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\(\left\{{}\begin{matrix}\overrightarrow{AB}=\left(1;-1\right)\\\overrightarrow{BC}=\left(-3;4\right)\end{matrix}\right.\)
\(\Rightarrow\overrightarrow{u}=3\overrightarrow{AB}+2\overrightarrow{BC}=\left(-3;5\right)\)
Gọi \(D\left(x;y\right)\Rightarrow\overrightarrow{DC}=\left(1-x;5-y\right)\)
Để ABCD là hbh \(\Leftrightarrow\overrightarrow{AB}=\overrightarrow{DC}\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-x=1\\5-y=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0\\y=6\end{matrix}\right.\)
\(\Rightarrow D\left(0;6\right)\)
Tham khảo
a,⇒C,A,Da,⇒C,A,D thẳngthẳng hàng⇒−−→CA+−−→CD=→0⇔−−→CA=−−→DChàng⇒CA→+CD→=0→⇔CA→=DC→
D(x;y)⇒−−→CA=−−→DC⇔{−1−x=2−2−y=0D(x;y)⇒CA→=DC→⇔{−1−x=2−2−y=0⇔{x=−1y=−2⇔{x=−1y=−2⇔{x=−3y=−2⇔{x=−3y=−2⇒D(−3;−2)⇒D(−3;−2)
b,E(xo;yo)⇒−−→AE=−−→BCb,E(xo;yo)⇒AE→=BC→⇔{xo−1=−3yo+2=−5⇔{xo−1=−3yo+2=−5⇔{xo=−2yo=−7⇔{xo=−2yo=−7⇒E(−2;−7)⇒E(−2;−7)
c,⇒G(xG;yG)⇒⎧⎪ ⎪⎨⎪ ⎪⎩xG=1+2−13=23yG=−2+3−23=−13c,⇒G(xG;yG)⇒{xG=1+2−13=23yG=−2+3−23=−13⇒G(23;−13)
bạn ơi bạn có thể viết rõ câu trả lời hơn được không vì nó khó hiểu quá
a.
\(\overrightarrow{u}=2\left(2;1\right)-\left(3;4\right)=\left(1;-2\right)\)
\(\overrightarrow{v}=3\left(3;4\right)-2\left(7;2\right)=\left(-5;8\right)\)
\(\overrightarrow{w}=5\left(7;2\right)+\left(2;1\right)=\left(37;11\right)\)
b.
\(\overrightarrow{x}=2\left(2;1\right)+\left(3;4\right)-\left(7;2\right)=\left(0;4\right)\)
\(\overrightarrow{z}=2\left(2;1\right)-3\left(3;4\right)+\left(7;2\right)=\left(2;-8\right)\)
c.
\(\overrightarrow{w}+\overrightarrow{a}=\overrightarrow{b}-\overrightarrow{c}\Rightarrow\overrightarrow{w}=\overrightarrow{b}-\overrightarrow{c}-\overrightarrow{a}\)
\(\Rightarrow\overrightarrow{w}=\left(3;4\right)-\left(7;2\right)-\left(2;1\right)=\left(-6;1\right)\)
a.
\(\left\{{}\begin{matrix}x_I=\dfrac{x_A+x_B}{2}=\dfrac{2-4}{2}=-1\\y_I=\dfrac{y_A+y_B}{2}=\dfrac{1+5}{2}=3\end{matrix}\right.\)
\(\Rightarrow I\left(-1;3\right)\)
b.
Do C thuộc trục hoành, gọi tọa độ C có dạng \(C\left(c;0\right)\)
Do D thuộc trục tung, gọi tọa độ D có dạng \(D\left(0;d\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AC}=\left(c-2;-1\right)\\\overrightarrow{DB}=\left(-4;5-d\right)\Rightarrow2\overrightarrow{DB}=\left(-8;10-2d\right)\end{matrix}\right.\)
Để \(\overrightarrow{AC}=2\overrightarrow{DB}\)
\(\Leftrightarrow\left\{{}\begin{matrix}c-2=-8\\-1=10-2d\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}c=-6\\d=\dfrac{11}{2}\end{matrix}\right.\)
Vậy \(C\left(-6;0\right)\) và \(D\left(0;\dfrac{11}{2}\right)\)
Giúp noè hehe :))
1/
a/ \(\overrightarrow{u}=\overrightarrow{a}+\overrightarrow{b}=\left(2-5;3+1\right)=\left(-3;4\right)\)
b/ \(\overrightarrow{v}=\overrightarrow{c}-5\overrightarrow{a}=\left(-4-10;11-15\right)=\left(-14;-4\right)\)
c/ \(\overrightarrow{c}=x\overrightarrow{a}+y\overrightarrow{b}\)
\(\Leftrightarrow\left(-4;11\right)=x\left(2;3\right)+y\left(-5;1\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-5y=-4\\3x+y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)
\(\Rightarrow\overrightarrow{c}=3\overrightarrow{a}+2\overrightarrow{b}\)
2/
a/ Để ABCD là hbh
\(\Leftrightarrow\overrightarrow{AB}=\overrightarrow{DC}\Leftrightarrow\left(x_B-x_A;y_B-y_A\right)=\left(x_C-x_D;y_C-y_D\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}3=4-x_D\\-2=-3-y_D\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_D=1\\y_D=-1\end{matrix}\right.\Rightarrow D\left(1;-1\right)\)
b/ E đối xứng vs A qua C
\(\Leftrightarrow\overrightarrow{EC}=\overrightarrow{CA}\)
\(\Leftrightarrow\left(x_C-x_E;y_C-y_E\right)=\left(x_A-x_C;y_A-y_C\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}4-x_E=-5\\-3-y_E=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_E=9\\y_E=-5\end{matrix}\right.\Rightarrow E\left(9;-5\right)\)
c/ A,B,M thẳng hàng<=> \(\overrightarrow{AB}=k\overrightarrow{BM}\Leftrightarrow\left(x_B-x_A;y_B-y_A\right)=k\left(x_M-x_B;y_M-y_B\right)\)
Có \(M\in Oy\Rightarrow x_M=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3=-2k\\-2=k\left(y_M-1\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}k=-\frac{3}{2}\\y_M=\frac{7}{3}\end{matrix}\right.\Rightarrow\left(0;\frac{7}{3}\right)\)
P/s: KT lại số lịu hộ tui nhoa, ko bít có soai dữ lịu chỗ nèo hong? =))