Cho  $ \Delta ABC$ có $A(5;3); B( - 1;2); C( - 4;5) $, viết phương trình đường thẳng đi qua $A$ và chia tam giác $ABC$ thành $2$ phần có tỉ số diện tích là $\frac{1}{2}.$
Gọi M(a;b) , ta có:  $ \left\{ \begin{array}{l}
\overrightarrow {BM}  = (a + 1;b - 2)\\
\overrightarrow {BC}  = \left( { - 3;3} \right)
\end{array} \right. $
Có 2 trường hợp thỏa mãn $ \left[ \begin{array}{l}
\overrightarrow {BM}  = \frac{1}{3}\overrightarrow {BC} \\
\overrightarrow {BM}  = \frac{2}{3}\overrightarrow {BC}
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 1 =  - 1\\
y - 2 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 =  - 2\\
y - 2 = 2
\end{array} \right.
\end{array} \right.$
$\Rightarrow \left[ \begin{array}{l}
M( - 2;3)\\
M( - 3;4)
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\overrightarrow {AM}  = ( - 7;0)\\
\overrightarrow {AM}  = ( - 8;1)
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
d:y - 3 = 0\\
d:x + 8y - 29 = 0
\end{array} \right. $

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