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Kẻ \(AH\perp BC\)
\(\Rightarrow S_{ABN}=\frac{1}{2}AH.BN\) ; \(S_{ANC}=\frac{1}{2}AH.NC\)
Mà \(S_{ABN}=3S_{ANC}\Rightarrow\frac{1}{2}AH.BN=\frac{3}{2}AH.NC\)
\(\Rightarrow BN=3NC\Rightarrow\overrightarrow{BN}=3\overrightarrow{NC}\)
Gọi \(N\left(a;b\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{BN}=\left(a-2;b-3\right)\\\overrightarrow{NC}=\left(-1-a;-2-b\right)\end{matrix}\right.\)
\(\Rightarrow\left(a-2;b-3\right)=3\left(-1-a;-2-b\right)\)
\(\Rightarrow\left\{{}\begin{matrix}a-2=-3-3a\\b-3=-6-3b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-\frac{1}{4}\\b=-\frac{3}{4}\end{matrix}\right.\) \(\Rightarrow N\left(-\frac{1}{4};-\frac{3}{4}\right)\)
Gọi G là trọng tâm tam giác\(\Rightarrow\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}\)
Đặt \(P=MA^2+MB^2+MC^2=\left(\overrightarrow{MG}+\overrightarrow{GA}\right)^2+\left(\overrightarrow{MG}+\overrightarrow{GB}\right)^2+\left(\overrightarrow{MG}+\overrightarrow{GC}\right)^2\)
\(=3MG^2+GA^2+GB^2+GC^2+2\overrightarrow{MG}\left(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}\right)\)
\(=3MG^2+GA^2+GB^2+GC^2\)
Do \(GA^2+GB^2+GC^2\) ko đổi nên \(P_{min}\) khi \(MG_{min}\Leftrightarrow M\) là chân đường vuông góc hạ từ G xuống BC
\(\Rightarrow\dfrac{CM}{BC}=\dfrac{2}{3}\Rightarrow\dfrac{BM}{BC}=\dfrac{1}{3}\)
\(\Rightarrow\dfrac{S_{ABM}}{S_{ABC}}=\dfrac{1}{3}\)
\(\overrightarrow{BC}=\left(16;4\right)=4\left(4;1\right)\) ; \(\overrightarrow{AC}=\left(2;2\right)=2\left(1;1\right)\)
Phương trình đường cao xuất phát từ A và vuông góc BC:
\(4\left(x-3\right)+1\left(y-2\right)=0\Leftrightarrow4x+y-14=0\)
Pt đường cao xuất phát từ B:
\(1\left(x+11\right)+1\left(y-0\right)=0\Leftrightarrow x+y+11=0\)
Tọa độ H là nghiệm: \(\left\{{}\begin{matrix}4x+y-14=0\\x+y+11=0\end{matrix}\right.\) \(\Rightarrow H\left(\dfrac{25}{3};-\dfrac{58}{3}\right)\)
a.
\(\left\{{}\begin{matrix}\overrightarrow{AB}=\left(-1;8\right)\\\overrightarrow{AC}=\left(3;6\right)\end{matrix}\right.\) mà \(\dfrac{-1}{3}\ne\dfrac{8}{6}\Rightarrow\overrightarrow{AB}\) và \(\overrightarrow{AC}\) không cùng phương hay A,B,C không thẳng hàng
\(\Rightarrow A,B,C\) là 3 đỉnh của 1 tam giác
b.
Theo công thức trung điểm: \(\left\{{}\begin{matrix}x_I=\dfrac{x_A+x_C}{2}=\dfrac{1+4}{2}=\dfrac{5}{2}\\y_I=\dfrac{y_A+y_C}{2}=\dfrac{-3+3}{2}=0\end{matrix}\right.\)
\(\Rightarrow C\left(\dfrac{5}{2};0\right)\)
Gọi G là trọng tâm tam giác, theo công thức trọng tâm:
\(\left\{{}\begin{matrix}x_G=\dfrac{x_A+x_B+x_C}{3}=\dfrac{1+0+4}{3}=\dfrac{5}{3}\\y_G=\dfrac{y_A+y_B+y_C}{3}=\dfrac{-3+5+3}{3}=\dfrac{5}{3}\\\end{matrix}\right.\) \(\Rightarrow G\left(\dfrac{5}{3};\dfrac{5}{3}\right)\)
c.
Gọi \(D\left(x;y\right)\Rightarrow\overrightarrow{DC}=\left(4-x;3-y\right)\)
ABCD là hình bình hành khi \(\overrightarrow{AB}=\overrightarrow{DC}\)
\(\Rightarrow\left\{{}\begin{matrix}4-x=-1\\3-y=8\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=5\\y=-5\end{matrix}\right.\)
\(\Rightarrow D\left(5;-5\right)\)
1.
\(\left\{{}\begin{matrix}x_I=\dfrac{x_A+x_B}{2}=-\dfrac{3}{2}\\y_I=\dfrac{y_A+y_B}{2}=1\end{matrix}\right.\) \(\Rightarrow I\left(-\dfrac{3}{2};1\right)\)
\(\left\{{}\begin{matrix}x_G=\dfrac{x_A+x_B+x_C}{3}=0\\y_G=\dfrac{y_A+y_B+y_C}{3}=0\end{matrix}\right.\) \(\Rightarrow G\left(0;0\right)\)
2.
\(\left\{{}\begin{matrix}\overrightarrow{CI}=\left(-\dfrac{9}{2};3\right)\\\overrightarrow{AG}=\left(-2;-3\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}CI=\sqrt{\left(-\dfrac{9}{2}\right)^2+3^2}=\dfrac{3\sqrt{13}}{2}\\AG=\sqrt{\left(-2\right)^2+\left(-3\right)^2}=\sqrt{13}\end{matrix}\right.\)
3.
Gọi \(D\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AB}=\left(-7;-4\right)\\\overrightarrow{DC}=\left(3-x;-2-y\right)\end{matrix}\right.\)
\(ABCD\) là hbh \(\Leftrightarrow\overrightarrow{AB}=\overrightarrow{DC}\)
\(\Leftrightarrow\left\{{}\begin{matrix}-7=3-x\\-4=-2-y\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=10\\y=2\end{matrix}\right.\)
\(\Rightarrow D\left(10;2\right)\)
4. Gọi \(H\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{CH}=\left(x-3;y+2\right)\\\overrightarrow{AH}=\left(x-2;y-3\right)\\\overrightarrow{BC}=\left(8;-1\right)\end{matrix}\right.\)
H là trực tâm \(\Leftrightarrow\left\{{}\begin{matrix}AH\perp BC\\CH\perp AB\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\overrightarrow{AH}.\overrightarrow{BC}=0\\\overrightarrow{CH}.\overrightarrow{AB}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}8\left(x-2\right)-1\left(y-3\right)=0\\-7\left(x-3\right)-4\left(y+2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}8x-y=13\\-7x-4y=-13\end{matrix}\right.\) \(\Rightarrow H\left(\dfrac{5}{3};\dfrac{1}{3}\right)\)
\(S_{ABN}=3S_{ANC}\) , mà \(S_{ABN}+S_{ANC}=S_{ABC}\)
\(\Rightarrow S_{ANC}=\dfrac{1}{4}S_{ABC}\Rightarrow\overrightarrow{NC}=\dfrac{1}{4}\overrightarrow{BC}\)
Gọi \(N\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{NC}=\left(-1-x;-2-y\right)\\\overrightarrow{BC}=\left(-3;-5\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}-1-x=-\dfrac{3}{4}\\-2-y=-\dfrac{5}{4}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\\y=-\dfrac{3}{4}\end{matrix}\right.\) \(\Rightarrow N\left(-\dfrac{1}{4};-\dfrac{3}{4}\right)\)