1. **Problem Statement:** Given triangle $ABC$ with points $A(4,3)$, $B(-1,-1)$, and $C(2,2)$, find: - The centroid $G$ of the triangle. - Midpoints $K$ and $H$ of segments $CB$ and $AB$ respectively. - Verify that lines $BK$, $AI$, and $CH$ intersect at a common point. - Compute vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$. - Calculate lengths $AB$, $AC$, dot product $\overrightarrow{AB} \cdot \overrightarrow{AC}$. - Find $\sin(\widehat{AB;AC})$ and $\cos(\widehat{AB;AC})$. - Define line $D$ passing through $A(4,0)$ with direction vector $\overrightarrow{n}$. 2. **Step 1: Find centroid $G$** The centroid $G$ of a triangle with vertices $A(x_1,y_1)$, $B(x_2,y_2)$, $C(x_3,y_3)$ is given by: $$G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$$ Substitute: $$G = \left(\frac{4 + (-1) + 2}{3}, \frac{3 + (-1) + 2}{3}\right) = \left(\frac{5}{3}, \frac{4}{3}\right)$$ 3. **Step 2: Find midpoints $K$ and $H$** - $K$ is midpoint of $CB$: $$K = \left(\frac{2 + (-1)}{2}, \frac{2 + (-1)}{2}\right) = \left(\frac{1}{2}, \frac{1}{2}\right)$$ - $H$ is midpoint of $AB$: $$H = \left(\frac{4 + (-1)}{2}, \frac{3 + (-1)}{2}\right) = \left(\frac{3}{2}, 1\right)$$ 4. **Step 3: Verify concurrency of lines $BK$, $AI$, and $CH$** - $B(-1,-1)$, $K(0.5,0.5)$ - $A(4,3)$, $I$ is not given explicitly, but assuming $I$ is midpoint of $BC$ or another point on $AI$. - $C(2,2)$, $H(1.5,1)$ Since $K$ and $H$ are midpoints, lines $BK$, $CH$, and $AI$ (median) concur at centroid $G$. 5. **Step 4: Compute vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$** $$\overrightarrow{AB} = B - A = (-1 - 4, -1 - 3) = (-5, -4)$$ $$\overrightarrow{AC} = C - A = (2 - 4, 2 - 3) = (-2, -1)$$ 6. **Step 5: Calculate lengths $AB$ and $AC$** $$AB = \sqrt{(-5)^2 + (-4)^2} = \sqrt{25 + 16} = \sqrt{41}$$ $$AC = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}$$ 7. **Step 6: Calculate dot product $\overrightarrow{AB} \cdot \overrightarrow{AC}$** $$\overrightarrow{AB} \cdot \overrightarrow{AC} = (-5)(-2) + (-4)(-1) = 10 + 4 = 14$$ 8. **Step 7: Calculate $\cos(\widehat{AB;AC})$ and $\sin(\widehat{AB;AC})$** Using the dot product formula: $$\cos(\theta) = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|AB||AC|} = \frac{14}{\sqrt{41} \times \sqrt{5}} = \frac{14}{\sqrt{205}}$$ Calculate $\sin(\theta)$ using: $$\sin(\theta) = \sqrt{1 - \cos^2(\theta)} = \sqrt{1 - \left(\frac{14}{\sqrt{205}}\right)^2} = \sqrt{1 - \frac{196}{205}} = \sqrt{\frac{9}{205}} = \frac{3}{\sqrt{205}}$$ 9. **Step 8: Define line $D$ passing through $A(4,0)$ with direction vector $\overrightarrow{n}$** Assuming $\overrightarrow{n}$ is normal vector to $AB$ or $AC$, for example: $$\overrightarrow{n} = (n_x, n_y)$$ Equation of line $D$: $$\overrightarrow{r} = (4,0) + t \overrightarrow{n}$$ **Final answers:** - Centroid $G = \left(\frac{5}{3}, \frac{4}{3}\right)$ - Midpoints $K = \left(\frac{1}{2}, \frac{1}{2}\right)$, $H = \left(\frac{3}{2}, 1\right)$ - Vectors $\overrightarrow{AB} = (-5,-4)$, $\overrightarrow{AC} = (-2,-1)$ - Lengths $AB = \sqrt{41}$, $AC = \sqrt{5}$ - Dot product $\overrightarrow{AB} \cdot \overrightarrow{AC} = 14$ - $\cos(\widehat{AB;AC}) = \frac{14}{\sqrt{205}}$, $\sin(\widehat{AB;AC}) = \frac{3}{\sqrt{205}}$ - Line $D$ passes through $A(4,0)$ with direction vector $\overrightarrow{n}$ as defined.