a:
A(2;4); B(-3;1); C(3;-1); D(x;y)
\(\overrightarrow{AB}=\left(-3-2;1-4\right)\)
=>\(\overrightarrow{AB}=\left(-5;-3\right)\)
\(\overrightarrow{DC}=\left(3-x;-1-y\right)\)
ABCD là hình bình hành
=>\(\overrightarrow{AB}=\overrightarrow{DC}\)
=>\(\left\{{}\begin{matrix}3-x=-5\\-1-y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=2\end{matrix}\right.\)
vậy: D(8;2)
b: Tọa độ trọng tâm G của ΔABC là:
\(\left\{{}\begin{matrix}x_G=\dfrac{x_A+x_B+x_C}{3}=\dfrac{2+\left(-3\right)+3}{3}=\dfrac{2}{3}\\y_G=\dfrac{y_A+y_B+y_C}{3}=\dfrac{4+1+\left(-1\right)}{3}=\dfrac{4}{3}\end{matrix}\right.\)
vậy: \(G\left(\dfrac{2}{3};\dfrac{4}{3}\right)\)
c:
Gọi H(x;y) là tọa độ trực tâm của ΔABC
A(2;4); B(-3;1); C(3;-1); H(x;y)
\(\overrightarrow{AH}=\left(x-2;y-4\right);\overrightarrow{BC}=\left(6;-2\right)\)
\(\overrightarrow{BH}=\left(x+3;y-1\right);\overrightarrow{AC}=\left(1;-5\right)\)
H là trực tâm của ΔABC nên AH\(\perp\)BC; BH\(\perp\)AC
=>\(\left\{{}\begin{matrix}\overrightarrow{AH}\cdot\overrightarrow{BC}=0\\\overrightarrow{BH}\cdot\overrightarrow{AC}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6\left(x-2\right)+\left(-2\right)\left(y-4\right)=0\\1\left(x+3\right)+\left(-5\right)\left(y-1\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}6x-12-2y+8=0\\x+3-5y+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-2y=12-8=4\\x-5y=-8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x-y=2\\x-5y=-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{9}{7}\\y=\dfrac{13}{7}\end{matrix}\right.\)
vậy: H(9/7;13/7)
Gọi O(x;y) là tọa độ tâm đường tròn ngoại tiếp ΔABC
O(x;y); A(2;4); B(-3;1); C(3;-1)
\(OA^2=\left(2-x\right)^2+\left(4-y\right)^2=\left(x-2\right)^2+\left(y-4\right)^2\)
\(OB^2=\left(-3-x\right)^2+\left(1-y\right)^2=\left(x+3\right)^2+\left(y-1\right)^2\)
\(OC^2=\left(3-x\right)^2+\left(-1-y\right)^2=\left(x-3\right)^2+\left(y+1\right)^2\)
Vì O là tâm đường tròn ngoại tiếp ΔABC nên OA=OB=OC
=>\(\left\{{}\begin{matrix}\left(x-2\right)^2+\left(y-4\right)^2=\left(x+3\right)^2+\left(y-1\right)^2\\\left(x+3\right)^2+\left(y-1\right)^2=\left(x-3\right)^2+\left(y+1\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-4x+4+y^2-8y+16=x^2+6x+9+y^2-2y+1\\x^2+6x+9+y^2-2y+1=x^2-6x+9+y^2+2y+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-4x-8y+20=6x-2y+10\\6x-2y+10=-6x+2y+10\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-10x+6y=-10\\12x-4y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{4}\\y=-\dfrac{15}{4}\end{matrix}\right.\)
vậy: O(-5/4;-15/4)
