a: A(3;1); B(5;3); C(-1;1)

\(AB=\sqrt{\left(5-3\right)^2+\left(3-1\right)^2}=\sqrt{2^2+2^2}=2\sqrt2\)

\(AC=\sqrt{\left(-1-3\right)^2+\left(1-1\right)^2}=\sqrt{\left(-4\right)^2}=4\)

\(BC=\sqrt{\left(-1-5\right)^2+\left(1-3\right)^2}=\sqrt{\left(-6\right)^2+\left(-2\right)^2}=2\sqrt{10}\)

=>ΔABC không là tam giác vuông cân

b: M(x;y); A(3;1); B(5;3); C(-1;1)

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

\(\overrightarrow{MA}-2\cdot\overrightarrow{MB}+4\cdot\overrightarrow{MC}=\overrightarrow{0}\)

=>3-x-2(5-x)+4(-1-x)=0 và 1-y-2(3-y)+4(1-y)=0

=>3-x-10+2x-4-4x=0 và 1-y-6+2y+4-4y=0

=>-3x+11=0 và -3y-1=0

=>-3x=-11 và 3y=-1

=>x=11/3 và y=-1/3

=>M(11/3;-1/3)

A(-1;1); B(3;9); C(2;4)

\(AB=\sqrt{\left(3+1\right)^2+\left(9-1\right)^2}=\sqrt{4^2+8^2}=\sqrt{16+64}=\sqrt{80}=4\sqrt5\)

\(AC=\sqrt{\left(2+1\right)^2+\left(4-1\right)^2}=\sqrt{3^2+3^2}=3\sqrt2\)

\(BC=\sqrt{\left(2-3\right)^2+\left(4-9\right)^2}=\sqrt{\left(-1\right)^2+\left(-5\right)^2}=\sqrt{26}\)

Xét ΔABC có \(cosBAC=\frac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)

\(=\frac{80+18-26}{2\cdot4\sqrt5\cdot3\sqrt2}=\frac{72}{24\sqrt{10}}=\frac{3}{\sqrt{10}}\)

=>\(\sin BAC=\sqrt{1-\left(\frac{3}{\sqrt{10}}\right)^2}=\frac{1}{\sqrt{10}}\)

Diện tích tam giác ABC là:

\(S_{ABC}=\frac12\cdot AB\cdot AC\cdot\sin BAC\)
\(=\frac12\cdot4\sqrt5\cdot3\sqrt2\cdot\frac{1}{\sqrt{10}}=\frac12\cdot4\cdot3=2\cdot3=6\)

Đề chính xác là \(\left|\overrightarrow{EB}-3\overrightarrow{EC}\right|\) đạt min đúng ko?

\(\overrightarrow{AB}=\left(2;-4\right)=2\left(1;-2\right)\) nên đường thẳng AB nhận \(\left(2;1\right)\) là 1 vtpt

Phương trình AB:

\(2\left(x+2\right)+1\left(y-1\right)=0\Leftrightarrow2x+y+3=0\)

Do E thuộc AB, đặt \(E\left(a;b\right)\Rightarrow2a+b+3=0\Rightarrow b=-2a-3\)

\(\Rightarrow E\left(a;-2a-3\right)\) \(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{EB}=\left(-a;2a\right)\\\overrightarrow{EC}=\left(1-a;2a+4\right)\end{matrix}\right.\)

\(\Rightarrow\overrightarrow{EB}-3\overrightarrow{EC}=\left(2a-3;-4a-12\right)\)

\(\Rightarrow\left|\overrightarrow{EB}-3\overrightarrow{EC}\right|=\sqrt{\left(2a-3\right)^2+\left(-4a-12\right)^2}=\sqrt{20a^2+84a+153}\)

\(=\sqrt{20\left(a+\dfrac{21}{10}\right)^2+\dfrac{324}{5}}\ge\sqrt{\dfrac{324}{5}}\)

Dấu = xảy ra khi \(a+\dfrac{21}{10}=0\Rightarrow a=-\dfrac{21}{10}\)

\(\Rightarrow E\left(-\dfrac{21}{10};\dfrac{6}{5}\right)\)

|(EB) ⃗ |+| 3(EC) ⃗ |

A(-1;1); B(1;-1); C(x;y)

\(\overrightarrow{AB}=\left(1+1;-1-1\right)=\left(2;-2\right)\) ; \(\overrightarrow{AC}=\left(x+1;y-1\right)\)

ΔABC vuông tại A

=>\(\overrightarrow{AB}\cdot\overrightarrow{AC}=0\)

=>2(x+1)+(-2)(y-1)=0

=>x+1-(y-1)=0

=>x+1-y+1=0

=>x-y+2=0

=>y=x+2

\(AB=\sqrt{2^2+\left(-2\right)^2}=\sqrt{4+4}=\sqrt8=2\sqrt2\)
\(AC=\sqrt{\left(x+1\right)^2+\left(y-1\right)^2}\)

AB=AC

=>\(\sqrt8=\sqrt{\left(x+1\right)^2+\left(y-1\right)^2}\)

=>\(\left(x+1\right)^2+\left(y-1\right)^2=8\)

=>\(\left(x+1\right)^2+\left(x+2-1\right)^2=8\)

=>\(2\left(x+1\right)^2=8\)

=>\(\left(x+1\right)^2=4\)

=>x+1=2 hoặc x+1=-2

=>x=1 hoặc x=-3

TH1: x=1

=>y=x+2=1+2=3

=>C(1;3)

TH2: x=-3

=>y=x+2=-3+2=-1

=>C(-3;-1)

\(\overrightarrow{CA}=\left(1-x_C;-2\right)\)

\(\overrightarrow{CB}=\left(-2-x_C;2\right)\)

\(\Leftrightarrow\left(x_C-1\right)\left(x_C+2\right)-4=0\)

\(\Leftrightarrow x_C^2+x_C-6=0\)

hay \(x_C=-3\)

Áp dụng công thức trọng tâm:

\(\left\{{}\begin{matrix}x_C=3x_G-x_A-x_B=-3\\y_C=3y_G-y_A-y_B=-4\end{matrix}\right.\) \(\Rightarrow C\left(-3;-4\right)\)

\(\Rightarrow\overrightarrow{CA}=\left(4;5\right)\) ; \(\overrightarrow{AB}=\left(1;2\right)\)

Đường cao d đi qua B vuông góc AC nên nhận \(\overrightarrow{CA}=\left(4;5\right)\) là 1 vtpt

Phương trình d: 

\(4\left(x-2\right)+5\left(y-3\right)=0\Leftrightarrow4x+5y-23=0\)

Đường cao d1 đi qua C vuông góc AB nên nhận (1;2) là 1 vtpt

Phương trình d1:

\(1\left(x+3\right)+2\left(y+4\right)=0\Leftrightarrow x+2y+11=0\)

H là giao điểm d và d1 nên tọa độ thỏa mãn:

\(\left\{{}\begin{matrix}4x+5y-23=0\\x+2y+11=0\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{101}{3}\\y=-\dfrac{67}{3}\end{matrix}\right.\) \(\Rightarrow H\left(\dfrac{101}{3};-\dfrac{67}{3}\right)\)

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