回復 1# shmilyho 的帖子
\(\begin{align}
& D\left( t,0 \right),C\left( t+3,0 \right) \\
& AD:2x-\left( t+6 \right)y=2t\ ,\quad t=\frac{2x-6y}{y+2} \\
& BC:6x-\left( t+3 \right)y=6t+18\ ,\quad t=\frac{6x-3y-18}{y+6} \\
& \frac{2x-6y}{y+2}=\frac{6x-3y-18}{y+6} \\
& 3{{y}^{2}}+4xy+12y-6=0 \\
\end{align}\)
