1. **Problem statement:** Find the equation and length of the altitude from vertex A in triangle ABC with vertices A(2,3), B(4,-1), and C(1,2). 2. **Recall:** The altitude from a vertex is a perpendicular line from that vertex to the opposite side. Here, the altitude from A is perpendicular to line BC. 3. **Find the slope of BC:** $$m_{BC} = \frac{y_C - y_B}{x_C - x_B} = \frac{2 - (-1)}{1 - 4} = \frac{3}{-3} = -1$$ 4. **Slope of altitude from A:** Since altitude is perpendicular to BC, its slope is the negative reciprocal of $m_{BC}$: $$m_{altitude} = -\frac{1}{m_{BC}} = -\frac{1}{-1} = 1$$ 5. **Equation of altitude from A:** Using point-slope form with point A(2,3) and slope 1: $$y - 3 = 1(x - 2)$$ Simplify: $$y = x + 1$$ 6. **Length of altitude:** Distance from A to line BC. Equation of line BC: Using point B(4,-1) and slope $-1$: $$y - (-1) = -1(x - 4) \Rightarrow y + 1 = -x + 4 \Rightarrow x + y - 3 = 0$$ Distance from point A$(x_0,y_0) = (2,3)$ to line $Ax + By + C = 0$ is: $$d = \frac{|A x_0 + B y_0 + C|}{\sqrt{A^2 + B^2}}$$ Here, $A=1$, $B=1$, $C=-3$: $$d = \frac{|1 \cdot 2 + 1 \cdot 3 - 3|}{\sqrt{1^2 + 1^2}} = \frac{|2 + 3 - 3|}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$$ **Final answers:** - Equation of altitude from A: $$y = x + 1$$ - Length of altitude from A: $$\sqrt{2}$$