Orthocenter Abc D2Bc49

1. **Problem Statement:** We need to find and graph the orthocenter of $\triangle ABC$. The orthocenter is the point where the three altitudes of a triangle intersect. 2. **Recall:** An altitude of a triangle is a perpendicular segment from a vertex to the line containing the opposite side. 3. **Step 1: Identify vertices coordinates.** Since the triangle is described but no exact coordinates are given, let's assign coordinates based on the description: - Let $A = (2,5)$ (top vertex) - Let $B = (2,1)$ (vertical segment $AB$) - Let $C = (6,2)$ (point to the right, so $BC$ slopes downward) 4. **Step 2: Find equations of sides:** - $AB$ is vertical: $x=2$ - $BC$: slope $m_{BC} = \frac{2-1}{6-2} = \frac{1}{4}$, equation $y-1=\frac{1}{4}(x-2)$ or $y=\frac{1}{4}x+\frac{1}{2}$ - $AC$: slope $m_{AC} = \frac{5-2}{2-6} = \frac{3}{-4} = -\frac{3}{4}$, equation $y-5 = -\frac{3}{4}(x-2)$ or $y = -\frac{3}{4}x + 6.5$ 5. **Step 3: Find altitudes:** - Altitude from $C$ is perpendicular to $AB$ (vertical), so horizontal line through $C$: $y=2$ - Altitude from $B$ is perpendicular to $AC$. Slope of $AC$ is $-\frac{3}{4}$, so altitude slope is $\frac{4}{3}$. Equation through $B(2,1)$: $$y-1=\frac{4}{3}(x-2)$$ 6. **Step 4: Find orthocenter as intersection of two altitudes:** - Solve system: $$y=2$$ $$y-1=\frac{4}{3}(x-2)$$ Substitute $y=2$: $$2-1=\frac{4}{3}(x-2) \Rightarrow 1=\frac{4}{3}(x-2)$$ Multiply both sides by $\cancel{3}$: $$\cancel{3} \times 1 = 4(x-2) \Rightarrow 3 = 4(x-2)$$ Divide both sides by $\cancel{4}$: $$\frac{3}{\cancel{4}} = x-2$$ $$x = 2 + \frac{3}{4} = \frac{11}{4} = 2.75$$ 7. **Step 5: Orthocenter coordinates:** $$H = \left(\frac{11}{4}, 2\right)$$ 8. **Step 6: Graph the triangle and orthocenter:** - Points: $A(2,5)$, $B(2,1)$, $C(6,2)$ - Orthocenter $H(2.75, 2)$ lies inside the triangle. **Final answer:** The orthocenter of $\triangle ABC$ is at $$\boxed{\left(\frac{11}{4}, 2\right)}$$.