Find the leg of an altitude in a triangle

Question

The vertices of $ABC$ are $A(8,5)$, $B(0,1)$ and $C(9, -2)$. Find the point where the altitude from $A$ intersects $BC$.

---

Progress: I have found the equation of the altitude from A to BC, and that is $3x-y=19$.

Answer 1

hint:

1). First find the equation of the line $BC$.

2). Suppose $P(h,k)$ is the point where the altitude meets $BC$. Then $P$ will satisfy the equation you found in step 1).

3). Now compute the slope of $AP$. Since $AP$ is perpendicular to $BC$. So the product of their slopes must be $-1$.

You will now have two equations in two unknowns $h,k$. Solve it.

Answer 2

Equation of $BC$ $$\frac{y-0}{x-1}=\frac{(-2)-1}{9-0}$$ $$\frac{y}{1}=\frac{1-x}{3}=t\Rightarrow y=t,x=1-3t$$

slope of $BC=\frac{-1}{3} \Rightarrow$ slope of $AL=3$ ($L$ is the point where the altitude from $A$ intersects $BC$)

since $L$ is on $BC$ $\frac{5-t}{8-(1-3t)}=3\Rightarrow t=-\frac{8}{5}$

then $L\equiv(\frac{29}{5},-\frac{8}{5})$