\(\left(d\right):\left\{{}\begin{matrix}x=2+4t\\y=-1+3t\end{matrix}\right.\)
a) Véc tơ chỉ phương \(\overrightarrow{u}=\left(4;3\right)\)
\(\left(2;-1\right)\in\left(d\right)\)
b) \(x=-2\Rightarrow2+4t=-2\Rightarrow t=1\Rightarrow y=-1+3.1=2\)
\(\Rightarrow K\left(-2;2\right)\in\left(d\right)\)
c) \(M\left(-1;3\right)\Rightarrow2+4t=-1\Rightarrow t=-\dfrac{3}{4}\Rightarrow y=-1+3.\left(-\dfrac{3}{4}\right)=-\dfrac{13}{4}\ne3\)
\(\Rightarrow M\left(-1;3\right)\notin\left(d\right)\)
\(N\left(4;\dfrac{1}{2}\right)\Rightarrow2+4t=4\Rightarrow t=\dfrac{1}{2}\Rightarrow y=-1+3.\dfrac{1}{2}=\dfrac{1}{2}\left(đúng\right)\)
\(\Rightarrow N\left(4;\dfrac{1}{2}\right)\in\left(d\right)\)
\(P\left(-2;5\right)\Rightarrow2+4t=-2\Rightarrow t=-1\Rightarrow y=-1+3.\left(-1\right)=-4\ne5\)
\(\Rightarrow P\left(-2;5\right)\notin\left(d\right)\)
\(Q\left(-6;-7\right)\Rightarrow2+4t=-6\Rightarrow t=-2\Rightarrow y=-1+3.\left(-2\right)=-7\left(đúng\right)\)
\(\Rightarrow Q\left(-6;-7\right)\in\left(d\right)\)
\(I\left(0;-\dfrac{5}{2}\right)\Rightarrow2+4t=0\Rightarrow t=-\dfrac{1}{2}\Rightarrow y=-1+3.\left(-\dfrac{1}{2}\right)=-\dfrac{7}{2}\ne-\dfrac{5}{2}\)
\(\Rightarrow I\left(0;-\dfrac{5}{2}\right)\notin\left(d\right)\)
d) \(\left(d\right)\cap\left(Ox\right)=A\left(x;0\right)\Rightarrow-1+3t=0\Rightarrow t=\dfrac{1}{3}\Rightarrow x=2+4.\dfrac{1}{3}=\dfrac{10}{3}\)
\(\Rightarrow A\left(\dfrac{10}{3};0\right)\)
\(\left(d\right)\cap\left(Oy\right)=B\left(0;y\right)\Rightarrow2+4t=0\Rightarrow t=-\dfrac{1}{2}\Rightarrow y=-1+3.\left(-\dfrac{1}{2}\right)=-\dfrac{5}{2}\)
\(\Rightarrow B\left(0;-\dfrac{5}{2}\right)\)
\(\Rightarrow AB=\sqrt{\left(0-\dfrac{10}{3}\right)^2+\left(-\dfrac{5}{2}-0\right)^2}=\sqrt{\dfrac{100}{9}+\dfrac{25}{4}}=\dfrac{5}{6}\)
cíu bé vs mn ơi
CÍU TUIIII
