a: (1): mx-y=2
=>y=mx-2
Tọa độ điểm cố định mà đường thẳng (1) luôn đi qua là:
\(\left\{{}\begin{matrix}x=0\\y=m\cdot0-2=-2\end{matrix}\right.\)
(2): (2-m)x+y=m
=>y=(m-2)x+m=mx-2x+m=m(x+1)-2x
Tọa độ điểm cố định mà đường thẳng (2) luôn đi qua là:
\(\left\{{}\begin{matrix}x+1=0\\y=-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-2\cdot\left(-1\right)=2\end{matrix}\right.\)
vậy: B(0;-2); C(-1;2)
b: Để (1) cắt (2) thì \(\dfrac{m}{2-m}\ne\dfrac{-1}{1}=-1\)
=>\(\dfrac{m}{m-2}\ne1\)
=>\(\dfrac{m-m+2}{m-2}\ne0\)
=>\(\dfrac{2}{m-2}\ne0\)
=>\(m-2\ne0\)
=>\(m\ne2\)
\(\left\{{}\begin{matrix}mx-y=2\\\left(2-m\right)x+y=m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}mx-y+\left(2-m\right)x+y=2+m\\mx-y=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-2x=m+2\\y=mx-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-m-2}{2}\\y=\dfrac{m\left(-m-2\right)}{2}-2=\dfrac{-m^2-2m-4}{2}\end{matrix}\right.\)
Vậy: \(A\left(\dfrac{-m-2}{2};\dfrac{-m^2-2m-4}{2}\right)\);B(0;-2); C(-1;2)
\(\overrightarrow{AB}=\left(\dfrac{m+2}{2};-2-\dfrac{-m^2-2m-4}{2}\right)\)
=>\(\overrightarrow{AB}=\left(\dfrac{m+2}{2};-2+\dfrac{m^2+2m+4}{2}\right)\)
=>\(\overrightarrow{AB}=\left(\dfrac{m+2}{2};\dfrac{m^2+2m}{2}\right)\)
\(\overrightarrow{AC}=\left(-1-\dfrac{-m-2}{2};2-\dfrac{-m^2-2m-4}{2}\right)\)
=>\(\overrightarrow{AC}=\left(-1+\dfrac{m+2}{2};2+\dfrac{m^2+2m+4}{2}\right)\)
=>\(\overrightarrow{AC}=\left(\dfrac{m}{2};\dfrac{m^2+2m+8}{2}\right)\)
Vì \(\widehat{BAC}=90^0\)
nên \(\overrightarrow{AB}\cdot\overrightarrow{AC}=0\)
=>\(\dfrac{m+2}{2}\cdot\dfrac{m}{2}+\dfrac{m^2+2m}{2}\cdot\dfrac{m^2+2m+8}{2}=0\)
=>\(m\left(m+2\right)+\left(m^2+2m\right)^2+8\left(m^2+2m\right)=0\)
=>\(\left(m^2+2m\right)^2+9\left(m^2+2m\right)=0\)
=>\(\left(m^2+2m\right)\left(m^2+2m+9\right)=0\)
=>\(m^2+2m=0\)
=>\(\left[{}\begin{matrix}m=0\left(nhận\right)\\m=-2\left(loại\right)\end{matrix}\right.\)
Khi m=0 thì \(\overrightarrow{AB}=\left(1;0\right);\overrightarrow{AC}=\left(0;4\right)\)
\(AB=\sqrt{1^2+0^2}=1;AC=\sqrt{0^2+4^2}=4\)
Vì ΔABC vuông tại A
nên \(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC=\dfrac{1}{2}\cdot1\cdot4=2\)
