Sitemap# Distance Formula

## Distance Formula problems exampled:

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The Distance Formula provides a method to determine the length of a line segment that is sloped, horizontal or vertical.

The Distance Formula is given as:

Dist(P1, P2) = √(Δx2 + Δy2)

“Dist” is an abbreviation for the word distance. P1 and P2 are each unique points on the same line segment. Δx is a horizontal distance and Δy is a vertical distance:

P1 = (x1, y1)

P2 = (x2, y2)

(P1 and P2 are Cartesian Coordinates)

Δx = Run = x2 − x1

Δy = Rise = y2 − y1

(Δx and Δy have same definition as Linear Equation Formula where y = mx + b, the slope m = Rise / Run or m = Δy / Δx)

Please refer to the Cartesian coordinate image (below) that illustrates P1, P2, Δx and Δy on a sloped line segment.

Geometric visual examination of Distance Formula:

For a right triangle hypotenuse length the Distance Formula equates to the Pythagoras Formula…

(Pythagorean Theorem is the math and geometry that proves Pythagoras Formula)

Pythagoras Formula:

C2 = A2 + B2

C = √( A2 + B2)

where,

A = Δx = x2 − x1

B = Δy = y2 − y1

Geometric visual of Pythagoras Formula:

We can now see that:

C2 = A2 + B2 = Δx2 + Δy2

C2 = A2 + B2 = (x2 − x1)2 + (y2 − y1)2

C2 = A2 + B2 = (x2 − x1)2 + (y2 − y1)2 =

Dist(P1, P2)2

√C2 = √Dist(P1, P2)2

C = Dist(P1, P2)

We can now conclude:

Pythagoras Formula = Distance Formula

Dist(P1, P2) = √((x2 − x1)2 + (y2 − y1)2)

Given the coordinates,

P1 = (6, 3)

P2 = (3, 6)

The distance between P1 and P2 is,

Δx = x2 − x1 = 3 − 6 = −3

Δy = y2 − y1 = 6 − 3 = 3

Dist(P1, P2) = √(−32 + 32) = √18 = 3√2

P1 = (6,3), P2 = (3, 6):

Given the coordinates,

P1 = (−3, 3)

P2 = (−1/2, −3)

The distance between P1 and P2 is,

Δx = x2 − x1 = −1/2 − (−3) =

−1/2 + 3 = 2 1/2

Δy = y2 − y1 = −3 − 3 = −6

Dist(P1, P2) = √((2 1/2)2 + (−6)2) = 6 1/2

P1 = (−3, 3), P2 = (−1/2, −3):

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