Triangle $ABC$ has vertices $A(2, 3),$ $B(1, -8)$ and $C(-3, 2).$ The line containing the altitude through $A$ intersects the point $(0, y).$ What is the value of $y$?

Answer 1

The altitude through $A$ is perpendicular to line $BC.$ First, we find the slope of the line containing $BC,$ which is $\frac{-8 - 3}{1 - (-3)} = \frac{-11}{4}.$ Therefore, the slope of the line containing the altitude through $A$ is $\frac{4}{11}.$ Since this line passes through $(2, 3),$ its equation is given by \[y - 3 = \frac{4}{11}(x - 2).\] Substituting $x = 0,$ we get \[y-3 = -\frac{8}{11} \quad\Rightarrow\quad y = \boxed{\frac{25}{11}}.\]