b: y=-3x-x+3=-4x+3
Tọa độ A là:
\(\left\{{}\begin{matrix}y=0\\-3x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\3x=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
Tọa độ B là:
\(\left\{{}\begin{matrix}y=0\\-4x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\-4x=-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=0\end{matrix}\right.\)
Tọa độ C là:
\(\left\{{}\begin{matrix}-3x+2=-4x+3\\y=-3x+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x+4x=3-2\\y=-3x+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1\\y=-3\cdot1+2=-3+2=-1\end{matrix}\right.\)
Vậy: \(A\left(\dfrac{2}{3};0\right);B\left(\dfrac{3}{4};0\right);C\left(1;-1\right)\)
c: \(AB=\sqrt{\left(\dfrac{3}{4}-\dfrac{2}{3}\right)^2+\left(0-0\right)^2}\)
\(=\sqrt{\left(\dfrac{9-8}{12}\right)^2}=\dfrac{1}{12}\)
\(AC=\sqrt{\left(1-\dfrac{2}{3}\right)^2+\left(-1-0\right)^2}=\sqrt{\left(\dfrac{1}{3}\right)^2+1^2}=\sqrt{1+\dfrac{1}{9}}=\sqrt{\dfrac{10}{9}}=\dfrac{\sqrt{10}}{3}\)
\(BC=\sqrt{\left(1-\dfrac{3}{4}\right)^2+\left(-1-0\right)^2}\)
\(=\sqrt{\left(\dfrac{1}{4}\right)^2+\left(-1\right)^2}\)
\(=\sqrt{\dfrac{1}{16}+1}=\sqrt{\dfrac{17}{16}}=\dfrac{\sqrt{17}}{4}\)
