6. What is the smallest area that triangle $O A B$ can have, if its sides $O A_{\text {and }} O B$ lie on the graph of the function $y=2|x|-x+1$, and the line $A B_{\text {passes through the point }} M(0 ; 2)$?
Solution.

$S_{\text {AOB }}=S_{\text {AOM }}+S_{\text {BOM }}, \quad A(a ; a+1), B(b ;-3 b+1)$,

$S_{AOM}=\frac{1}{2} OM \cdot a, \quad S_{BOM}=\frac{1}{2} OM \cdot(-b), \quad OM=1$,

$S_{AOB}=\frac{a-b}{2}$.

The line $AB_{\text {passes through }}$ point $\mathrm{M}$, its equation is $y=k x+2$. Express the variables $a$ and $\mathrm{b}$ in terms of the parameter $\mathrm{k}$, substituting the coordinates of points $A$ and $B$ into the equation of the line $AB$: $a=k a+1, \quad a=\frac{1}{1-k},-3 b=k b+1, b=-\frac{1}{k+3} . \quad$ Express the area of the triangle

$$
S_{AOB}=\frac{1}{2}\left(\frac{1}{1-k}+\frac{1}{k+3}\right)=\frac{2}{3-2 k-k^{2}}=\frac{2}{4-(k+1)^{2}}
$$

$$
4-(k+1)^{2} \leq 4
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_9e353629715b7c8c09e0g-09.jpg?height=632&width=665&top_left_y=375&top_left_x=1301)

$$
S_{AOB}=\frac{2}{4-(k+1)^{2}} \geq \frac{1}{2} . \text { The minimum value } \min S_{\text {AOB }}=\frac{1}{2} \text { when } k=-1 .
$$

Answer: 0.5 .