Question

In the triangle ABC with vertices A (2, 3), B (4, −1) and C (1, 2), find the equation and the length of the altitude from the vertex A.

Answer 1

Equation of side BC:

$$\begin{gathered} y+1=\frac{2+1}{1-4}\left(x-4\right) \\ \Rightarrow x+y-3=0 \end{gathered}$$

The equation of the altitude that is perpendicular to $x+y-3=0$ is $x-y+\lambda =0$.

Line $x-y+\lambda =0$ passes through (2, 3).

$\therefore 2-3+\lambda =0\Rightarrow \lambda =1$

Thus, the equation of the altitude from the vertex A (2, 3) is $x-y+1=0$.

Let d be the length of the altitude from A (2, 3).

$$\begin{gathered} d=\left|\frac{2+3-3}{\sqrt{1^2+1^2}}\right| \\ \Rightarrow d=\sqrt{2} \end{gathered}$$

Hence, the required distance is $\sqrt{2}$.