Question

What is the distance between the points (2 and one-half, negative 8) and (2 and one-half, 3) Sukant’s work is shown below?

StartAbsoluteValue 2 and one-half EndAbsoluteValue + StartAbsoluteValue 2 and one-half EndAbsoluteValue = 2 and one-half + 2 and one-half = 5. The distance is 5 units.

What error did Sukant make?

Answer 1

He should have used the y-coordinates

Explanation:

I don't have one.

Answer 2

Given Information:

Point 1 = ( 2½, -8)

Point 2 = ( 2½, 3)

Required Information:

Distance between two points = d = ?

Answer:

Distance between two points =
`d = 11`
`units`

Explanation:

The distance between two points is calculated using

`d = \sqrt{(x_(2) - x_(1))^2 + (y_(2) - y_(1))^2}` eq. 1

We have (x₁, y₁) = (2½, -8) and (x₂, y₂) = (2½, 3)

Substitute the above points into the eq. 1

`d = √((2\textonehalf - 2\textonehalf)^2 + (3 - (-8))^2)`

`d = √((2\textonehalf - 2\textonehalf)^2 + (3 + 8)^2)`

`d = √((0)^2 + (11)^2)`

`d = √(0 + 121)`

`d = √(121)`

`d = 11`
`units`

it is very unclear as what error did Sukant make due to unclear typing!

But I guess, Sukant simply added the x-coordinates together that is

d = 2½ + 2½ = 5 units

Note: 2½ = 5/2 = 2.5