Distance between 2 points

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. This is just the Pythagorean theorem: the two points are opposite ends of the slanted side of a right triangle.

In the figure the horizontal gap between $A(1, 1)$ and $B(4, 5)$ is $3$ and the vertical gap is $4$. These are the two legs, so the distance is the hypotenuse: $\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

The order of the points does not matter. For $(-2, 3)$ and $(1, -1)$, the horizontal gap is $1 - (-2) = 3$ and the vertical gap is $-1 - 3 = -4$, giving $\sqrt{3^2 + (-4)^2} = \sqrt{25} = 5$. Squaring removes the sign, so you get the same answer whichever point you start from.

Find the distance between $(1, 2)$ and $(4, 6)$.
Horizontal gap $= 4 - 1 = 3$, vertical gap $= 6 - 2 = 4$.
Distance $= \sqrt{3^2 + 4^2} = \sqrt{25} = 5$.

Find the distance between $(0, 0)$ and $(6, 8)$.
Horizontal gap $= 6$, vertical gap $= 8$.
Distance $= \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$.

A circle has a diameter with endpoints $(4, -26)$ and $(5, -8)$. How long is the diameter?
Horizontal gap $= 5 - 4 = 1$, vertical gap $= -8 - (-26) = 18$.
Length $= \sqrt{1^2 + 18^2} = \sqrt{1 + 324} = \sqrt{325}$.