3.12. Given the equations of one side $A B$ of the rhombus $A B C D$: $x-3 y+10=0$, and one of its diagonals: $x+4 y-4=0$. The diagonals of the rhombus intersect at point $P(0 ; 1)$. Find the equations of the other sides of the rhombus, the distance from point $P$ to side $A B$, and the internal angles of the rhombus.
Solution. Let's make a schematic drawing (Fig. 3.6).

1. The equation of the diagonal $AC$ can be obtained from the condition of orthogonality of $AC$ and $BD$ and the fact that $BD$ passes through the point $P$,

$$
(x-0) / 1=(y-1) / 4, 4 x-y+1=0
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_f5a9bb7960904462604bg-121.jpg?height=206&width=506&top_left_y=131&top_left_x=366)

Fig. 3.6

2. The coordinates of points $A$ and $B$ will be found from the solutions of the corresponding systems of equations:

$$
\begin{aligned}
& \left\{\begin{array}{l}
x-3 y+10=0 \\
4 x-y+1=0
\end{array} \Rightarrow A(7 / 11 ; 39 / 11)\right. \\
& \left\{\begin{array}{l}
x-3 y+10=0 \\
x+4 y-4=0
\end{array} \Rightarrow B(-4 ; 2)\right.
\end{aligned}
$$

3. Since the diagonals of the rhombus are bisected by the point of intersection, we will find the coordinates of points $C\left(x_{C} ; y_{C}\right)$ and $D\left(x_{D} ; y_{D}\right)$ using the formulas for dividing a segment in a given ratio:

$$
\begin{aligned}
x_{P} & =\frac{x_{A}+x_{C}}{2} ; 0=\frac{7 / 11+x_{C}}{2} ; y_{P}=\frac{y_{A}+y_{C}}{2} ; \\
1 & =\frac{39 / 11+y_{C}}{2} ; C(-7 / 11 ;-17 / 11), \\
x_{P} & =\frac{x_{B}+x_{D}}{2} ; 0=\frac{-4+x_{D}}{2} ; y_{P}=\frac{y_{B}+y_{D}}{2} ; \\
1 & =\frac{2+y_{D}}{2} ; D(4 ; 0) .
\end{aligned}
$$

4. The equations of the sides $BC, CD$, and $AD$ will be found as the equations of lines passing through two given points (formula (3.8)):

$$
\begin{aligned}
& B C: \frac{x+4}{-7 / 11+4}=\frac{y-2}{-17 / 11-2}, \frac{x+4}{37 / 11}=\frac{y-2}{-39 / 11} \\
& \quad 39 x+37 y+82=0 . \\
& C D: \frac{x-4}{-7 / 11-4}=\frac{y-0}{-17 / 11-0}, \frac{x-4}{-51 / 11}=\frac{y}{-17 / 11}, \\
& \quad x-3 y-4=0 . \\
& A D: \frac{x-4}{7 / 11-4}=\frac{y-0}{39 / 11-0}, \frac{x-4}{-37 / 11}=\frac{y}{39 / 11} \\
& \quad 39 x+37 y-156=0 .
\end{aligned}
$$

5. The distance from the point $P(0 ; 1)$ to the line $AB$ with the equation $x-3 y+10=0$ will be found as follows:

$$
d=\frac{|0-3 \cdot 1+10|}{\sqrt{1^{2}+(-3)^{2}}}=\frac{7}{\sqrt{10}}=\frac{7 \sqrt{10}}{10} \approx 2.21
$$

6. The angle between the two lines $AB$ and $BC$ is determined from the formula (3.13):

$$
\begin{aligned}
& \cos (\angle A B C)=\frac{1 \cdot 39-3 \cdot 37}{\sqrt{1^{2}+(-3)^{2}} \sqrt{39^{2}+37^{2}}}=-\frac{72}{170} \approx 0.4235 \\
& \angle A B C=115^{\circ}, \angle B A D=\angle B C D=65^{\circ} .
\end{aligned}
$$