Find the perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0).

**Distance** between two points P$(x$₁, $y$₁) and Q$(x$₂, $y$₂):

**PQ = $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$**

This is just the Pythagoras theorem applied to the right triangle formed with horizontal and vertical legs.

**Example**: Distance between (3, 4) and (7, 1).
= $\sqrt{(7-3{)}^2+(1-4{)}^2}=\sqrt{16+9}=\sqrt{25}=5$.

**Special distance from origin**: distance of $(x,y)$ from (0, 0) is $\sqrt{x^2+y^2}$.