+1768 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 $y$?
0 The altitude will have the negative inverse of the slope of \(\overline{BC}\) becasue it is perpedicular. Now with having a point\(A\) on the line and its slope, we know everything about the line therefore:
\(\frac{2-(-8)}{-3-1}=-\frac{5}{2} \)
negative inverse=\(\frac{2}{5}\)
using point-slope form:
\(y-y_1=x(x-x_1)\\ y-3=x(\frac{2}{5}-2) \text{ <----lets plug in (0,y)}\\ y-3=0\\ \boxed{y=3}\)
