6. What is the smallest area that a right triangle can have, with its hypotenuse containing the point \( M(1; 3) \), and its legs lying on the lines \( y = x \) and \( y = -x \)?

(12 points)
Solution.

$A B: \quad y=k x+d, \quad M \in A B \Rightarrow d=3-k$

$A(a ;-a) \in A B \Rightarrow-a=k a+3-k \Rightarrow a=\frac{3-k}{k+1}$,

$B(b ; b) \in A B \Rightarrow b=k b+3-k \Rightarrow b=\frac{k-3}{k-1}$,

![](https://cdn.mathpix.com/cropped/2024_05_06_3205747dae2ad15e28c2g-08.jpg?height=420&width=526&top_left_y=1383&top_left_x=1416)

$S_{A B O}=\frac{1}{2} d \cdot(b-a)=\frac{3-k}{2}\left(\frac{k-3}{k-1}-\frac{k-3}{k+1}\right)=\frac{(k-3)^{2}}{1-k^{2}}$,

$S'=\frac{2(k-3)(1-3 k)}{\left(k^{2}-1\right)^{2}}=0, k_{\min }=\frac{1}{3}, \quad S_{\text {min }}=8$.

Answer: 8.