In the article "Four Methods for Finding a Class of Linear Equations" in Issue 5, 1983 of *Intermediate Mathematics*, an example is given: "A line is drawn through a point $(1, -2)$ inside the conic section $C: 14x^2 + 24xy + 21y^2 - 4x + 18y - 139 = 0$, such that the chord intercepted by the line is bisected by point $M$. Find the equation of this chord." The article introduces four methods, which are very good and broaden the mind. Now, I will supplement with another method.

保留源文本的换行和格式，这里直接输出翻译结果。请注意，最后一句“保留源文本的换行和格式，这里直接输出翻译结果。”是中文说明，不包含在翻译内容中。
Solution 5: Let the equation of the required chord be
$$
\begin{array}{l}
\left\{\begin{array}{l}
x=1+t \cos \alpha, \\
y=-2+t \sin \alpha
\end{array} \text { ( } t\right. \text { is a parameter), substitute into the } \\
\text { quadratic curve equation } 14(1+\mathrm{t} \cos \omega)^{\circ}+21(1+\mathrm{t} \cos \alpha \cdot \\
(-2+t \sin 0 .)+2 .!(-2+i \sin \alpha)^{2}-4(1 \\
+t \cos \alpha)+18(-2+t \sin \alpha)-139=0 \text {, and simplify } \\
\end{array}
$$
$$
\begin{array}{l}
\text { to get }\left(14 \cos ^{2} \alpha+24 \cos \alpha \sin \alpha+21 \sin ^{2} \alpha\right) \\
\cdot t^{2}-6(4 \cos \alpha+7 \sin \alpha) t-129=0 \text {. Its two roots }
\end{array}
$$
are $t_{1} 、 t_{2}$.
$\because \mathrm{M}$ is the midpoint of the chord, $\therefore t_{1}+t_{2}=0$, hence $4 \cos \alpha+7 \sin \alpha=0$.
$$
\therefore \operatorname{tg} \theta=-\frac{4}{7} \text {, } N k=-\frac{4}{7} \text {. }
$$

The equation of the required line is $4 \mathrm{x}+7 \mathrm{y}+10=0$.
(Author's affiliation: Zhou Tie Middle School, Yixing County, Jiangsu Province)