Compute distance between the 2 points (1- √2, -1) & (2+√2, 4). Show work.?

Use the distance formula, which is based on the Pythagorean Theorem.

Distance between (x₁, y₁) and (x₂, y₂)
= √[(x₁ - x₂)² + (y₁ + y₂)²]

Hence, distance between the 2 points (1- √2, -1) and (2 + √2, 4)
= √{[(2 + √2) - (1 - √2)]² + [4 - (-1)]²}
= √ [(2 + √2 - 1 + √2)² + (4 + 1)²]
= √ [(1 + 2√2)² + (5)²}
= √[(1 + 4√2 + 8) + 25]
= √(34 + 4√2)

Since this forms a right triangle, it's a Pythagorean equation problem

d² = x² + y²

x is Δx like (2+√2) - (1- √2)
= 2 +√2 -1+ √2
= 1 + 2√2

y is Δy like 4-(-1) = 5

Square them, add together, then take the root.
I get √[34 + 4√(2)]
≈6.29737

For the two points (a,b) & (c,d), a formula for the distance is √( (c-a)² + (d-b)² ).
Corresponding this to your two points, what are

a=?
b=?
c=?
d=?

Then, what are c-a = ? and d-b = ?

Do these and we can take it from there...!

'' Show work please ''

d = √[ [(1 - √2) - (2 + √2)]^2 + [ (- 1) - 4]^2]

= √[ ( 1 - √2 - 2 - √2)^2 + ( - 5)^2]

= √[ (-1 - 2√2)^2 + 25]

= √[ 9 + 4√2 + 25]

= √{ 34 + 4√2]

= 6.297368835

= 6.3

d² = ( 2√2 + 1)² + 5²
d² = 8 + 1 + 4√2 + 25
d² = 34 + 4√2
d = [ 34 + 4√2 ]^(1/2)
d = 2^(1/2) [ 17 + 2√2 ]^(1/2)