FINDING LENGTH OF A LINE SEGMENT

Problem 1 :

What is the approximate length of RS with endpoints R(2, 3) and S(4, -1)?

Solution :

Given, R(2, 3) and S(4, -1)

x₁ = 2,  y₁ = 3, x₂ = 4,  y₂ = -1

RS = √[(x₂ – x₁)² + (y₂ – y₁)²]  

= √[(4 – 2)² + ((-1) – 3)²]

= √[(2)² + (-4)²]

= √[4 + 16]

= √20

= 4.5

So, the approximate length is 4.5 units.

Problem 2 :

What is the approximate length of AB with endpoints A(-3, 2) and B(1, -4)?

Solution :

Given, A(-3, 2) and B(1, -4)

x₁ = -3,  y₁ = 2, x₂ = 1,  y₂ = -4

RS = √[(x₂ – x₁)² + (y₂ – y₁)²] 

= √[(1 – (-3))² + (-4 – 2)²]

= √[(4)² + (-6)²]

= √[16 + 36]

= √52

= 7.2

So, the approximate length is 7.2 units.

Problem 3 :

The endpoints of MN are M(-3, -9) and N(4, 8). What is the approximate length of MN?

Solution :

Given, M(-3, -9) and N(4, 8)

x₁ = -3,  y₁ = -9, x₂ = 4,  y₂ = 8

MN = √[(x₂ – x₁)² + (y₂ – y₁)²] 

= √[(4 – (-3))² + (8 – (-9)²]

= √[(7)² + (17)²]

= √[49 + 289]

= √338

= 18.4

So, the approximate length is 18.4 units.

Find the length of the segment. Round to the nearest tenth of a unit.

Problem 4 :

Solution :

Given, P(1, 2) and Q(5, 4)

x₁ = 1,  y₁ = 2, x₂ = 5,  y₂ = 4

PQ = √[(x₂ – x₁)² + (y₂ – y₁)²] 

= √[(5 – 1)² + (4 – 2)²]

= √[(4)² + (2)²]

= √[16 + 4]

= √20

= 4.5

So, the approximate length is 4.5 units.

Problem 5 :

Solution :

Given, Q(-3, 5) and Q(2, 3)

x₁ = -3,  y₁ = 5, x₂ = 2,  y₂ = 3

PQ = √[(x₂ – x₁)² + (y₂ – y₁)²] 

= √[(2 – (-3))² + (3 – 5)²]

= √[(5)² + (-2)²]

= √[25 + 4]

= √29

= 5.385

So, the approximate length is 5.385 units.

Problem 6 :

Solution :

Given, S(-1, 2) and T(3, -2)

x₁ = -1,  y₁ = 2, x₂ = 3,  y₂ = -2

PQ = √[(x₂ – x₁)² + (y₂ – y₁)²] 

= √[(3 – (-1))² + ((-2) – 2)²]

= √[(4)² + (-4)²]

= √[16 + 16]

= √32

= 5.7

So, the approximate length is 5.7 units.

Problem 7 :

Calculate the perimeter of triangle ABC,

Solution :

Length of side AB = √[(x₂ – x₁)² + (y₂ – y₁)²] 

AB = √[(-1 + 3)² + (5 + 1)²] 

= √[2² + 6²] 

= √[4 + 36]

= √40

AB = 6.32

BC = √[(4 + 1)² + (-2 - 5)²] 

= √[5² + (-7)²] 

= √[25 + 49]

= √74

BC = 8.6

CA = √[(4 + 3)² + (-2 + 1)²] 

= √[7² + (-1)²] 

= √[49 + 1]

= √50

= 7.07

Perimeter of triangle ABC = AB + BC + CA

= 6.32 + 8.6 + 7.07

= 21.99

Approximately 22 units.

Problem 8 :

Two wolves spot a deer in a field. The positions of the animals are shown. Which wolf is closer to the deer?

Solution :

Wolf A (8, 16), Wolf B(16, 4) and Deer (48, 32)

Distance between wolf A and deer = √[(48 - 8)² + (32 - 16)²] 

= √[40² + 16²] 

= √(1600 + 256)

= √1856

= 43.08

Distance between wolf B and deer = √[(48 - 16)² + (32 - 4)²] 

= √[32² + 28²] 

= √(1024 + 784)

= √1808

= 42.52

Wolf B is closer to deer.

Problem 9 :

A theater is 3 miles east and 1 mile north of a bus stop. A museum is 4 miles west and 3 miles south of the bus stop. Estimate the distance between the theater and the museum

Solution :

Position of theater is at (-3, 1)

Position of museum (4, -3)

Distance between theater and museum = √[(4 + 3)² + (-3 - 1)²] 

= √[7² + (-4)²] 

= √[49 + 16]

= √65

= 8.06 miles

Problem 10 :

The endpoints of line segment AB are A(2x, y − 1) and B(y + 3, 3x + 1). The midpoint of line segment AB is M (−7/2, −8) . What is the length of line segment AB?

Solution :

Distance between AM = distance between MB

Midpoint of AB = M

(2x + y + 3)/2, (y - 1 + 3x + 1)/2 = (-7/2, -8)

(2x + y + 3)/2, (y + 3x)/2 = (-7/2, -8)

Equating x and y coordinates, we get

(2x + y + 3)/2 = -7/2

2x + y + 3 = -7

2x + y = -7 - 3

2x + y = -10 ---------(1)

(y + 3x)/2 = -8

3x + y = -16 ---------(2)

(1) - (2)

2x - 3x = -10 + 16

-x = 6

x = -6

Applying x = -6 in (1), we get

2(-6) + y = -10

-12 + y = -10

y = -10 + 12

y = 2

A(2x, y − 1) and B(y + 3, 3x + 1)

A(-12, 1) and B(5, -17)

Distance between AB = √[(5 + 12)² + (-17 - 1)²] 

= √[17² + (-18)²] 

= √[289 + 324] 

= √613

= 24.7 units