50. Two lines $y_{1}=k_{1} x-1, y_{2}=k_{2} x+2$ intersect on the $x$-axis, then the area of the triangle formed by the line $y=\frac{k_{1}}{k_{2}} x-\frac{k_{2}}{k_{1}}$ and the two coordinate axes is —. $\qquad$
Answer: 4

Solution: Let $y_{1}=0$, we get

Let $y_{2}=0$, we get

According to the problem,
$$
x=\frac{1}{k_{1}},
$$

Let $y_{2}=0$, we get
$$
\begin{array}{c}
x=-\frac{2}{k_{2}}, \\
\frac{1}{k_{1}}=-\frac{2}{k_{2}}, \\
\frac{k_{1}}{k_{2}}=-\frac{1}{2}, \frac{k_{2}}{k_{1}}=-2 .
\end{array}
$$

Therefore,
$$
\begin{array}{c}
\frac{1}{k_{1}}=-\frac{2}{k_{2}} \\
\frac{k_{1}}{k_{2}}=-\frac{1}{2}, \frac{k_{2}}{k_{1}}=-2 .
\end{array}
$$

The line $y=-\frac{1}{2} x+2$ intersects the two coordinate axes at points $(4,0)$ and $(0,2)$, respectively. The area of the triangle formed by the line $y=\frac{k_{1}}{k_{2}} x-\frac{k_{2}}{k_{1}}$ and the two coordinate axes is $\frac{1}{2} \times 4 \times 2=4$.