[Coordinate method on the plane $]$ [ Orthocenter and orthotriangle ]

Given points $A(5, -1), B(4, -8), C(-4, -4)$. Find the coordinates of the intersection point of the altitudes of triangle $ABC$.
Applying the condition of perpendicularity of two lines ($k_{1} \cdot k_{2}=-1$), find the equations of the lines on which two altitudes of the triangle lie. Then find the coordinates of the intersection point of these lines by solving the corresponding system of equations.

## Solution

Let's find the equation of the line BC through two points:

$$
\frac{y-(-8)}{-4-(-8)}=\frac{x-4}{-4-4}, \text { or } y=-\frac{1}{2} x-6
$$

Then its slope $k_{1}=-\frac{1}{2}$. If $k_{2}$ is the slope of the line containing the altitude $AP$, then $k_{1} \cdot k_{2}=-1$. Therefore,

$$
k_{2}=-\frac{1}{k_{1}}=2
$$

The equation of the line containing the altitude $AP$ of triangle $ABC$ will be found using point $A(5, -1)$ and the slope $k_{2}=2$:

$$
y+1=2(x-5) \text {, or } y=2 x-11.
$$

Let's find the equation of the line $AC$ through two points:

$$
\frac{y-(-1)}{-4-(-1)}=\frac{x-5}{-4-5}, \text { or } y=\frac{1}{3} x-\frac{8}{3}
$$

Then its slope $k_{3}=\frac{1}{3}$. If $k_{4}$ is the slope of the line containing the altitude $BQ$, then $k_{4} \cdot k_{3}=-1$. Therefore,

$$
k_{4}=-\frac{1}{k_{3}}=-3
$$

The equation of the line containing the altitude $BQ$ of triangle $ABC$ will be found using point $B(4, -8)$ and the slope $k_{4}=-3$:

$$
y+8=-3(x-4), \text { or } y=-3 x+4
$$

The coordinates of the point $H$ of intersection of the altitudes of triangle $ABC$ will be found by solving the system of equations defining the lines $AP$ and $BQ$:

$$
\left\{\begin{array}{l}
y=2 x-11 \\
y=-3 x+4
\end{array}\right.
$$

We get: $x=3, y=-5$.

![](https://cdn.mathpix.com/cropped/2024_05_06_93426a107eaf3f76289cg-02.jpg?height=431&width=560&top_left_y=1694&top_left_x=752)

## Answer

$(3, -5)$.