Question

helppp 

Answer 1

a: 

 

b: Tọa độ A là:

\(\left\{{}\begin{matrix}x=0\\y=-0+4=4\end{matrix}\right.\)

Tọa độ B là:

\(\left\{{}\begin{matrix}x=0\\y=0-4=-4\end{matrix}\right.\)

Tọa độ C là:

\(\left\{{}\begin{matrix}-x+4=x-4\\y=x-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2x=-8\\y=x-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=4-4=0\end{matrix}\right.\)

Vậy: A(0;4); B(0;-4); C(4;0)

c: \(AB=\sqrt{\left(0-0\right)^2+\left(-4-4\right)^2}=8\)

\(AC=\sqrt{\left(4-0\right)^2+\left(0-4\right)^2}=4\sqrt{2}\)

\(BC=\sqrt{\left(4-0\right)^2+\left(0+4\right)^2}=4\sqrt{2}\)

Xét ΔABC có \(CA^2+CB^2=AB^2\)

nên ΔCAB vuông tại C

=>\(S_{CAB}=\dfrac{1}{2}\cdot CA\cdot CB=\dfrac{1}{2}\cdot4\sqrt{2}\cdot4\sqrt{2}=\dfrac{1}{2}\cdot32=16\left(cm^2\right)\)