8,9

Find the distance between the parallel lines $y=-3 x+5$ and $y=-3 x-4$.

#
The first method.

Since the coordinates of point $A(0 ; 5)$ satisfy the equation $y=-3 x+5$, this point lies on the first line. Since the lines are parallel, the distance between them is equal to the distance from point $A$ to the second line. Let's write the equation of this line in general form $(y+3 x+4=0)$ and use the formula for the distance between a point and a line:

$$
d=\frac{|5+3 \cdot 0+4|}{\sqrt{1^{2}+3^{2}}}=\frac{9}{\sqrt{10}}
$$

The second method.

Let the first line intersect the $O Y$ axis at point $A(0 ; 5)$, and the second line at point $B(0 ;-4)$. Then $A B=5$ - $(-4)=9$. If $\alpha$ is the acute angle between each of these lines and the $O X$ axis, then $\operatorname{tg} \alpha=3$. Then

$$
\cos \alpha=\frac{1}{\sqrt{1+\operatorname{tg}^{2} \alpha}}=\frac{1}{\sqrt{10}}
$$

Let $C$ be the projection of point $B$ onto the line $y=-3 x+5$. Then $\angle A B C=\alpha$, and the desired distance between the lines is equal to the length of the segment $B C$.

From the right triangle $A B C$, we find that

$$
B C=A B \cos \alpha=9 \cdot \frac{1}{\sqrt{10}}=\frac{9}{\sqrt{10}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_1b17058527caf5b5d058g-34.jpg?height=574&width=412&top_left_y=-1&top_left_x=826)

## Answer