Question

[Nguyễn Hà Vy]

 

Answer 1

[Nguyễn Lê Phước Thịnh]

a:

b: Tọa độ A là:

\(\left\{{}\begin{matrix}2x-2=-x-2\\y=2x-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x=0\\y=2x-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=2\cdot0-2=-2\end{matrix}\right.\)

Tọa độ B là:

\(\left\{{}\begin{matrix}y=0\\2x-2=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=0\\2x=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)

Tọa độ C là:

\(\left\{{}\begin{matrix}y=0\\-x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x+2=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-2\\y=0\end{matrix}\right.\)

Vậy: A(0;-2); B(1;0); C(-2;0)

\(AB=\sqrt{\left(1-0\right)^2+\left(0+2\right)^2}=\sqrt{5}\)

\(AC=\sqrt{\left(-2-0\right)^2+\left(0+2\right)^2}=2\sqrt{2}\)

\(BC=\sqrt{\left(-2-1\right)^2+\left(0-0\right)^2}=3\)

Xét ΔABC có \(cosBAC=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\dfrac{5+8-9}{2\cdot\sqrt{5}\cdot2\sqrt{2}}=\dfrac{4}{4\sqrt{10}}=\dfrac{1}{\sqrt{10}}\)

=>\(sinBAC=\sqrt{1-\left(\dfrac{1}{\sqrt{10}}\right)^2}=\dfrac{3}{\sqrt{10}}\)

Diện tích tam giác BAC là:

\(S_{BAC}=\dfrac{1}{2}\cdot AB\cdot AC\cdot sinBAC=\dfrac{1}{2}\cdot\sqrt{5}\cdot2\sqrt{2}\cdot\dfrac{3}{\sqrt{10}}\)

\(=3\)