Đáp án:
\(\begin{array}{l}a)\,D\left( {12;4} \right)\\b)\,\,M\left( {0;\frac{{10}}{3}} \right)\\c)\,\,N\left( {\frac{{35}}{4};0} \right)\end{array}\)
Giải thích các bước giải:
a) Để ABCD là hình bình hành
\(\begin{array}{l} \Rightarrow \overrightarrow {AB} = \overrightarrow {DC} \Rightarrow \left\{ \begin{array}{l} - 2 - 1 = 9 - {x_D}\\6 - 2 = 8 - {y_D}\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l} - 3 = 9 - {x_D}\\4 = 8 - {y_D}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}{x_D} = 12\\{y_D} = 4\end{array} \right.\\ \Rightarrow D\left( {12;4} \right)\end{array}\)
b) Gọi \(M\left( {0;m} \right) \in Oy\)
Để B, M, A thẳng hàng
\( \Rightarrow \exists k \in R\,\,\left( {k \ne 0} \right):\,\,\overrightarrow {AM} = k\overrightarrow {AB} \)
\( \Rightarrow \left\{ \begin{array}{l} - 1 = k.\left( { - 3} \right)\\m - 2 = k.4\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}k = \frac{1}{3}\\m - 2 = \frac{4}{3}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}k = \frac{1}{3}\\m = \frac{{10}}{3}\end{array} \right.\)
Vậy \(M\left( {0;\frac{{10}}{3}} \right)\) .
c) Gọi \(N\left( {n;0} \right) \in Ox\).
Tam giác ANC cân tại N
\(\begin{array}{l} \Rightarrow NA = NC \Rightarrow N{A^2} = N{C^2}\\ \Rightarrow {\left( {1 - n} \right)^2} + {2^2} = {\left( {9 - n} \right)^2} + {8^2}\\ \Leftrightarrow {n^2} - 2n + 5 = {n^2} - 18n + 145\\ \Leftrightarrow 16n = 140\\ \Leftrightarrow n = \frac{{35}}{4}\end{array}\)
Vậy \(N\left( {\frac{{35}}{4};0} \right)\).
