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a/ Để tứ giác ADCB là hbh
\(\Leftrightarrow\overrightarrow{AD}=\overrightarrow{BC}\Leftrightarrow\left(x_D-x_A;y_D-y_A\right)=\left(x_C-x_B;y_C-y_B\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_D-1=4+2\\y_D-2=4-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_D=7\\y_D=0\end{matrix}\right.\Rightarrow D\left(7;0\right)\)
b/ Có phải đề bài là:
\(2\overrightarrow{EA}-4\overrightarrow{EB}+\overrightarrow{EC}=\overrightarrow{0}?\)
\(\Rightarrow2\left(x_A-x_E;y_A-y_E\right)-4\left(x_B-x_E;y_B-y_E\right)+\left(x_C-x_E;y_C-y_E\right)=0\)
\(\Leftrightarrow2\left(1-x_E;2-y_E\right)-4\left(-2-x_E;6-y_E\right)+\left(4-x_E;4-y_E\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}2-2x_E+8+4x_E+4-x_E=0\\4-2y_E-24+4y_E+4-y_E=0\end{matrix}\right.\)
Bạn tự giải nốt
a) Gọi \(D\left(x;y\right)\)
\(2\overrightarrow{DA}=\left(20-2x;10-2y\right)\\ 3\overrightarrow{DB}=\left(9-3x;6-3y\right)\\ -\overrightarrow{DC}=\overrightarrow{CD}=\left(x-6;y+5\right)\)
\(\Rightarrow\left\{{}\begin{matrix}20-2x+9-3x+x-6=0\\10-2y+6-3y+y+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{23}{4}\\y=\dfrac{21}{4}\end{matrix}\right.\)
b)\(\overrightarrow{AF}=\left(-15;3\right)\\\overrightarrow{AB}=\left(-7;-3\right) \\ \overrightarrow{AC}=\left(-4;-10\right)\\\overrightarrow{AF}=a\overrightarrow{AB}+bAC\Rightarrow\left\{{}\begin{matrix}-7a-4b=-15\\-3a-10b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{81}{29}\\b=-\dfrac{33}{29}\end{matrix}\right.\)
Hok nhanh phết, chưa j đã đến phần toạ độ vecto r
1/ \(\overrightarrow{MB}=\left(x_B-x_M;y_B-y_M\right)=\left(2-x_M;3-y_M\right)\)
\(\Rightarrow2\overrightarrow{MB}=\left(4-2x_M;6-2y_M\right)\)
\(\overrightarrow{3MC}=\left(3x_C-3x_M;3y_C-3y_M\right)=\left(-3-3x_M;6-3y_M\right)\)
\(\Rightarrow2\overrightarrow{MB}+3\overrightarrow{MC}=\left(4-2x_M-3-3x_M;6-2y_M+6-3y_M\right)=0\)
\(\Leftrightarrow\left(1-5x_M;12-5y_M\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-5x_M=0\\12-5y_M=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_M=\frac{1}{5}\\y_M=\frac{12}{5}\end{matrix}\right.\Rightarrow M\left(\frac{1}{5};\frac{12}{5}\right)\)
2/ \(\overrightarrow{m}=2\left(1;2\right)+3\left(3;4\right)=\left(2+9;4+12\right)=\left(11;16\right)\)
3/ \(\overrightarrow{AB}=\left(x_B-x_A;y_B-y_A\right)=\left(-5-3;4+2\right)=\left(-8;6\right)\)
\(\overrightarrow{AC}=\left(x_C-x_A;y_C-y_A\right)=\left(\frac{1}{3}-3;0+2\right)=\left(-\frac{8}{3};2\right)\)
\(\Rightarrow x=\frac{\overrightarrow{AB}}{\overrightarrow{AC}}=\frac{\left(-8;6\right)}{\left(-\frac{8}{3};2\right)}=3\)
Câu 4 tương tự
Câu 5 vt lại đề bài nhé bn, nghe nó vô lý sao á, m,n ở đâu ra vậy, cả A,B,C nx
Vì ABCD là hình bình hành
nên vecto AB=vecto DC
=>\(\left\{{}\begin{matrix}x_C-x_D=x_B-x_A\\y_C-y_D=y_B-y_A\end{matrix}\right.\Leftrightarrow D\left(-4;1\right)\)
\(\overrightarrow{EA}=\left(-1-x;-y\right)\)
\(\overrightarrow{EB}=\left(3-x;1-y\right)\)
\(\overrightarrow{EC}=\left(-x;2-y\right)\)
Theo đề, ta có: \(\left\{{}\begin{matrix}-1-x+3\left(3-x\right)-2\left(-x\right)=0\\-y+3\left(1-y\right)-2\left(2-y\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-1-x+9-3x+2x=0\\-y+3-3y-4+2y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2x+8=0\\-2y-1=0\end{matrix}\right.\Leftrightarrow E\left(4;-\dfrac{1}{2}\right)\)