113. Given a triangle with vertices $A(4 ; 6), B(-3 ; 0)$, $C(2 ;-3)$. Find the angles of the triangle, the equation of the bisector $A D$, the equation of the altitude $C E$, and the point of their intersection.
Solution. The angular coefficients of the lines $AB, BC, AC$ (Fig. 19) will be found using formula (6) §5:

$$
\begin{gathered}
k_{A B}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{0-6}{-3-4}=\frac{6}{7} ; k_{B C}=\frac{-3-0}{2-(-3)}=-\frac{3}{5} ; k_{A C}= \\
=\frac{-3-6}{2-4}=\frac{9}{2} .
\end{gathered}
$$

Now we will find the angles of the triangle using formula (1):

$$
\begin{gathered}
\operatorname{tg} A=\frac{k_{A C}-k_{A B}}{1+k_{A C} \cdot k_{A B}}=\frac{\frac{9}{2}-\frac{6}{7}}{1+\frac{9}{2} \cdot \frac{6}{7}}=\frac{3}{4}, A=\operatorname{arctg} 0.75 \approx 36^{\circ} 52' \\
\operatorname{tg} B=\frac{-\frac{3}{5}-\frac{6}{7}}{1+\left(-\frac{3}{5}\right) \frac{6}{7}}=3, B=\operatorname{arctg} 3 \approx 71^{\circ} 34' \\
\operatorname{tg} C=\frac{-\frac{3}{5}-\frac{9}{2}}{1+\left(-\frac{3}{5}\right) \frac{9}{2}}=3, C=\operatorname{arctg} 3
\end{gathered}
$$

Therefore, $B=C \approx 71^{\circ} 34'$, i.e., $\triangle ABC$ is isosceles.

To find the equation of the bisector of angle $A$, we first need to write the equations of the sides $AB$ and $AC$ of the given triangle.

Equation of $AB$:

$$
\frac{y-6}{0-6}=\frac{x-4}{-3-4}
$$

or

$$
6 x-7 y+18=0
$$

Equation of $AC$:

$$
\frac{y-6}{-3-6}=\frac{x-4}{2-4}
$$

or

$$
9 x-2 y-24=0
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_1b9a1c1720bae4af5bc5g-033.jpg?height=376&width=429&top_left_y=1225&top_left_x=701)

Fig. 19

Let point $M(X ; Y)$ lie on the bisector $AD$ (where $X$ and $Y$ are the current coordinates of the bisector), then it will be equidistant from the sides $AB$ and $AC$ of angle $A$.

The distance $d_{1}$ from point $M(X ; Y)$ to side $AB$, considering that the origin and point $M$ are on the same side of line $AB\left(\hat{\delta}_{1}<0\right)$, can be written as:

$$
d_{1}=-\delta_{1}=-\frac{6 X-7 Y+18}{-\sqrt{36+49}}
$$

Similarly, the distance $d_{2}$ from point $M$ to line $AC$:

$$
d_{2}=-\delta_{2}=-\frac{9 X-2 Y-24}{\sqrt{81+4}}
$$

Therefore, the equation of the bisector $AD$:

$$
d_{1}=d_{2}, \text { i.e., }-\frac{6 X-7 Y+18}{-\sqrt{85}}=-\frac{9 X-2 Y-24}{\sqrt{85}}
$$

or after simplifications:

$$
5 X-3 Y-2=0
$$

Now we will write the equation of $CE$. By condition, line $CE$ is perpendicular to line $AB$, so $k_{CE}=-\frac{1}{k_{AB}}$, i.e., $k_{CE}=-\frac{7}{6}$.

Considering also that line $CE$ passes through point $C(2 ;-3)$, we will write the desired equation of line $CE$:

$$
y-(-3)=-\frac{7}{6}(x-2) \text { or } 7 x+6 y+4=0
$$

We will find the intersection point $F$ of lines $AD$ and $CE$, for this we will solve the system of equations:

![](https://cdn.mathpix.com/cropped/2024_05_22_1b9a1c1720bae4af5bc5g-034.jpg?height=129&width=768&top_left_y=1062&top_left_x=207)

i.e., $x=0$, therefore, $y=-\frac{2}{3}$.

Thus, the desired intersection point of the specified lines $\boldsymbol{F}\left(0 ;-\frac{2}{3}\right)$.