Problem 9.6. The graph of the line $y=k x+l$ intersects the $O x$ axis at point $B$, the $O y$ axis at point $C$, and the graph of the function $y=1 / x$ at points $A$ and $D$, as shown in the figure. It turns out that $A B=B C=C D$. Find $k$, given that $O C=3$.

![](https://cdn.mathpix.com/cropped/2024_05_06_1f4981519479532effc1g-05.jpg?height=621&width=622&top_left_y=83&top_left_x=416)
Answer: 18.

![](https://cdn.mathpix.com/cropped/2024_05_06_1f4981519479532effc1g-05.jpg?height=616&width=618&top_left_y=802&top_left_x=418)

Fig. 6: to the solution of problem 9.6

Solution. Draw a line through point $D$ parallel to the $O y$ axis; let it intersect the $O x$ axis at point $D_{1}$ (Fig. 6). It is easy to see that segment $OC$ is the midline of triangle $B D D_{1}$, so $D D_{1}=2 \cdot O C=2 \cdot 3=6$.

Point $D$ lies on the graph of $y=1 / x$ and has an ordinate of 6, so its coordinates are $\left(\frac{1}{6} ; 6\right)$. The line $y=k x+l$ passes through points $C(0 ; 3)$ and $D\left(\frac{1}{6} ; 6\right)$, therefore,

$$
k=\frac{6-3}{\frac{1}{6}-0}=18
$$

Another solution. Notice that $l=O C=3$, i.e., the equation of the line is $y=k x+3$. Then the coordinates of points $A$ and $D$ are the solutions to the system of equations

$$
\left\{\begin{array}{l}
y=k x+3 \\
y=\frac{1}{x}
\end{array}\right.
$$

By equating the right-hand sides of the equations, this system reduces to solving the quadratic equation

$$
k x^{2}+3 x-1=0
$$

Let $x_{0}$ be the abscissa of point $D$. Then $-2 x_{0}$ is the abscissa of point $A$. Both of these numbers are roots of the given quadratic equation. From Vieta's theorem, it follows that

$$
\left\{\begin{array}{l}
x_{0}-2 x_{0}=-\frac{3}{k} \\
x_{0} \cdot\left(-2 x_{0}\right)=-\frac{1}{k}
\end{array}\right.
$$

This is equivalent to

$$
\left\{\begin{aligned}
x_{0} & =\frac{3}{k} \\
x_{0}^{2} & =\frac{1}{2 k}
\end{aligned}\right.
$$

From this, we derive that $9 / k^{2}=1 / 2 k$, i.e., $18 / k=1$ and $k=18$.