1. **Problem statement:** Given points A(2,3), B(-1,-4), and C(0,-2), find: a. The distance |AB| b. The slope of line segment AB c. The midpoint of AB d. The angle between AB and AC e. The point two-thirds of the way from A to B 2. **Formulas and rules:** - Distance between two points $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$ is $$|P_1P_2|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$ - Slope of line segment between $P_1$ and $P_2$ is $$m=\frac{y_2-y_1}{x_2-x_1}$$ - Midpoint of segment $P_1P_2$ is $$\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$$ - Angle $\theta$ between vectors $\vec{u}$ and $\vec{v}$ is $$\cos\theta=\frac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|}$$ where $\vec{u}\cdot\vec{v}$ is the dot product - A point $P$ fraction $t$ along $P_1P_2$ is $$P=(x_1+t(x_2-x_1),y_1+t(y_2-y_1))$$ 3. **Calculations:** a. Distance |AB|: $$|AB|=\sqrt{(-1-2)^2+(-4-3)^2}=\sqrt{(-3)^2+(-7)^2}=\sqrt{9+49}=\sqrt{58}$$ b. Slope of AB: $$m=\frac{-4-3}{-1-2}=\frac{-7}{-3}=\frac{\cancel{-7}}{\cancel{-3}}=\frac{7}{3}$$ c. Midpoint of AB: $$\left(\frac{2+(-1)}{2},\frac{3+(-4)}{2}\right)=\left(\frac{1}{2},\frac{-1}{2}\right)$$ d. Angle between AB and AC: Vectors: $$\vec{AB}=(-1-2,-4-3)=(-3,-7)$$ $$\vec{AC}=(0-2,-2-3)=(-2,-5)$$ Dot product: $$\vec{AB}\cdot\vec{AC} = (-3)(-2)+(-7)(-5)=6+35=41$$ Magnitudes: $$|\vec{AB}|=\sqrt{(-3)^2+(-7)^2}=\sqrt{9+49}=\sqrt{58}$$ $$|\vec{AC}|=\sqrt{(-2)^2+(-5)^2}=\sqrt{4+25}=\sqrt{29}$$ Angle: $$\cos\theta=\frac{41}{\sqrt{58}\sqrt{29}}=\frac{41}{\sqrt{1682}}$$ $$\theta=\cos^{-1}\left(\frac{41}{\sqrt{1682}}\right)$$ e. Point two-thirds from A to B ($t=\frac{2}{3}$): $$P=\left(2+\frac{2}{3}(-1-2),3+\frac{2}{3}(-4-3)\right)=\left(2+\frac{2}{3}(-3),3+\frac{2}{3}(-7)\right)$$ $$=\left(2-2,3-\frac{14}{3}\right)=\left(0,\frac{9}{3}-\frac{14}{3}\right)=\left(0,-\frac{5}{3}\right)$$