گام به گام کاردرکلاس صفحه 31 درس 1 حسابان یازدهم (جبر و معادله)

نقطه M وسط خط AB بدست می آوریم. اگر خط AB در نقطه M بر خط PM عمود باشد و اندازه خطوط PA و PB باهم برابر بودند، آنگاه نقطه P واقع بر عمود منصف AB خواهد بود:

\(\begin{array}{*{20}{l}}\begin{array}{l}\left\{ {\begin{array}{*{20}{l}}{{x_M} = \frac{{{x_A} + {x_B}}}{2} = \frac{{0 + 6}}{2} = 3}\\{{y_M} = \frac{{{y_A} + {y_B}}}{2} = \frac{{15 + \left( { - 3} \right)}}{2} = 6}\end{array}} \right.\\ \Rightarrow M\left| {\begin{array}{*{20}{l}}3\\6\end{array}} \right.\end{array}\\\begin{array}{l}\\\left\{ {\begin{array}{*{20}{l}}{{m_{PM}} = \frac{{{y_P} - {y_M}}}{{{x_P} - {x_M}}} = \frac{{11 - 6}}{{ - 12 - 3}} = \frac{5}{{ - 15}} = - \frac{1}{3}}\\{{m_{AB}} = \frac{{{y_A} - {y_B}}}{{{x_A} - {x_B}}} = \frac{{ - 3 - 15}}{{0 - 6}} = \frac{{ - 18}}{{ - 6}} = 3}\end{array}} \right.\\\\ \Rightarrow {m_{PM}} \times {m_{AB}} = - 1 \Rightarrow PM \bot AB\end{array}\\\begin{array}{l}\\\\\left\{ {\begin{array}{*{20}{l}}{PA = \sqrt {{{\left( {{x_A} - {x_P}} \right)}^2} + {{\left( {{y_A} - {y_P}} \right)}^2}} }\\{PB = \sqrt {{{\left( {{x_B} - {x_P}} \right)}^2} + {{\left( {{y_B} - {y_P}} \right)}^2}} }\end{array}} \right.\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{PA = \sqrt {{{\left( {0 - \left( { - 12} \right)} \right)}^2} + {{\left( { - 3 - 11} \right)}^2}} }\\{PB = \sqrt {{{\left( {6 - \left( { - 12} \right)} \right)}^2} + {{\left( {15 - 11} \right)}^2}} }\end{array}} \right.\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{PA = \sqrt {{{12}^2} + {{\left( { - 14} \right)}^2}} = 2\sqrt {85} }\\{PB = \sqrt {{{18}^2} + {{\left( 4 \right)}^2}} = 2\sqrt {85} }\end{array}} \right.\\\\ \Rightarrow PA = PB\end{array}\end{array}\)