1: A(2;0); B(-3;4); C(1;-5)

Tọa độ vecto AB là:

\(\left\{{}\begin{matrix}x=-3-2=-5\\y=4-0=4\end{matrix}\right.\)

=>\(\overrightarrow{AB}=\left(-5;4\right)\)

Tọa độ vecto AC là:

\(\left\{{}\begin{matrix}x=1-2=-1\\y=-5-0=-5\end{matrix}\right.\)

Vậy: \(\overrightarrow{AC}=\left(-1;-5\right)\)

\(\overrightarrow{AB}=\left(-5;4\right)\)

Vì \(\left(-1\right)\cdot\left(-5\right)=5< >-20=-5\cdot4\)

nên A,B,C không thẳng hàng

=>A,B,C là ba đỉnh của một tam giác

2: Tọa độ trọng tâm G của ΔABC là:

\(\left\{{}\begin{matrix}x=\dfrac{2-3+1}{3}=\dfrac{0}{3}=0\\y=\dfrac{0+4-5}{3}=-\dfrac{1}{3}\end{matrix}\right.\)

3:

\(\overrightarrow{AB}=\left(-5;4\right);\overrightarrow{DC}=\left(1-x;-5-y\right)\)

ABCD là hình bình hành

nên \(\overrightarrow{AB}=\overrightarrow{DC}\)

=>\(\left\{{}\begin{matrix}1-x=-5\\-5-y=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=1+5=6\\y=-5-4=-9\end{matrix}\right.\)

Vậy: D(6;-9)

4: \(\overrightarrow{MA}=\left(2-x;-y\right);\overrightarrow{MB}=\left(-3-x;4-y\right);\overrightarrow{MC}=\left(1-x;-5-y\right)\)

\(2\overrightarrow{MA}+\overrightarrow{MB}+3\overrightarrow{MC}=\overrightarrow{0}\)

=>\(\left\{{}\begin{matrix}2\left(2-x\right)+\left(-3-x\right)+3\left(1-x\right)=0\\2\left(-y\right)+\left(4-y\right)+3\left(-5-y\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4-2x-3-x+3-3x=0\\-2y+4-y-15-3y=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-6x+4=0\\-6y-11=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-6x=-4\\-6y=11\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{11}{6}\end{matrix}\right.\)

vậy: \(M\left(\dfrac{2}{3};-\dfrac{11}{6}\right)\)

5:

A(2;0); B(-3;4); C(1;-5); N(x;y)

A là trọng tâm của ΔBNC

=>\(\left\{{}\begin{matrix}x_A=\dfrac{x_B+x_N+x_C}{3}\\y_A=\dfrac{y_B+y_N+y_C}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2=\dfrac{-3+1+x}{3}\\0=\dfrac{4-5+y}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2=6\\y-1=0\end{matrix}\right.\)

=>x=8 và y=1

Vậy: N(8;1)

6: A là trung điểm của BE

=>\(\left\{{}\begin{matrix}x_B+x_E=2\cdot x_A\\y_B+y_E=2\cdot y_A\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-3+x_E=2\cdot2=4\\4+y_E=2\cdot0=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x_E=7\\y_E=-4\end{matrix}\right.\)

Vậy: E(7;-4)

Chọn D.

Ta có: tứ giác ABCD là hình bình hành khi

Ta có: tứ giác ABCD là hình bình hành khi 

C,D thuộc (P) nên \(C\left(x_1;x_1^2-4x_1-5\right);D\left(x_2;x_2^2-4x_2-5\right)\)

ABCD là hình bình hành

=>\(\overrightarrow{AB}=\overrightarrow{DC}\)

\(\overrightarrow{AB}=\left(3;-4\right);\overrightarrow{DC}=\left(x_1-x_2;x_1^2-4x_1-5-x_2^2+4x_2+5\right)\)

=>\(\left\{{}\begin{matrix}x_1-x_2=3\\x_1^2-x_2^2-4x_1+4x_2=-4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x_1-x_2=3\\\left(x_1-x_2\right)\left(x_1+x_2-4\right)=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1-x_2=3\\x_1+x_2-4=-\dfrac{4}{x_1-x_2}=-\dfrac{4}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x_1-x_2=3\\x_1+x_2=-\dfrac{4}{3}+4=\dfrac{12}{3}-\dfrac{4}{3}=\dfrac{8}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x_1=3+\dfrac{8}{3}=\dfrac{17}{3}\\x_1+x_2=\dfrac{8}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{17}{6}\\x_2=\dfrac{8}{3}-\dfrac{17}{6}=-\dfrac{1}{6}\end{matrix}\right.\)

Khi x=17/6 thì \(y=x^2-4x-5=\left(\dfrac{17}{6}\right)^2-4\cdot\dfrac{17}{6}-5=-\dfrac{299}{36}\)

Khi x=-1/6 thì \(y=\left(-\dfrac{1}{6}\right)^2-4\cdot\dfrac{-1}{6}-5=\dfrac{1}{36}+\dfrac{2}{3}-5=-\dfrac{155}{36}\)

Vậy: \(C\left(\dfrac{17}{6};-\dfrac{299}{36}\right);D\left(-\dfrac{1}{6};-\dfrac{155}{36}\right)\)

\(\overrightarrow{AB}=\left(-3;7\right)\)

\(\overrightarrow{DC}=\left(1-x_D;5-y_D\right)\)

Để ABCD là hbh thì 

\(\left\{{}\begin{matrix}1-x_D=-3\\5-y_D=7\end{matrix}\right.\Leftrightarrow D\left(2;-2\right)\)