a: Để (d) cắt (d') tại một điểm trên trục tung thì
\(\left\{{}\begin{matrix}4\ne\dfrac{4}{3}\\m-1=15-3m\end{matrix}\right.\)
=>\(m-1=15-3m\)
=>m+3m=15+1
=>4m=16
=>m=4
b:Thay m=4 vào (d), ta được:
\(y=4\cdot x+4-1=4x+3\)
Thay m=4 vào (d'), ta được:
\(y=\dfrac{4}{3}x+15-3\cdot4=\dfrac{4}{3}x+3\)
Tọa độ A 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 B là:
\(\left\{{}\begin{matrix}y=0\\\dfrac{4}{3}x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\\dfrac{4}{3}x=-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=0\\x=-3:\dfrac{4}{3}=-\dfrac{9}{4}\end{matrix}\right.\)
c:
Tọa độ C là:
\(\left\{{}\begin{matrix}4x+3=\dfrac{4}{3}x+3\\y=4x+3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=4\cdot0+3=3\end{matrix}\right.\)
A(-3/4;0); B(-9/4;0); C(0;3)
\(AB=\sqrt{\left(-\dfrac{9}{4}+\dfrac{3}{4}\right)^2+\left(0-0\right)^2}=\sqrt{\left(-\dfrac{6}{4}\right)^2+0^2}=1,5\)
\(AC=\sqrt{\left(0+\dfrac{3}{4}\right)^2+\left(3-0\right)^2}=\dfrac{3\sqrt{17}}{4}\)
\(BC=\sqrt{\left(0-\dfrac{-9}{4}\right)^2+\left(3-0\right)^2}=3,75\)
Chu vi tam giác ABC là:
\(C_{ABC}=BA+AC+BC=3,75+1,5+\dfrac{3\sqrt{17}}{4}=\dfrac{21+3\sqrt{17}}{4}\)
Xét ΔABC có
\(cosBAC=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)
\(=\dfrac{1,5^2+\left(\dfrac{3}{4}\sqrt{17}\right)^2-3,75^2}{2\cdot1,5\cdot\dfrac{3\sqrt{17}}{4}}=\dfrac{-\sqrt{17}}{17}\)
=>\(sinBAC=\sqrt{1-cos^2BAC}=\sqrt{1-\dfrac{1}{17}}=\dfrac{4}{\sqrt{17}}\)
Diện tích tam giác ABC là:
\(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC\cdot sinBAC\)
\(=\dfrac{1}{2}\cdot\dfrac{4}{\sqrt{17}}\cdot1,5\cdot\dfrac{3}{4}\sqrt{17}=\dfrac{9}{4}\)
