
(i) (0,0)
___
(ii) √10
√
(ii) (x – 4)2 + (y – 4)2 = 16
O(0, 0) A(3, 1)
x2 y2
x1 y1
(iii) m : y = 8
|OA| = √ (3 − 0)2 + (1 − 0)2
_________________
___
n:x=8
= √ 10
(iv) (8,8)
(iv) Yes, as distance from OA is
= radius.
(v) 64 square units
Q. 10.
___________________
(iii) |OA| = (x2 − x1)2 + (y2 − y1)2
(v)
(i) (y + 2)2 + y2 = 2
y2 + 4y + 4 + y2 − 2 = 0
2y2 + 4y + 2 = 0
(2y + 2)(y + 1) = 0
y = −1
10
Q. 9.
)
Q. 11.
(0,0)
or y = −1
Only one point of contact
⇒ tangent.
Active Maths 3 Book 2 – Strands 1–5 – Ch 11 Solutions
7
(vi) l: y = x + 2
Q. 3.
x=0 y=2
(0,2)
y = 0 x = −2
___
(−2,0)
= √ 20
(ii) (x + 2)2 + (y + 1)2 = 20
(0,2)
(vii)
(0,0)
(–2,0)
x2
s
+ (x +
2)2
(i) 4
Q. 5.
(i) 3
Q. 6.
(i) r = 2
Q. 7.
(i)
= 10
(ii)
x = −3
y=1+2=3
x = −3
( − 1 +2 3, 4 −2 2 )
______ ______
___________________
− 1)2 + (−2 − 1)2
√(3
______
___
√13
(iii) (x − 1)2 + (y − 1)2 = 13
Q. 8.
y=x+2
x=1
(ii) (x − 2)2 + (y − 2)2 = 4
√4 + 9
+ 4x − 6 = 0
(x − 1)(x + 3) = 0
l
(ii) (x − 3)2 + (y + 1)2 = 9
= (1, 1)
x2 + 2x − 3 = 0
x=1
(ii) (x − 2)2 + (y − 4)2 = 16
Q. 4.
x2 + x2 + 4x + 4 − 10 = 0
2x2
___________________
+ 2)2 + (−3 + 1)2
√(2 _______
= √ 16 + 4
(1,3)
(–3,–1)
(i)
(1, 3)
(
−2 + 4 −3 + 7
(i) _______, _______
2
2
(1, 2)
)
(ii) Radius (1, 2)
(−2, −3)
(−2 −
√_______
+ (−3 − 2)2
____________________
1)2
y = −3 + 2
= −1 (−3, −1)
√9 + 25
___
√34
Exercise 11.3
Q. 1.
(iii) (x − 1)2 + (y − 2)2 = 34
(i) (x − 1)2 + (y − 4)2 = 25
Q. 9.
(iii) (x − 6)2 + (y + 5)2 = 1
(iii) (x − 1)2 + (y − 2)2 = 10
(iv) (x − 0)2 + (y − 7)2 = 16
(vi)
x2
+
y2
Q. 10.
= 64
+ (y − 1)2 = 36
12
12
(viii) x − __ + y + __ = 81
4
2
2
(ix) (x − 1) + (y − 1.5)2 = 49
1
(x) (x + 3)2 + (y + 8)2 = __
4
(vii) (x −
(
Q. 2.
(i)
8)2
) (
)
_________________
− 1)2 + (6 − 1)2
√(5 _______
= √42 + 52
________
= √16 + 25
___
= √41
(ii) (x − 1)2 + (y − 1)2 = 41
8
______
√9 + 1
___
√10
(ii) (x + 2)2 + (y − 3)2 = 9
(v) (x + 2)2 + (y + 1)2 = 4
___________________
(ii) Radius = √ (− 2 − 1)2 + (3 − 2)2
Active Maths 3 Book 2 – Strands 1–5 – Ch 11 Solutions
(–1,3)
Centre
point
(
(5,3)
−1 + 5 3 + 3
Centre point _______, ______
2
2
= (2, 3)
)
_________________
Radius = √ (5 − 2)2 + (3 − 3)2
__
= √9
=3
Ans = (x − 2)2 + (y − 3)2 = 9
Q. 11.
(i) Centre (2, 3) r = 5
(ii) E = (−7, 0)
(ii) (−2,5) r = 6
F = (7, 0)
(iii) (1,−3) r = 7
(iii) G = (0, 7)
H = (0, −7)
(iv) (−5,−8) r = 1
(v) (3,4) r = 8
Q. 12.
(i) (0, 0)
Q. 3.
r = 10
(ii) (0, 8) r = 7
( )
(0, −2) and (0, 6)
(ii) x-intercepts
(−6, 0) and (8, 0)
(0, 12) and (0, −4)
Intercepts
(i) (−2, 0) (2, 0)
x
(0, 2) (0, −2)
y
(i) (3, 0)
(ii) (−3, 0) (3, 0)
x
(ii) 1 unit
(0,−3) (0, 3)
y
(iii) (1, 0)
(iii) (−4, 0) (4, 0)
x
(0, 4) (0,−4)
y
(iv) (−6, 0) (6, 0)
x
(0, 6) (0,−6)
y
(v) 4 –
p(1)2
= (4 – p) square units
(i) x2 + y2 = 1
(x – 4)2 + y2 = 1
(x – 2)2 + y2 = 9
Q. 5. To get x-intercept let y = 0
(ii) 1 area = p(1)2 = p
∴ answer = 3p square units
(iii) p(3)2 = 9p square units
(iv) 9p – 3p = 6p square units
Exercise 11.4
Q. 1.
y-intercepts
Q. 4.
(iv) 4 square units
Q. 14.
(−3, 0) and (4, 0)
y-intercepts
(iii) (2, 0) r = 9
1 1
1
(iv) __, __ r = __
2 3
3__
(v) (2, 3) r = √ 5
Q. 13.
(i) x-intercepts
(i) a = − 4
b=4
c=4
(i) x2 − 4x + 4 + 9 = 34
x2 − 4x − 21 = 0
(x + 3)(x − 7) = 0
x = −3 x = 7
(−3,0)
and
(7,0)
(ii) x2 + 6x + 9 + 4 = 40
x2 + 6x − 27 = 0
(x − 3)(x + 9) = 0
(3,0)
and
(−9,0)
(iii) x2 + 10x + 25 + 64 − 164 = 0
d = −4
x2 + 10x − 75 = 0
(ii) A(−4, 0)
(x + 15)(x − 5) = 0
x = −15 x = 5
B(4, 0)
(iii) C(0, 4)
D(0, −4)
Q. 2.
(−15,0)
g=7
(5,0)
(iv) x2 + 2x + 1 + 1 − 50 = 0
(x − 6)(x + 8) = 0
(i) e = −7
f=7
and
x = 6 x = −8
(6,0)
and
(−8,0)
h = −7
Active Maths 3 Book 2 – Strands 1–5 – Ch 11 Solutions
9
(v) x2 − 8x + 16 + 9 − 58 = 0
Q. 7.
(i) c: (x − 2)2 + (y − 1)2 = 13
x2 − 8x + 25 − 58 = 0
y-intercept
(x + 3)(x − 11) = 0
4+
y2
− 2y + 1 − 13 = 0
y2 − 2y − 8 = 0
x = −3 x = 11
(−3,0)
(vi)
x2
and
(11,0)
(y + 2)(y − 4) = 0
− 20x + 100 + 121 = 125
x2
y = −2
− 20x + 96 = 0
B(0,−2)
(x − 12)(x − 8)
and
(
x=0
Q. 6. y-intercept
(i) 9 +
(8,0)
y2
Radius = |BC|
__
= √9
(y + 5)(y − 13) = 0
(ii) 4 +
y2
=3
(0, 13)
Equation (x − 0)2 + (y − 1)2 = 32
− 10y + 25 − 104 = 0
y2
− 10y − 75 = 0
(y + 5)(y − 15) = 0
x2 + (y − 1)2 = 9
Q. 8.
and
(x + 3)(x − 7) = 0
x = −3 x = 7
y2 + 2y − 35 = 0
A(−3,0)
(y + 7)(y − 5) = 0
( )
Radius |AC|
(0,5)
_____________
= √ (2 + 3)2 + (0)2
(iv) 1 + y2 + 2y + 1 − 10 = 0
y2 + 2y − 8 = 0
=5
n: (x − 2)2 + y2 = 25
(y − 2)(y + 4) = 0
y=2 y=−4
(0,2)
and
(0,−4)
and B(7,0)
4
(ii) Centre __, 0 = C(2, 0)
2
y = −7 y = 5
and
y=0
x2 − 4x − 21 = 0
(0,15)
(iii) 36 + y2 + 2y + 1 − 72 = 0
(0,−7)
(i) x-intercept
x2 − 4x + 4 + 9 = 34
y = −5 y = 15
(0,−5)
B(0,−2) C(0,1)
= √ (0 − 0)2 + (1 + 2)2
− 8y − 65 = 0
(0, −5) and
)
_________________
− 8y + 16 − 90 = 0
y2
y=4
and A(0,4)
(ii) Centre point
0 + 0 −______
2+4
______
,
2
2
C(0,1)
x = 12 x = 8
(12,0)
x=0
Q. 9. C4 (1 + 2)2 + (−2 −5)2 = 58
32 + (−7)2 = 58
(v) 9 + y2 − 8y + 16 − 58 = 0
9 + 49 = 58
y2 − 8y − 33 = 0
58 = 58
Ans: D is on c4.
(y − 11)(y + 3) = 0
y = 11 y = −3
(0, 11)
and
Q. 10. Is (−2 + 4)2 + (0 − 1)2 = 9?
(0, −3)
Is 22 + (− 1)2 = 9?
4+1<9
5<9
∴ A is inside the circle.
10
Active Maths 3 Book 2 – Strands 1–5 – Ch 11 Solutions
Q. 11. Is (7 − 1)2 + (3 − 2)2 = 9?
(v) If [AB] is diameter, find B.
62 + 12 > 9
A(3, −2) → R(−1, 1) → B(−5, 4)
37 > 9
∴ B(7,3) lies outside the circle.
Q. 3.
(ii) Radius of s = 2
c3 X = outside
Distance (1, 0) to (3, 0)
c4 Y = on
_____________
= √(3 − 1)2 + (0)2
c5 Z = inside
__
= √4
Revision Exercises
(
−7 + 1 0 + 0
(i) _______, ______
2
2
A (−3,0)
=2
Radius of q = 1
)
_____________
√(3 − 2)2 + (0)2
=1
_____________
(1 +
√___
3)2
+
(0)2
Yes. Radius of s is twice radius of q.
√16
x2 + y2 = 1
(iii) p:
r=4
(iii) (x + 3)2 + (y − 0)2 = 16
(iv) y-intercept x = 0
y2 = 16 − 9
__
(0, √ 7 )
q:
(x − 2)2 + y2 = 1
s:
(x − 1)2 + y2 = 4
Centre
Q. 4.
9 + y2 = 16
Q. 2.
centre (1, 0)
s
c2 W = on
(ii)
centre (0, 0)
q centre (2, 0)
Q. 12. c1 V = inside
Q. 1.
(i) p
y2 = 7
(i) t
(0,3)
2
s
(0,0)
1
u (0,−3)
__
(0,−√ 7 )
Radius
2
(ii) t
x2 + (y − 3)2 = 4
(i) R (−1, 1)
s
x2 + y2 = 1
(ii) r = 5
u x2 + (y + 3)2 = 4
(iii)
(iii) Circumference = 2pr
(–1,1)
Cs : 2p(1)
5
Ans = 2p
(iv) Ct
2pr
2p (2)
Ans = 4p
(v) 4p : 2p
(iv) A(3,−2)
2:1
(3 + 1)2 + (−2 − 1)2 = 25
(vi) Yes
16 + 9 = 25
25 = 25
∴ A is on circle.
Q. 5.
(i) r = 4
(ii) (8, 8)
(iii) Radius of u = 4
Active Maths 3 Book 2 – Strands 1–5 – Ch 11 Solutions
11
(iv) (x − 8)2 + (y − 8)2 = 16
y = −1 y = 3
y = −1 y = 3
(v) Area : 64 square units
l
(vi) Area of circle
x = −3 x = 1
A=p
r2
(−3,−1)
A = p (4)2
and
(ii) c: (x + 2)2 + (y − 2)2 = 2
= 16p
(iii)
l
x=y−2
c (y − 2 + 2)2 + (y − 2)2 = 2
= 50.26548246
y2 + y2 − 4y + 4 − 2 = 0
≈ 50.27
2y2 − 4y + 2 = 0
Area of square = 64.
y2 − 2y + 1 = 0
Area of square is greater.
Q. 6.
(i)
(y − 1)(y − 1) = 0 y = 1
x−y+2=0
l
s
(1,3)
Ans: (−1, 1) x = −1
x2 + y2 = 10
l x=y−2
s (y −
2)2
+ y2 = 10
y2 − 4y + 4 + y2 = 10
2y2 − 4y − 6 = 0
y2 − 2y − 3 = 0
(y + 1)(y − 3) = 0
Q. 7.
(i) Slope AB
Slope AC
Q. 8.
5+1
______
5 − 3 ___
6 3
2
______
= __ = __
=
2+2 4 2
2 − 5 −3
3 ___
6
2
__
×
= ___ = −1
2 −3 −6
⇒ AB ⊥ AC
x2 + 16x2 − 64x + 64 − 25 = 0
17x2 − 64x + 39 = 0
( − 2 +2 5, − 1 +2 3 )
3
D ( , 1)
2
(v) Radius |BD| = ( + 2 ) + (1 + 1)
√ 23
______
(ii)
(i) x2 + y2 = 25
(17x − 13)(x − 3)
13
x = ___ x = 3
17
So, x = 3 y = 4(3) − 8
______
__
y=4
_________________
__
√
√
2
2
_______
49
= ___ + 4
4
___
65
= ___
4
2
3
65
2
(vi) x − __ + ( y − 1 ) = ___
4
2
(
)
(
A(3,4)
13
84
3 + ___ 4 − ___
17
17
_______ _______
,
2
2
32 ___
8
___
,−
17 17
32
8
1
(iii) n: y + ___ = − __ x − ___
4
17
17
32
32
___
___
4y +
= −x +
17
17
21y + 32 = −17x +32
(ii) Midpoint
(
(
)
(
17x + 21y = 0
(iv) 17(0) + 21(0) = 0
0=0
12
Active Maths 3 Book 2 – Strands 1–5 – Ch 11 Solutions
)
)
)
Q. 9.
___________
___
(ii) r = √ (1)2 + (−7)2
(i) r = √17
(ii) t
___
√50
x = 17 − 4y
(17 − 4y)2 + y2 = 17
k
Equation: x2 + y2 = 50
289 − 136y + 16y2 + y2 = 17
(iii) p2 + p2 = 50
17y2 − 136y + 272 = 0
2p2 = 50
y2 − 8y + 16 = 0
p2 = 25
(y − 4)(y − 4) = 0
p = ±5
(iv) 32 + n2 < 50
y=4
n2 < 49
⇒ Only one point of intersection
n<7
⇒ Tangent
(iii)
Ans: n = 6
(1,4)
Q. 11.
t
(i) r = 3
(ii) (x − 7)2 + (y − 3)2 = 9
(iii) 7 − 2(3) − 1 = 0
(0,0)
7−6−1=0
0=0
(iv) x − 2(6) − 1 = 0
k
(iv)
x − 13 = 0
(1,4)
x = 13
t
Q. 12.
(i)
_________________
(4 − 2)2 + (5 − 4)2
√_______
√22 + 12
__
(0,0)
= √5
_________________
(ii) |AB| = √(4 − 0)2 + (5 − 5)2
___
= √ 16
k
l
Point of tangency for l is (−1,−4)
1
Slope = −__
4
1
(y + 4) = −__(x + 1)
4
4y + 16 = −x −1
l: x + 4y + 17 = 0
Q. 10.
(i)
=4
1
1
Area = __ bh = __ (4)(3)
2
2
= 6 square units
__
2
(iii) pr2 = p ( √ 5 ) = 15.70796327
≅ 15.71 square
units
6
100
(iv) ______ × ____ = 38.19%
15.71
1
(0,0)
(1,–7)
Active Maths 3 Book 2 – Strands 1–5 – Ch 11 Solutions
13