1. **Stating the problem:** We are given two lines: $$y = \sqrt{3}x$$ and $$y = -\sqrt{3}x + 6$$ We need to find: - The intersection points $A$ of these two lines. - The intersection point $B$ of the line $y = -\sqrt{3}x + 6$ with the x-axis. - The distances $OA$, $OB$, and $AB$, where $O$ is the origin $(0,0)$. 2. **Finding point A (intersection of the two lines):** Set the two equations equal since at $A$ both $y$ values are the same: $$\sqrt{3}x = -\sqrt{3}x + 6$$ Add $\sqrt{3}x$ to both sides: $$\sqrt{3}x + \sqrt{3}x = 6$$ $$2\sqrt{3}x = 6$$ Divide both sides by $2\sqrt{3}$: $$x = \frac{6}{2\sqrt{3}} = \frac{6}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{6\sqrt{3}}{2 \times 3} = \frac{6\sqrt{3}}{6} = \sqrt{3}$$ Substitute $x = \sqrt{3}$ into $y = \sqrt{3}x$: $$y = \sqrt{3} \times \sqrt{3} = 3$$ So, point $A$ is: $$A(\sqrt{3}, 3)$$ 3. **Finding point B (intersection of $y = -\sqrt{3}x + 6$ with x-axis):** At the x-axis, $y=0$, so: $$0 = -\sqrt{3}x + 6$$ Solve for $x$: $$\sqrt{3}x = 6$$ $$x = \frac{6}{\sqrt{3}} = \frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}$$ So, point $B$ is: $$B(2\sqrt{3}, 0)$$ 4. **Calculating distances:** - Distance $OA$ from origin $O(0,0)$ to $A(\sqrt{3}, 3)$: $$OA = \sqrt{(\sqrt{3} - 0)^2 + (3 - 0)^2} = \sqrt{3 + 9} = \sqrt{12} = 2\sqrt{3}$$ - Distance $OB$ from origin $O(0,0)$ to $B(2\sqrt{3}, 0)$: $$OB = \sqrt{(2\sqrt{3} - 0)^2 + (0 - 0)^2} = \sqrt{4 \times 3} = \sqrt{12} = 2\sqrt{3}$$ - Distance $AB$ between points $A(\sqrt{3}, 3)$ and $B(2\sqrt{3}, 0)$: $$AB = \sqrt{(2\sqrt{3} - \sqrt{3})^2 + (0 - 3)^2} = \sqrt{(\sqrt{3})^2 + (-3)^2} = \sqrt{3 + 9} = \sqrt{12} = 2\sqrt{3}$$ **Final answers:** - $A(\sqrt{3}, 3)$ - $B(2\sqrt{3}, 0)$ - $OA = OB = AB = 2\sqrt{3}$