Trong mặt phẳng với hệ toạ độ $Oxy$, cho hình chữ nhật $ABCD$ có cạnh $AB: x -2y -1 =0$, đường chéo $BD: x- 7y +14 = 0$ và đường chéo $AC$ đi qua điểm $M(2;1)$. Tìm toạ độ các đỉnh của hình chữ nhật.
$BD \cap AB = B(7;3)$, pt đg thẳng BC: 2x + y – 17 = 0 do $BC \bot AB$
$A \in AB \Rightarrow A(2a + 1;a),C \in BC \Rightarrow C(c;17 - 2c),a \ne 3,c \ne 7  do  A,C\not\equiv B$,
I =$\left( {\frac{{2a + c + 1}}{2};\frac{{a - 2c + 17}}{2}} \right)$ là trung điểm của AC, BD (tâm hình chữ nhật).
I$ \in BD \Leftrightarrow 3c - a - 18 = 0 \Leftrightarrow a = 3c - 18 \Rightarrow A(6c - 35;3c - 18)$
M, A, C thẳng hàng $\Rightarrow \overrightarrow {MA} ,\overrightarrow {MC} \Rightarrow c^2 – 13c +42 =0 \Rightarrow \left[ \begin{array}{l}
c = 7(loai)\\
c = 6
\end{array} \right. $
$c = 6 \Rightarrow A(1;0), C(6;5) , D(0;2), B(7;3)$.

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