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Cartesian Plane Worksheet

Cartesian Plane Worksheet
  • Page 1
 1.  
What is the distance of the point (6, 8) from the origin?
a.
10
b.
102
c.
9
d.
11


Solution:

Distance of (6, 8) from the origin (0, 0) = (0 - 6)2+(0 - 8)2
[Distance formula.]

= 36 + 64

= 100

= 10


Correct answer : (1)
 2.  
What is the distance of the point (6a, a2 - 9) from the origin?
a.
(a2 - 3) units
b.
(a2 - 6) units
c.
(a2 + 9) units
d.
(a2 - 9) units


Solution:

Distance of (6a, a2 - 9) from the origin (0, 0) = (0 - 6a)2+[0 - (a2 - 9)]2
[Distance formula.]

= 36a2+(a2 - 9)2

= 36a2+a4 + 81 - 18a2

= a4+18a2+81

= (a2+9)2

= (a2 + 9) units


Correct answer : (3)
 3.  
If the distance from the origin to the point (k, k + 2) is 10 units, then find the value of k.
a.
-6, 8
b.
6, - 8
c.
-6, - 8
d.
6, 8


Solution:

The distance from the origin to (k, k + 2) = 10

(k-0)2+(k+2-0)2 = 10
[Distance formula.]

k2 + (k + 2)2 = 100
[Square on both sides.]

k2 + k2 + 4k + 4 = 100

2k2 + 4k - 96 = 0

k2 + 2k - 48 = 0
[Simplify.]

(k + 8)(k - 6) = 0
[Factor.]

k = 6, - 8.
[Solve for k.]


Correct answer : (2)
 4.  
What is the distance between the points (3, 3) and (12, 15)?
a.
225 units
b.
15 units
c.
144 units
d.
81 units


Solution:

The distance between the points (3, 3) and (12, 15) = (12 - 3)2+(15 - 3)2
[Use the distance formula.]

= 81 + 144

= 225

= 15 units.


Correct answer : (2)
 5.  
The distance between the points (4, 4) and (10, k) is 10 units. What are the values of k?
a.
12, 4
b.
-12, -4
c.
4, -12
d.
12, -4


Solution:

The distance between the two points (4, 4) and (10, k) = 10

(10-4)2+(k-4)2 = 10

36 + (k - 4)2 = 100
[Square on both sides.]

(k - 4)2 = 64

k - 4 = 8 , -8

k = 12, -4.
[Solve for k.]


Correct answer : (4)
 6.  
If θ is a real number, then find the distance between the points (8sin 5θ, 8cos 5θ) and (- 8cos 5θ, 8sin 5θ).
a.
8
b.
82 units
c.
92 units
d.
128


Solution:

The distance between the points (8sin 5θ, 8cos 5θ) and (- 8cos 5θ, 8sin 5θ)
= (- 8cos 5θ - 8sin 5θ)2+(8sin 5θ - 8cos 5θ)2

= 2(64cos² 5θ+64sin² 5θ)
[Expand and simplify.]

= 128
[Use sin² 5θ + cos² 5θ = 1.]

= 82 units


Correct answer : (2)
 7.  
Find the distance between the points (7, 0) and (8, tan 7θ) for all real values of θ.
a.
- sec 7θ
b.
|sec 7θ |
c.
± sec 7θ
d.
sec 7θ


Solution:

The distance between the points (7, 0) and (8, tan 7θ) = (8 - 7)2+(tan 7θ - 0)2
[Distance formula.]

= 1+tan2 7θ

= sec2 7θ
[Use 1 + tan2 7θ = sec2 7θ.]

= | sec 7θ |
[Use x2 = | x |.]


Correct answer : (2)
 8.  
The distance of the point (cot 8θ, 9) from (0, 10) is 2 units where 8θ is an acute angle. Find the value of θ.
a.
π7
b.
π9
c.
π32
d.
π8


Solution:

The distance of the point (cot 8θ, 1) from the origin = 2

(0 - cot 8θ)2+(0 - 1)2 = 2
[Distance formula.]

cot2 8θ + 1 = 2
[Square on both sides.]

cot2 8θ = 1

cot 8θ = 1
[Solve for cot 8θ.]

8θ = π4
[Solve for 8θ.]

θ = π32
[Solve for θ.]


Correct answer : (3)
 9.  
If a is any real number, then what is the distance from (4, 0) to (0, a)?
a.
> 4 units
b.
≥ 4 units
c.
< 4 units
d.
= 4 units


Solution:

The distance from (4, 0) to (0, a) is (0 - 4)2+(a - 0)2.
[Distance formula.]

= 16+a2

≥ 4 units
[As 16+a2 ≥ 4.]


Correct answer : (2)
 10.  
If A = (k, k), B = (3 + k, 4 + k), C = (4 + k, 3 + k) are any three points of a plane, then for all the real values of k which of the following is correct?
a.
AB < AC
b.
AB = AC
c.
2AB = AC
d.
AB > AC


Solution:

A = (k, k), B = (3 + k, 4 + k), C = (4 + k, 3 + k) where k is a real number.

AB = (3 + k - k)2+(4 + k - k)2 = 9 + 16 = 5
[Distance formula.]

AC = (4 + k - k)2+(3 + k - k)2 = 16 + 9 = 5
[Distance formula.]

So for all real values of k, AB = AC.


Correct answer : (2)

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