3.11. The equation of one side of the square \(ABCD-AB\) is \(x + 3y - 5 = 0\). Form the general equation of the other three sides of the square, given that \(P(-1; 0)\) is the point of intersection of its diagonals.

![](https://cdn.mathpix.com/cropped/2024_05_22_f5a9bb7960904462604bg-119.jpg?height=282&width=266&top_left_y=364&top_left_x=147)

Fig. 3.5
The solution is as follows. Let's make a schematic drawing (Fig. 3.5).

1. Find the coordinates of point $K$, the intersection of $AB$ with the perpendicular $PK$ dropped from point $P$ to $AB$, i.e., having the current vector $\overrightarrow{PM}=\{x+1 ; y-0\}$ and perpendicular to the line $AB$ (parallel to its normal vector $\vec{n}=\{1 ; 3\}$). Writing the condition of collinearity of $\overrightarrow{PM}$ and $\vec{n}$, we get the equation of the perpendicular $PK$:

$$
(x+1) / 1=y / 3, 3x-y+3=0
$$

Multiply the equation of $AB$ by 3 and subtract the equation of $PK$ from the result, then we get $10y=18, y_{K}=1.8, x_{K}=-0.4$. Thus, $K(-0.4 ; 1.8)$.

2. Determine the coordinates of points $A\left(x_{A} ; y_{A}\right)$ and $B\left(x_{B} ; y_{B}\right)$. First, we calculate $|KP|^{2}$:

$$
|KP|^{2}=(-1-(-0.4))^{2}+(0-1.8)^{2}=3.6 \text{. }
$$

Now write the equality of the squares of the distances

$$
|AK|^{2}=|BK|^{2}=|PK|^{2}
$$

substituting into them the expression for $x=5-3y$ found from the equation of the line $AB$. We have

$$
(5-3y+0.4)^{2}+(y-1.8)^{2}=3.6 \cdot 10y^{2}-36y+28.8=0
$$

Solving the quadratic equation, we get $y_{1}=2.4 ; y_{2}=1.2$. Substitute the found $y_{1}$ and $y_{2}$ into the equation of the line $AB$ and obtain the corresponding abscissas of points $A$ and $B$:

$$
x_{A}=x_{1}=-2.2, \quad y_{A}=y_{1}=2.4 ; \quad x_{B}=x_{2}=1.4, \quad y_{B}=y_{2}=1.2 \text{. }
$$

Thus, $A(-2.2 ; 2.4), B(1.4 ; 1.2)$.

3. Find the coordinates of points $C$ and $D$, using the formulas for the division of a segment in a given ratio and the fact that point $P$ is the midpoint of $AC$ and $BD$:

$$
\begin{aligned}
& x_{P}=\frac{x_{A}+x_{C}}{2}=-1 ; y_{P}=\frac{y_{A}+y_{C}}{2}=0 ; \quad C(-0.2 ;-2.4) \\
& x_{P}=\frac{x_{B}+x_{D}}{2}=-1 ; y_{P}=\frac{y_{B}+y_{D}}{2}=0 ; \quad D(-3.4 ;-1.2)
\end{aligned}
$$

4. Find the equation of the line $AD$ from the condition of collinearity of the current vector of this line

$$
\overline{AM}=\{x+2.2 ; y-2.4\}
$$

and the normal vector of the line $AB \vec{n}=\{1 ; 3\}$:

$$
(x+2.2) / 1=(y-2.4) / 3
$$

Transform the equation of $AD$ to the general form: $3x-y+9=0$.

5. Proceeding similarly to the previous point, we get the equation of $BC$.

$$
\begin{aligned}
\overrightarrow{BM} & =\{x-1.4 ; y-1.2\} \| \vec{n}=\{1 ; 3\}, \\
(x-1.4) / 1 & =(y-1.2) / 3 \text{ or } 3x-y-3=0 .
\end{aligned}
$$

6. Find the equation of $CD$ as the equation of a line passing through two points (formula (3.8)):

$$
(x-0.2) /(-3.4-0.2)=(y-(-2.4)) /(-1.2-(-2.4))
$$

In the denominators of the fractions in the last equality, the coordinates of the direction vector $\vec{a}=\{-3.6 ; 1.2\}$ can be replaced by the coordinates of the collinear vector $\ddot{a}^{\prime}=\{-3 ; 1\}$, which is obtained from $\vec{a}$ by dividing by 1.2. The equation of $CD$ then takes the form $(x-0.2) /-3=(y-(-2.4)) / 1$ or $x+3y+7=0$ - the general equation of the line.