Bài 6:
a: Tọa độ trọng tâm của ΔOAB là:
\(\left\{{}\begin{matrix}x=\dfrac{0+\left(-2\right)+5}{3}=\dfrac{5-2}{3}=1\\y=\dfrac{0+\left(-2\right)+\left(-4\right)}{3}=\dfrac{-6}{3}=-2\end{matrix}\right.\)
b: Để G là trọng tâm của ΔABC thì
\(\left\{{}\begin{matrix}x_A+x_B+x_C=3\cdot x_G\\y_A+y_B+y_C=3\cdot y_G\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_C+\left(-2\right)+5=3\cdot2=6\\y_C+\left(-2\right)+\left(-4\right)=3\cdot0=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x_C=6-5+2=1+2=3\\y_C=6\end{matrix}\right.\)
vậy: C(3;6)
Bài 8:
a: C là trọng tâm của ΔABD
=>\(\left\{{}\begin{matrix}x_A+x_B+x_D=3\cdot x_C\\y_A+y_B+y_C=3\cdot y_c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_D=3\cdot2-\left(-4\right)-2=8\\y_D=3\cdot\left(-2\right)-1-4=-6-5=-11\end{matrix}\right.\)
Vậy: D(8;-11)
b: ABCE là hình bình hành
=>\(\overrightarrow{AB}=\overrightarrow{EC}\)
=>\(\left\{{}\begin{matrix}2-x=2-\left(-4\right)\\-2-y=4-1=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2-x=2+4\\-2-y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2-2-4=-4\\y=-2-3=-5\end{matrix}\right.\)
Bài 3 : \(\overrightarrow{a}=\left(3;5\right);\overrightarrow{b}=\left(2;-3\right);\overrightarrow{c}=\left(-1;8\right)\)
a) \(\overrightarrow{u}=3\overrightarrow{a}+m\overrightarrow{b}=\left(3.3+2m;3.5-3m\right)=\left(2m+9;15-m\right)\)
\(\overrightarrow{v}=\overrightarrow{c}=\left(-1;8\right)\)
\(\overrightarrow{u}\) cùng phương \(\overrightarrow{v}\) khi và chỉ khi
\(\dfrac{2m+9}{-1}=\dfrac{15-m}{8}\)
\(\Leftrightarrow16m+72=-15+m\)
\(\Leftrightarrow15m=-87\)
\(\Leftrightarrow m=-\dfrac{87}{15}\)
c) \(\overrightarrow{u}=\overrightarrow{a}+\overrightarrow{b}=\left(3+2;5-3\right)=\left(5;2\right)\)
\(\overrightarrow{v}=m\overrightarrow{a}=\left(3m;5m\right)\)
\(\overrightarrow{u}\) cùng phương \(\overrightarrow{v}\) khi và chỉ khi
\(\dfrac{5}{3m}=\dfrac{2}{5m}\)
\(\Leftrightarrow25m=6m\)
\(\Leftrightarrow19m=0\)
\(\Leftrightarrow m=0\)
Bài 2:
a, \(\overrightarrow{a}\left(3;5\right)\Rightarrow\left|\overrightarrow{a}\right|=\sqrt{3^2+5^2}=\sqrt{34}\)
\(\overrightarrow{b}\left(2;-1\right)\Rightarrow\left|\overrightarrow{b}\right|=\sqrt{2^2+\left(-1\right)^2}=\sqrt{5}\)
\(\overrightarrow{a}.\overrightarrow{b}=3.2+5.\left(-1\right)=1\)
b, \(\left\{{}\begin{matrix}x_u=2.3+3.2=12\\y_u=2.5+3.\left(-1\right)=7\end{matrix}\right.\)
\(\overrightarrow{u}\left(12;7\right)\) \(\Rightarrow\overrightarrow{u}=\sqrt{12^2+7^2}=\sqrt{193}\)
c, \(\left\{{}\begin{matrix}x_v=3-2+12=13\\y_v=5-\left(-1\right)+7=13\end{matrix}\right.\)
\(\overrightarrow{v}\left(13;13\right)\) \(\Rightarrow\left|\overrightarrow{v}\right|=\sqrt{13^2+13^2}=13\sqrt{2}\)
d, \(\overrightarrow{u}\left(2\overrightarrow{a}-\overrightarrow{b}\right)=12.4+7.11=125\) (do \(\left(2\overrightarrow{a}-\overrightarrow{b}\right)=\left(4;11\right)\))
\(\overrightarrow{u}.\overrightarrow{v}=12.13+7.13=247\)
e, \(2\left(\overrightarrow{a}+\overrightarrow{b}\right)-\overrightarrow{u}=\left(-2;5\right)\Rightarrow\left|2\left(\overrightarrow{a}+\overrightarrow{b}\right)-\overrightarrow{u}\right|=\sqrt{\left(-2\right)^2+5^2}=\sqrt{29}\)
f, \(\left(3\overrightarrow{a}-\overrightarrow{u}\right)=\left(-3;8\right);\left(\overrightarrow{v}+4\overrightarrow{b}\right)=\left(21;9\right)\)
\(\left(3\overrightarrow{a}-\overrightarrow{u}\right).\left(\overrightarrow{v}+4\overrightarrow{b}\right)=-3.21+8.9=9\)
BT7. Gọi (x;y) là tọa độ của D
a) ABCD là hbh khi và chỉ khi \(\overrightarrow{AB}=\overrightarrow{DC}\Rightarrow\left(-3;1\right)=\left(1-x;1-y\right)\Rightarrow\left\{{}\begin{matrix}1-x=-3\\1-y=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\y=0\end{matrix}\right.\)
Vậy D(4;0) thỏa mãn yêu cầu bài toán
b) ABDC là hbh khi và chỉ khi
\(\overrightarrow{AB}=\overrightarrow{CD}\Rightarrow\left(-3;1\right)=\left(x-1;y-1\right)\Rightarrow\left\{{}\begin{matrix}x-1=-3\\y-1=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2\\y=2\end{matrix}\right.\)
Vậy D(-2;2) thỏa yêu cầu bài toán
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