درسنامه کامل ریاضی هفتم فصل 8 بردار و مختصات

1 ناحیه چهارم                        2 ناحیه سوم                       3 ناحیه اول

4 مرکز مختصات                     5 ناحیه دوم                        6 محور x های منفی

7 محور y های مثبت               8 ناحیه اول (بر روی نیم ساز ناحیه اول و سوم)

هنگامی که مختصات نقطه ای بر روی محور y ها باشد، مؤلفه x آن برابر صفر خواهد بود:

\(m + 6 = 0 \Rightarrow m = - 6 \Rightarrow \left[ {\begin{array}{*{20}{c}} \circ \\{13}\end{array}} \right]\)

\(\begin{array}{l} - 4a + 1 = 3a + 8 \Rightarrow - 4a - 3a = 8 - 1\\\\ \Rightarrow - 7a = 7 \Rightarrow a = - 1 \Rightarrow \left[ \begin{array}{l}5\\5\end{array} \right]\end{array}\)

نقطه در ناحیه اول قرار دارد.

\(\begin{array}{l}\left. \begin{array}{l}A = \left[ \begin{array}{l}5\\4\end{array} \right] = \left[ \begin{array}{l}{x_A}\\{y_A}\end{array} \right]\\\\B = \left[ \begin{array}{l} - 3\\6\end{array} \right] = \left[ \begin{array}{l}{x_B}\\{y_B}\end{array} \right]\end{array} \right\} \Rightarrow M = \left[ \begin{array}{l}{x_M}\\{y_M}\end{array} \right]\\\\\left. \begin{array}{l}{x_M} = \frac{{{x_A} + {x_B}}}{2} = \frac{{5 + ( - 3)}}{2} = \frac{2}{2} = 1\\\\{y_M} = \frac{{{y_A} + {y_B}}}{2} = \frac{{6 + 4}}{2} = \frac{{10}}{2} = 5\end{array} \right\} \Rightarrow M = \left[ \begin{array}{l}1\\5\end{array} \right]\end{array}\)

فقط بردار \(\overrightarrow O = \left[ {\begin{array}{*{20}{c}} \circ \\ \circ \end{array}} \right]\) ابتدا و انتهای آن روی هم قرار دارند؛ بنابراین:

\(\begin{array}{l}\left[ \begin{array}{l}9 - 3x\\1 - \frac{y}{3}\end{array} \right] = \left[ {\begin{array}{*{20}{c}} \circ \\ \circ \end{array}} \right] \Rightarrow \left\{ \begin{array}{l}9 - 3x = 0 \Rightarrow 9 = 3x \Rightarrow x = 3\\\\1 - \frac{y}{3} = 0 \Rightarrow 1 = \frac{y}{3} \Rightarrow y = 3\end{array} \right.\\\\ \Rightarrow x + y = 6\end{array}\)

\(\begin{array}{l}1)\,\,\left[ \begin{array}{l} - 2\\4\end{array} \right]\,\,\,\,\,\,y \to \,\,\,\,\,\,\left[ \begin{array}{l}2\\4\end{array} \right]\\\\2)\,\,\left[ \begin{array}{l}5\\ - 3\end{array} \right]\,\,\,\,\,\,\,x \to \,\,\,\,\left[ \begin{array}{l}5\\3\end{array} \right]\\\\3)\,\,\left[ \begin{array}{l} - 15\\ - 20\end{array} \right]\,\,\,\,\,\,\,\,\,y \to \,\,\,\,\,\left[ \begin{array}{l}15\\ - 20\end{array} \right]\\\\4)\,\,\left[ \begin{array}{l} - 15\\ - 20\end{array} \right]\,\,\,\,\,\,\,x \to \,\,\,\,\,\left[ \begin{array}{l} - 15\\20\end{array} \right]\\\\5)\,\,\left[ \begin{array}{l}15\\ - 20\end{array} \right]\,\,\,\,\,\,\,\left[ \begin{array}{l} \circ \\ \circ \end{array} \right] \to \,\,\,\,\left[ \begin{array}{l} - 15\\20\end{array} \right]\\\\6)\,\,\left[ \begin{array}{l} - 10\\ + 3\end{array} \right]\,\,\,\,\,\,\,\left[ \begin{array}{l} \circ \\ \circ \end{array} \right] \to \,\,\,\,\left[ \begin{array}{l} + 10\\ - 3\end{array} \right]\end{array}\)

ابتدا بردار \(\overrightarrow {AC} \) را بدست می آوریم:

\(\begin{array}{l}\overrightarrow {BA} = \overrightarrow {OA} - \overrightarrow {OB} = \left[ {\begin{array}{*{20}{c}}{ - 4}\\3\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{ - 2}\\1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 2}\\2\end{array}} \right]\\\\2\overrightarrow {BA} + \overrightarrow {AC} = \left[ {\begin{array}{*{20}{c}}{ - 5}\\4\end{array}} \right] \Rightarrow 2\left[ {\begin{array}{*{20}{c}}{ - 2}\\2\end{array}} \right] + \overrightarrow {AC} = \left[ {\begin{array}{*{20}{c}}{ - 5}\\4\end{array}} \right]\\\\ \Rightarrow \left[ {\begin{array}{*{20}{c}}{ - 4}\\4\end{array}} \right] + \overrightarrow {AC} = \left[ {\begin{array}{*{20}{c}}{ - 5}\\4\end{array}} \right] \Rightarrow \overrightarrow {AC} = \left[ {\begin{array}{*{20}{c}}{ - 5}\\4\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{ - 4}\\4\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 1}\\ \circ \end{array}} \right]\\\\\overrightarrow {AC} = \overrightarrow {OC} - \overrightarrow {OA} \Rightarrow \left[ {\begin{array}{*{20}{c}}{ - 1}\\ \circ \end{array}} \right] = \overrightarrow {OC} - \left[ {\begin{array}{*{20}{c}}{ - 4}\\3\end{array}} \right]\\\\ \Rightarrow \overrightarrow {OC} = \left[ {\begin{array}{*{20}{c}}{ - 1}\\ \circ \end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 4}\\3\end{array}} \right] \Rightarrow \overrightarrow {OC} = \left[ {\begin{array}{*{20}{c}}{ - 5}\\3\end{array}} \right] \Rightarrow C = \left[ {\begin{array}{*{20}{c}}{ - 5}\\3\end{array}} \right]\end{array}\)

\(\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{4x + 3}\\{7 + y}\end{array}} \right]\, = \left[ {\begin{array}{*{20}{c}}{x - 3}\\{10 - 2y}\end{array}} \right]\\\\4x + 3 = x - 3 \Rightarrow 3x = - 3 \Rightarrow x = - 1\\\\7 + y = 10 - 2y \Rightarrow 3y = - 3 \Rightarrow y = - 1\\\\ \Rightarrow \frac{{3x - 1}}{{4y + 2}} = \frac{{3( - 1) - 1}}{{4( - 1) + 2}} = \frac{{ - 4}}{{ - 2}} = 2\end{array}\)