Question

Trong mặt phẳng tọa độ Oxy, cho tam giác ABC có A(3;4),B(2;1), C(-1;-2).Gọi M(x;y) là điểm trên đường thẳng BC sao cho \(S_{ABC}=4S_{AMB}\). Tính giá trị : \(x^2-y^2\)

Answer 1

Do \(M\in BC\Rightarrow d\left(A;BC\right)=d\left(A;BM\right)\Rightarrow\left[{}\begin{matrix}\overrightarrow{BC}=4\overrightarrow{BM}\\\overrightarrow{BC}=-4\overrightarrow{BM}\end{matrix}\right.\)

\(\overrightarrow{BC}=\left(-3;-3\right)\) ; \(\overrightarrow{BM}=\left(x-2;y-1\right)\)

TH1: \(\overrightarrow{BC}=4\overrightarrow{BM}\Rightarrow\left(-3;-3\right)=\left(4x-8;4y-4\right)\)

\(\Rightarrow\left\{{}\begin{matrix}4x-8=-3\\4y-4=-3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{5}{4}\\y=\frac{1}{4}\end{matrix}\right.\) \(\Rightarrow x^2-y^2=\frac{3}{2}\)

TH2: \(\overrightarrow{BC}=-4\overrightarrow{BM}\Rightarrow\left(-3;-3\right)=\left(-4x+8;-4y+4\right)\)

\(\Rightarrow\left\{{}\begin{matrix}-4x+8=-3\\-4y+4=-3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{11}{4}\\y=\frac{7}{4}\end{matrix}\right.\) \(\Rightarrow x^2-y^2=\frac{9}{2}\)

Answer 2

Vì M∈BC⇒d(A;BC)=d(A;BM)⇒⎡⎣−−→BC=4−−→BM−−→BC=−4−     −→BMM∈BC⇒d(A;BC)=d(A;BM)⇒[BC→=4BM→BC→=−4BM→

 ; 

Trường hợp 1: 

⇒{4x−8=−34y−4=−3⇒{4x−8=−34y−4=−3 ⇒{x=54y=14⇒{x=54y=14 

Trường hợp 2: 

    ⇒{−4x+8=−3−4y+4=−3⇒{−4x+8=−3−4y+4=−3 ⇒{x=114y=74⇒{x=114y=74