a: Tọa độ A là;
\(\left\{{}\begin{matrix}y=0\\2x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=0\end{matrix}\right.\)
Tọa độ C là;
\(\left\{{}\begin{matrix}-\dfrac{1}{2}x+1=0\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{1}{2}x=-1\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
Tọa độ M là:
\(\left\{{}\begin{matrix}2x+4=-\dfrac{1}{2}x+1\\y=2x+4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{2}x=-3\\y=2x+4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-3:\dfrac{5}{2}=-\dfrac{6}{5}\\y=2\cdot\dfrac{-6}{5}+4=\dfrac{-12}{5}+\dfrac{20}{5}=\dfrac{8}{5}\end{matrix}\right.\)
A(-2;0); C(2;0); M(-1,2;1,6)
\(MA=\sqrt{\left(-2+1.2\right)^2+\left(0-1,6\right)^2}=\dfrac{4\sqrt{5}}{5}\)
\(MC=\sqrt{\left(2+1,2\right)^2+\left(0-1,6\right)^2}=\dfrac{8\sqrt{5}}{5}\)
\(AC=\sqrt{\left(-2-2\right)^2+\left(0-0\right)^2}=4\)
Vì \(MA^2+MC^2=AC^2\)
nên ΔMAC vuông tại M
b: \(S_{MAC}=\dfrac{1}{2}\cdot MA\cdot MC=\dfrac{1}{2}\cdot\dfrac{4\sqrt{5}}{5}\cdot\dfrac{8\sqrt{5}}{5}=\dfrac{16}{5}\)
