جواب فعالیت صفحه 66 درس 4 ریاضی دوازدهم تجربی (مشتق)

\(\begin{array}{l}A = \left| \begin{array}{l} - 1\\\\0\end{array} \right.\,\,\,\,\,,\,\,\,\,\,B = \left| \begin{array}{l}2\\\\2\end{array} \right.\\\\{m_{AB}} = \frac{{2 - 0}}{{2 - ( - 1)}} = \frac{2}{3}\\\\ \Rightarrow {m_{AB}} > 0\end{array}\)

\(\begin{array}{l}A = \left| \begin{array}{l} - 2\\\\2\end{array} \right.\,\,\,\,\,,\,\,\,\,\,B = \left| \begin{array}{l}0\\\\0\end{array} \right.\\\\{m_{AB}} = \frac{{0 - 2}}{{0 - ( - 2)}} = \frac{{ - 2}}{2} = - 1\\\\ \Rightarrow {m_{AB}} < 0\end{array}\)

\(\begin{array}{l}A\left| {\begin{array}{*{20}{c}}{ - 1}\\{ - 2}\end{array}} \right.\;,\;O\left| {\begin{array}{*{20}{c}}0\\0\end{array}} \right. \Rightarrow m = \frac{{0 - ( - 2)}}{{0 - ( - 1)}} = 2 = {m_{{d_3}}}\\B\left| {\begin{array}{*{20}{c}}{ - 3}\\0\end{array}} \right.\;,\;C\left| {\begin{array}{*{20}{c}}0\\{ - 2}\end{array}} \right. \Rightarrow m = \frac{{ - 2 - 0}}{{0 - ( - 3)}} = - \frac{2}{3} = {m_{{d_4}}}\end{array}\)

\(\begin{array}{l}D\left| {\begin{array}{*{20}{c}}{ - 2}\\0\end{array}} \right.\;,\;E\left| {\begin{array}{*{20}{c}}3\\2\end{array}} \right. \Rightarrow m = \frac{{3 - ( - 2)}}{{2 - 0}} = \frac{5}{2} = {m_{{d_1}}}\\F\left| {\begin{array}{*{20}{c}}0\\3\end{array}} \right.\;,\;G\left| {\begin{array}{*{20}{c}}1\\0\end{array}} \right. \Rightarrow m = \frac{{0 - 3}}{{1 - 0}} = - 3 = {m_{{d_2}}}\end{array}\)