[Coordinate method on the plane]

Given points $A(6 ; 1), B(-5 ;-4), C(-2 ; 5)$. Form the equation of the line on which the height of triangle $A B C$, drawn from vertex $A$, lies.

#
Apply the condition of perpendicularity of two lines ( $k_{1} \cdot k_{2}=-1$ ).

## Answer

$$
x+3 y-9=0
$$

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

$$
\frac{y-(-4)}{5-(-4)}=\frac{x-(-5)}{-2-(-5)}, \text { or } y=3 x+11
$$

Then its slope $k_{1}=3$. If $k_{2}$ is the slope of the line perpendicular to the given one, then $k_{1} \cdot$ $k_{2}=-1$. Therefore,

$$
k_{2}=-\frac{1}{k_{1}}=-\frac{1}{3}
$$

The equation of the desired line, containing the height $A H$ of triangle $A B C$, will be found using point $A(6 ; 1)$ and the slope $k_{2}=-\frac{1}{3}$ :

$$
y-1=-\frac{1}{3}(x-6)
$$

Write the equation of this line in general form:

$$
x+3 y-9=0
$$