7. Find the angle between two lines:
1) $y=-2 x, y=3 x+5$
2) $4 x+2 y-5=0,6 x+3 y+1=0$
3) $\sqrt{3} x-y-2=0, \quad \sqrt{3} x+y-1=0$
4) $2 x+3 y-1=0,3 x-2 y+1=0$
5) $3 x+4 y-5=0,5 x-2 y+7=0$
6) $y=2 x+3, y-4=0$
7) $2 x-4 y+7=0, x-5=0$.
Solution.

1) $k_{1}=-2, k_{2}=3$, therefore, by formula (1) we get (Fig. $18, a)$ :

$$
\operatorname{tg} \varphi=\frac{k_{2}-k_{1}}{1+k_{1} k_{2}}=\frac{3-(-2)}{1+(-2) \cdot 3}=-1, \varphi=135^{\circ}
$$

2) $k_{1}=-2, k_{2}=-2$, therefore,

$$
\operatorname{tg} \varphi=\frac{-2-(-2)}{1+(-2) \cdot(-2)}=0, \varphi=0^{\circ}
$$

i.e., these lines are parallel;
3) $k_{1}=\sqrt{3}, k_{2}=-\sqrt{3}$, therefore, by the formula we get

$$
\operatorname{tg} \varphi=\frac{-\sqrt{3}-\sqrt{3}}{1+(-\sqrt{3)} \cdot \sqrt{3}}=\sqrt{3}, \varphi=60^{\circ}
$$

4) $k_{1}=-\frac{2}{3} ; k_{2}=\frac{3}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_1b9a1c1720bae4af5bc5g-031.jpg?height=387&width=356&top_left_y=401&top_left_x=145)
a)

![](https://cdn.mathpix.com/cropped/2024_05_22_1b9a1c1720bae4af5bc5g-031.jpg?height=378&width=461&top_left_y=398&top_left_x=597)
8)

It is easy to see that $k_{1} k_{2}=-1$, therefore, these two lines form an angle of $90^{\circ}$, i.e., they are perpendicular to each other;

5) solve both equations for $y$. We get the equations of the lines with the slopes

$$
y=-\frac{3}{4} x+\frac{5}{4}, y=\frac{5}{2} x+\frac{7}{2}
$$

From this, it is clear that $k_{1}=-\frac{3}{4}, k_{2}=\frac{5}{2}$, therefore,

$$
\operatorname{tg} \varphi=\frac{\frac{5}{2}\left(-\frac{3}{4}\right)}{1+\left(-\frac{3}{4}\right) \cdot \frac{5}{2}}=-\frac{26}{7} \approx-3.714, \varphi \approx 105^{\circ} 4^{\prime}
$$

6) $k_{1}=2, k_{2}=0$, therefore,

$$
\operatorname{tg} \varphi=\frac{0-2}{1+2 \cdot 0}=-2, \varphi \approx 116^{\circ} 34^{\prime}
$$

7) $k_{1}=\operatorname{tg} \alpha_{1}=\frac{1}{2}$, hence $\alpha_{1}=\operatorname{arctg} \frac{1}{2}$ (Fig. 18, b).

The line $x-5=0$ is parallel to the Oy axis or perpendicular to the Ox axis, therefore, $\alpha_{2}=90^{\circ}$. Thus,

$$
\varphi=\alpha_{2}-\alpha_{1}=90^{\circ}-\operatorname{arctg} \frac{1}{2} \approx 90^{\circ}-26^{\circ} 34^{\prime}=63^{\circ} 26^{\prime}
$$