# Formula for Distance Between Two Points

First, consider one-dimensional case. If a and b - are two points of coordinate line, then the distance between them rho (a;b) is expressed by the formula rho (a;b)=|a-b| . It is clear, that rho(a;b)=rho(b;a) . So, rho(-2;5)=|-2-5|=|-7|=-(-7)=7.

Example 1. Find all such points x that |x-1|=3.

Points x, distance to which from point 1 equals 3 satisfy given equation. These are points are -2 and 4. So, the equation has two roots: -2 and 4.

Example 2. Find all such points x that |x+1|<=2:

Points x, distance to which from point -1 is less or equal to 2 satisfy given equation. These are points of segment [-3,1].

Example 3. Find all such points x that |x+1|>2.

Points x, distance to which from point -1 is greater than 2 satisfy given equation. These are points of two open rays: from -oo to -3 and from 1 to +oo. We can write the answer as (-oo,-3)uu(1,+oo).

Now, let's consider two-dimensional case. Suppose we are given two points with coordinates A(x_1,y_1) and B(x_2,y_2).

We need to find distance between them. Let's draw these points on coordinate plane. We already know from one-dimensional case that |BC|=|x_2-x_1| and |AC|=|y_2-y_1|.

Now, from right-angled triangle ABC we find using Pythagorean theorem that |AB|=sqrt(|BC|^2+|AC|^2)=sqrt((x_2-x_1)^2+(y_2-y_1)^2).

Distance between points A(x_1,y_1) and B=(x_2,y_2) is d=sqrt((x_2-x_1)^2+(y_2-y_1)^2).

Example 4. Find distance between points A(2,-3) and B(-7,-1).

Such sort of example is pretty easy, just pay attention to signs.

d=sqrt(((-7)-2)^2+((-1)-(-3))^2)=sqrt((-9)^2+(2)^2)=sqrt(81+4)=sqrt(85).

Example 5. Find radius of circle with center A(-1,3) if point B(3,5) lies on circle.

Radius of circle is distance between center and any point on circle. Therefore, r=sqrt((3-(-1))^2+(5-3)^2)=sqrt((4)^2+(2)^2)=sqrt(16+4)=sqrt(20)=2sqrt(5).