1. The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ This formula comes from the Pythagorean theorem, where the distance is the hypotenuse of a right triangle formed by the differences in $x$ and $y$ coordinates. 2. Example 1: Find the distance between $(1, 2)$ and $(4, 6)$. $$d = \sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$ 3. Example 2: Distance between $(-1, -1)$ and $(2, 3)$. $$d = \sqrt{(2 - (-1))^2 + (3 - (-1))^2} = \sqrt{3^2 + 4^2} = 5$$ 4. Example 3: Distance between $(0, 0)$ and $(5, 12)$. $$d = \sqrt{(5 - 0)^2 + (12 - 0)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$$ 5. Example 4: Distance between $(3, 7)$ and $(3, 2)$. $$d = \sqrt{(3 - 3)^2 + (2 - 7)^2} = \sqrt{0 + (-5)^2} = 5$$ 6. Example 5: Distance between $(-4, 0)$ and $(0, -3)$. $$d = \sqrt{(0 - (-4))^2 + (-3 - 0)^2} = \sqrt{4^2 + (-3)^2} = 5$$ 7. Example 6: Distance between $(1, 1)$ and $(4, 5)$. $$d = \sqrt{(4 - 1)^2 + (5 - 1)^2} = \sqrt{3^2 + 4^2} = 5$$ 8. Example 7: Distance between $(2, 3)$ and $(7, 3)$. $$d = \sqrt{(7 - 2)^2 + (3 - 3)^2} = \sqrt{5^2 + 0} = 5$$ 9. Example 8: Distance between $(-2, -3)$ and $(-2, 2)$. $$d = \sqrt{(-2 - (-2))^2 + (2 - (-3))^2} = \sqrt{0 + 5^2} = 5$$ 10. Example 9: Distance between $(0, 0)$ and $(0, 0)$. $$d = \sqrt{(0 - 0)^2 + (0 - 0)^2} = 0$$ 11. Example 10: Distance between $(1, 4)$ and $(5, 1)$. $$d = \sqrt{(5 - 1)^2 + (1 - 4)^2} = \sqrt{4^2 + (-3)^2} = 5$$