Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: A(3;-5); B(-2;2); C(4;1)
\(\overrightarrow{BC}=\left(4+2;1-2\right)=\left(6;-1\right)\) ; \(\overrightarrow{AC}=\left(4-3;1+5\right)=\left(1;6\right)\) ; \(\overrightarrow{BA}=\left(3+2;-5-2\right)=\left(5;-7\right)\)
b: Vì \(\frac65<>\frac{-1}{-7}\)
nên \(\overrightarrow{BC};\overrightarrow{BA}\) không tạo thành một đường thẳng
=>B,C,A không thẳng hàng
c: Tọa độ I là:
\(\begin{cases}x_{I}=\frac12\cdot\left(x_{A}+x_{C}\right)=\frac12\cdot\left(3+4\right)=\frac72\\ y_{I}=\frac12\cdot\left(y_{A}+y_{C}\right)=\frac12\cdot\left(-5+1\right)=\frac12\cdot\left(-4\right)=-2\end{cases}\)
Tọa độ J là:
\(\begin{cases}x_{J}=\frac12\cdot\left(x_{A}+x_{B}\right)=\frac12\cdot\left(3-2\right)=\frac12\\ y_{J}=\frac12\cdot\left(y_{A}+y_{B}\right)=\frac12\cdot\left(-5-2\right)=-\frac72\end{cases}\)
d: Tọa độ trọng tâm G là:
\(\begin{cases}x_{G}=\frac13\cdot\left(x_{A}+x_{B}+x_{C}\right)=\frac13\cdot\left(3-2+4\right)=\frac13\cdot5=\frac53\\ y_{G}=\frac13\cdot\left(y_{A}+y_{B}+y_{C}\right)=\frac13\cdot\left(-5-2+1\right)=\frac13\cdot\left(-6\right)=-2\end{cases}\)
=>G(5/3;-2)
e: A là trọng tâm của ΔHBC
=>\(\begin{cases}x_{H}+x_{B}+x_{C}=3\cdot x_{A}\\ y_{H}+y_{B}+y_{C}=3\cdot y_{A}\end{cases}\Rightarrow\begin{cases}x_{H}+\left(-2\right)+4=3\cdot3=9\\ y_{H}+\left(-2\right)+1=3\cdot\left(-5\right)=-15\end{cases}\)
=>\(\begin{cases}x_{H}=9-4+2=9-2=7\\ y_{H}=-15+2-1=-13-1=-14\end{cases}\)
=>H(7;-14)
g: A(3;-5); B(-2;-2); C(4;1); D(x;y)
\(\overrightarrow{AB}=\left(-2-3;-2+5\right)=\left(-5;3\right)\) ; \(\overrightarrow{DC}=\left(4-x;1-y\right)\)
ABCD là hình bình hành
=>\(\overrightarrow{AB}=\overrightarrow{DC}\)
=>4-x=-5 và 1-y=3
=>x=9 và y=1-3=-2
=>D(9;-2)
1/ Có G là trọng tâm tam giác ABC
Vì \(C\in Oy;G\in Ox\Rightarrow x_C=0;y_G=0\)
\(\Rightarrow\left\{{}\begin{matrix}x_G=\frac{x_A+x_B+x_C}{3}\\y_G=\frac{y_A+y_B+y_C}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_G=\frac{1+5+0}{3}\\0=\frac{-1-3+y_C}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_G=2\\y_C=4\end{matrix}\right.\Rightarrow C\left(0;4\right);G\left(2;0\right)\)
2/ \(\overrightarrow{AE}=3\overrightarrow{AB}-2\overrightarrow{AC}\)
\(\Rightarrow\left(x_E-x_A;y_E-y_A\right)=3\left(x_B-x_A;y_B-y_A\right)-2\left(x_C-x_A;y_C-y_A\right)\)
\(\Leftrightarrow\left(x_E-2;y_E-5\right)=3\left(-1;-4\right)-2\left(1;-2\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_E-2=-3-2\\y_E-5=-12+4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_E=-3\\y_E=-3\end{matrix}\right.\Rightarrow E\left(-3;-3\right)\)
3/ \(\overrightarrow{OA}=\overrightarrow{BC}\Rightarrow\left(x_A-x_O;y_A-y_O\right)=\left(x_C-x_B;y_C-y_B\right)\)
\(\Leftrightarrow\left(-2;1\right)=\left(x_C-4;y_C-5\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x_C-4=-2\\y_C-5=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_C=2\\y_C=6\end{matrix}\right.\Rightarrow C\left(2;6\right)\)
P/s: Kt lại số lịu hộ tui nhoa, nhỡ may soai thì tiu :)
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.
\(\overrightarrow{w}+\overrightarrow{a}=\overrightarrow{b}-\overrightarrow{c}\Rightarrow\overrightarrow{w}=\overrightarrow{b}-\overrightarrow{c}-\overrightarrow{a}\)
\(\Rightarrow\overrightarrow{w}=\left(3;4\right)-\left(7;2\right)-\left(2;1\right)=\left(-6;1\right)\)
\(a,\overrightarrow{AB}=\left(2;10\right)\)
\(\overrightarrow{AC}=\left(-5;5\right)\)
\(\overrightarrow{BC}=\left(-7;-5\right)\)
\(b,\) Thiếu dữ kiện
\(c,Cos\left(\overrightarrow{AB},\overrightarrow{AC}\right)=\dfrac{\left|2\left(-5\right)+10.5\right|}{\sqrt{2^2+10^2}.\sqrt{\left(-5\right)^2+5^2}}=\dfrac{2\sqrt{13}}{13}\)
\(\Rightarrow\left(\overrightarrow{AB},\overrightarrow{AC}\right)=56^o18'\)
\(Cos\left(\overrightarrow{AB},\overrightarrow{BC}\right)=\dfrac{\left|2\left(-7\right)+10\left(-5\right)\right|}{\sqrt{2^2+10^2}.\sqrt{\left(-7\right)^2+\left(-5\right)^2}}\)
\(\Rightarrow\left(\overrightarrow{AB},\overrightarrow{BC}\right)=43^o9'\)
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 \(D\left(0;\dfrac{11}{2}\right)\)