9.
a.
Ta có: \(\overrightarrow{AB}=\left(6;2\right)\); \(\overrightarrow{AC}=\left(4;-6\right)\)
Do \(\dfrac{6}{4}\ne\dfrac{2}{-6}\) nên 2 vecto \(\overrightarrow{AB};\overrightarrow{AC}\) không cùng phương
Hay 3 điểm A;B;C không thẳng hàng
b.
Gọi G là trọng tâm ABC, theo công thức trọng tâm:
\(\left\{{}\begin{matrix}x_G=\dfrac{x_A+x_B+x_C}{3}=\dfrac{-2+4+2}{3}=\dfrac{4}{3}\\y_G=\dfrac{y_A+y_B+y_C}{3}=\dfrac{3+5-3}{3}=\dfrac{5}{3}\\\end{matrix}\right.\)
Vậy \(G\left(\dfrac{4}{3};\dfrac{5}{3}\right)\)
c.
\(AB=\sqrt{6^2+2^2}\approx6\)
\(AC=\sqrt{4^2+\left(-6\right)^2}\approx7\)
\(\overrightarrow{BC}=\left(-2;-8\right)\Rightarrow BC=\sqrt{\left(-2\right)^2+\left(-8\right)^2}\approx8\)
\(cos\widehat{A}=\dfrac{\overrightarrow{AB}.\overrightarrow{AC}}{\left|\overrightarrow{AB}\right|.\left|\overrightarrow{AC}\right|}=\dfrac{6.4+2.\left(-6\right)}{\sqrt{6^2+2^2}.\sqrt{4^2+\left(-6\right)^2}}\approx0,263\)
\(\Rightarrow\widehat{A}\approx75^0\)
\(cos\widehat{B}=\dfrac{\overrightarrow{BA}.\overrightarrow{BC}}{\left|\overrightarrow{BA}\right|.\left|\overrightarrow{BC}\right|}=\dfrac{-\overrightarrow{AB}.\overrightarrow{BC}}{\left|\overrightarrow{AB}\right|.\left|\overrightarrow{BC}\right|}=\dfrac{-\left(6.\left(-2\right)+2.\left(-8\right)\right)}{\sqrt{6^2+2^2}.\sqrt{\left(-2\right)^2+\left(-8\right)^2}}\approx0,537\)
\(\Rightarrow\widehat{B}\approx58^0\)
\(\widehat{C}=180^0-A-B=47^0\)
10.
Theo công thức trọng tâm:
\(\left\{{}\begin{matrix}x_I=\dfrac{x_A+x_B+x_C}{3}\\y_I=\dfrac{y_A+y_B+y_C}{3}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x_C=3x_I-x_A-x_B=3.1-3-\left(-1\right)=1\\y_C=3y_I-y_A-y_B=3.\left(-1\right)-\left(-1\right)-2=-4\end{matrix}\right.\)
\(\Rightarrow C\left(1;-4\right)\)
Gọi D(x;y) \(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AB}=\left(-4;3\right)\\\overrightarrow{DC}=\left(1-x;-4-y\right)\end{matrix}\right.\)
ABCD là hbh khi và chỉ khi \(\overrightarrow{AB}=\overrightarrow{DC}\)
\(\Rightarrow\left\{{}\begin{matrix}1-x=-4\\-4-y=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=5\\y=-7\end{matrix}\right.\)
\(\Rightarrow D\left(5;-7\right)\)
Theo t/c hbh, O là trung điểm AC
\(\Rightarrow\left\{{}\begin{matrix}x_O=\dfrac{x_A+x_C}{2}=\dfrac{3+1}{2}=2\\y_O=\dfrac{y_A+y_C}{2}=\dfrac{-1+\left(-4\right)}{2}=-\dfrac{5}{2}\end{matrix}\right.\)
\(\Rightarrow O\left(2;-\dfrac{5}{2}\right)\)
