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 $y$?
+14933 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?
Hello Guest!
\(m_{BC}=\frac{y_B-y_C}{x_B-x_C}=\frac{-8-2}{1-(-3)}=-\frac{10}{4}=-\frac{5}{2}\\ m_{h_A}=-\frac{1}{m_{BC}}=\frac{2}{5}\\ f_{h_A}(x)=m_{h_a}(x-x_A)+y_A\\ f_{h_A}(x)=\frac{2}{5}x-\frac{4}{5}+3\\ f_{h_A}(x)=\frac{2}{5}x\color{blue}+\frac{11}{5}\)
The value of y is \(\color{blue}\frac{11}{5}\).
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