1. **State the problem:** Find the coordinates of the orthocenter of the triangle with vertices at points $A(0,0)$, $B(8,2)$, and $C(2,3)$. The orthocenter is the point where the three altitudes of the triangle intersect. 2. **Recall the formula and rules:** - An altitude is a perpendicular line from a vertex to the opposite side. - To find the orthocenter, find equations of two altitudes and solve their system. 3. **Find slope of side $BC$:** $$m_{BC} = \frac{3 - 2}{2 - 8} = \frac{1}{-6} = -\frac{1}{6}$$ 4. **Find slope of altitude from $A$ (perpendicular to $BC$):** $$m_{A} = -\frac{1}{m_{BC}} = -\frac{1}{-\frac{1}{6}} = 6$$ 5. **Equation of altitude from $A(0,0)$:** $$y = 6x$$ 6. **Find slope of side $AC$:** $$m_{AC} = \frac{3 - 0}{2 - 0} = \frac{3}{2}$$ 7. **Find slope of altitude from $B$ (perpendicular to $AC$):** $$m_{B} = -\frac{1}{m_{AC}} = -\frac{1}{\frac{3}{2}} = -\frac{2}{3}$$ 8. **Equation of altitude from $B(8,2)$:** Use point-slope form: $$y - 2 = -\frac{2}{3}(x - 8)$$ Simplify: $$y = -\frac{2}{3}x + \frac{16}{3} + 2 = -\frac{2}{3}x + \frac{16}{3} + \frac{6}{3} = -\frac{2}{3}x + \frac{22}{3}$$ 9. **Find intersection of altitudes (orthocenter):** Set $$6x = -\frac{2}{3}x + \frac{22}{3}$$ Multiply both sides by 3 to clear denominators: $$3 \times 6x = 3 \times \left(-\frac{2}{3}x + \frac{22}{3}\right)$$ $$18x = -2x + 22$$ Add $2x$ to both sides: $$18x + 2x = 22$$ $$20x = 22$$ Divide both sides by 20: $$x = \frac{22}{20} = \frac{11}{10}$$ 10. **Find $y$ coordinate:** $$y = 6x = 6 \times \frac{11}{10} = \frac{66}{10} = \frac{33}{5}$$ **Final answer:** The orthocenter is at $$\left(\frac{11}{10}, \frac{33}{5}\right)$$.