1. The problem is to find the distance between two points in a plane using the distance formula. 2. The distance formula between points $A(x_1,y_1)$ and $B(x_2,y_2)$ is: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ This formula comes from the Pythagorean theorem. 3. Example 1: Find the distance between $A(1,2)$ and $B(4,6)$. $$d = \sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$ 4. Example 2: Find the distance between $A(-1,-1)$ and $B(2,3)$. $$d = \sqrt{(2 - (-1))^2 + (3 - (-1))^2} = \sqrt{3^2 + 4^2} = 5$$ 5. Example 3: Find the distance between $A(0,0)$ and $B(5,12)$. $$d = \sqrt{(5 - 0)^2 + (12 - 0)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$$ 6. Example 4: Find the distance between $A(3,4)$ and $B(3,4)$. $$d = \sqrt{(3 - 3)^2 + (4 - 4)^2} = \sqrt{0 + 0} = 0$$ 7. Example 5: Find the distance between $A(-2,7)$ and $B(1,3)$. $$d = \sqrt{(1 - (-2))^2 + (3 - 7)^2} = \sqrt{3^2 + (-4)^2} = 5$$ 8. Example 6: Find the distance between $A(6,8)$ and $B(2,4)$. $$d = \sqrt{(2 - 6)^2 + (4 - 8)^2} = \sqrt{(-4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$$ 9. Example 7: Find the distance between $A(0,-3)$ and $B(4,0)$. $$d = \sqrt{(4 - 0)^2 + (0 - (-3))^2} = \sqrt{16 + 9} = 5$$ 10. Example 8: Find the distance between $A(-5,-5)$ and $B(-1,-1)$. $$d = \sqrt{(-1 - (-5))^2 + (-1 - (-5))^2} = \sqrt{4^2 + 4^2} = 4\sqrt{2}$$ 11. Example 9: Find the distance between $A(7,1)$ and $B(10,5)$. $$d = \sqrt{(10 - 7)^2 + (5 - 1)^2} = \sqrt{3^2 + 4^2} = 5$$ 12. Example 10: Find the distance between $A(-3,0)$ and $B(0,-4)$. $$d = \sqrt{(0 - (-3))^2 + (-4 - 0)^2} = \sqrt{3^2 + (-4)^2} = 5$$