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$a. \begin{cases}d_1 : mx+y+2=0 \\ d_2 :x+my+m+1=0 \end{cases} \Rightarrow \begin{cases}d_1 :mx+y=-2 \\ d_2 :x+my=-m-1 \end{cases} $
$D=\left| \begin{array}{l}
m\\
2
\end{array} \right.\left. \begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,1\\
\,\,\,\,\,\,\,\,\,m
\end{array} \right|=m^2-1$
$d_x=\left| \begin{array}{l}
- 2\\
- m - 1
\end{array} \right.\left. \begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,1\\
\,\,\,\,\,\,\,\,\,m
\end{array} \right|=-2m+m+1=1-m$
$D_y=\left| \begin{array}{l}
m\\
1
\end{array} \right.\left. \begin{array}{l}
\,\,\,\,\,\,\,\,\, - 2\\
\, - m - 1
\end{array} \right|=-m^2-m+2$
Xét $D=0\Leftrightarrow m^2-1=0\Leftrightarrow m=\pm 1$
$m=1: \begin{cases}D=0 \\ D_x=0 \\D_y=0\end{cases} : d_1\equiv d_2$
$m=-1 : \begin{cases}D=0 \\ D_x=2\neq 0 \end{cases} : d_1//d_2$
$m\neq \pm 1: D\neq 0 : d_1$ cắt $d_2$
$b. \begin{cases}d_1 :mx+(3-m)y=-3 \\ d_2 : x+(m+1)y=-3 \end{cases} $
$D=\left| \begin{array}{l}
m\\
1
\end{array} \right.\left. \begin{array}{l}
\,\,\,\,\,\,\,\,3 - m\\
\,\,\,\,\,\,\,\,\,m + 1
\end{array} \right|=m^2+2m-3$
$D_x=\left| \begin{array}{l}
- 3\\
- 3
\end{array} \right.\left. \begin{array}{l}
\,\,\,\,\,\,\,\,3 - m\\
\,\,\,\,\,\,\,\,\,m + 1
\end{array} \right|=-6m+6$
$D_y=\left| \begin{array}{l}
m\\
1
\end{array} \right.\left. \begin{array}{l}
\,\,\,\,\,\,\,\, - 3\\
\,\,\,\,\,\,\,\, - 3
\end{array} \right|=-3m+3$
Xét $D=0\Leftrightarrow m^2+2m-3=0\Leftrightarrow m=1$ hoặc $m=-3$
$m=1 :\begin{cases}D=0 \\ D_x=0\\ D_y=0 \end{cases} : d_1\equiv d_2$
$m=-3 :\begin{cases}D=0 \\ D_x=24 \end{cases} : d_1//d_2$
$m\in R$\{$-3;1$} :$d_1$ cắt $d_2$
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Đăng bài 07-06-12 11:13 AM
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