6. What is the smallest area that a right triangle can have, with its hypotenuse containing the point \( M(1 ; 0) \), and its legs lying on the lines \( y = -2 \) and \( x = 0 \)?

(12 points)
Solution. $A B: \quad y=k x+b, \quad M \in A B \Rightarrow b=-k$

$$
\begin{aligned}
& A(a ;-2) \in A B \Rightarrow-2=k a+b \Rightarrow a=1-\frac{2}{k} \\
& S_{A B C}=\frac{1}{2} B C \cdot C A=\frac{1}{2}(b+2) a=\frac{1}{2}(2-k)(1-2 / k) \\
& S^{\prime}=\frac{(2-k)(2+k)}{2 k^{2}}=0, k_{\min }=-2, \quad S_{\min }=4
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_3205747dae2ad15e28c2g-02.jpg?height=460&width=583&top_left_y=2009&top_left_x=1299)

## Answer: 4.