جواب تمرین صفحه 9 درس 1 ریاضی یازدهم تجربی (هندسۀ تحلیلی و جبر)

\(\begin{array}{l}L:\quad 2x - y = 1 \Rightarrow y = 2x - 1 \Rightarrow {m_L} = 2\\\\T:\quad y = 2x - 3 \Rightarrow {m_T} = 2\\\\\Delta :\quad x + 2y = 0 \Rightarrow y = - \frac{1}{2}x \Rightarrow {m_\Delta } = - \frac{1}{2}\\\\ \Rightarrow \left\{ \begin{array}{l}{m_L} = {m_T}\\\\{m_L} \cdot {m_\Delta } = - 1\end{array} \right. \Rightarrow \quad \left\{ \begin{array}{l}L\parallel T\\\\L \bot \Delta \\\\T \bot \Delta \end{array} \right.\end{array}\)

\(\begin{array}{l}\left. \begin{array}{l}A\left( {14\;,\;3} \right)\\B\left( {10\;,\; - 13} \right)\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{x_M} = \frac{{{x_A} + {x_B}}}{2} = \frac{{14 + 10}}{2} = 12\\{y_M} = \frac{{{y_A} + {y_B}}}{2} = \frac{{3 + \left( { - 13} \right)}}{2} = - 5\end{array} \right. \Rightarrow M\left( {12\;,\; - 5} \right)\\O\left( {0\;,\;0} \right)\quad \Rightarrow OM = \sqrt {{{\left( {{x_M} - {x_O}} \right)}^2} + {{\left( {{y_M} - {y_O}} \right)}^2}} = \sqrt {{{\left( {12 - 0} \right)}^2} + {{\left( {\left( { - 5} \right) - 0} \right)}^2}} = \sqrt {169} = 13\end{array}\)

\(\begin{array}{l}\left. \begin{array}{l}A\left( {1\;,\;2} \right)\\B\left( {2\;,\;5} \right)\\C\left( {4\;,\;1} \right)\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}AB = \sqrt {{{\left( {{x_A} - {x_B}} \right)}^2} + {{\left( {{y_A} - {y_B}} \right)}^2}} = \sqrt {{{\left( {1 - 2} \right)}^2} + {{\left( {2 - 5} \right)}^2}} = \sqrt {1 + 9} = \sqrt {10} \\AC = \sqrt {{{\left( {{x_A} - {x_C}} \right)}^2} + {{\left( {{y_A} - {y_C}} \right)}^2}} = \sqrt {{{\left( {1 - 4} \right)}^2} + {{\left( {2 - 1} \right)}^2}} = \sqrt {9 + 1} = \sqrt {10} \end{array} \right. \Rightarrow AB = AC\quad \left( i \right)\\\left\{ \begin{array}{l}{m_{AB}} = \frac{{{y_A} - {y_B}}}{{{x_A} - {x_B}}} = \frac{{2 - 5}}{{1 - 2}} = 3\\{m_{AC}} = \frac{{{y_A} - {y_C}}}{{{x_A} - {x_C}}} = \frac{{2 - 1}}{{1 - 4}} = - \frac{1}{3}\end{array} \right. \Rightarrow {m_{AB}} \cdot {m_{AC}} = - 1 \Rightarrow AB \bot AC\quad \left( {ii} \right)\\\mathop \Rightarrow \limits^{\left( i \right)\;,\;\left( {ii} \right)} \end{array}\)

مثلث ABC متساوی الساقین قائم الزاویه است

الف

\(\begin{array}{l}A\left( {2\;,\; - 2} \right)\\\\B\left( {6\;,\;4} \right)\\\\ \Rightarrow 2r = AB = \\\\\sqrt {{{\left( {{x_A} - {x_B}} \right)}^2} + {{\left( {{y_A} - {y_B}} \right)}^2}} = \\\\\sqrt {{{\left( {2 - 6} \right)}^2} + {{\left( {\left( { - 2} \right) - 4} \right)}^2}} = \\\\\sqrt {16 + 36} = \sqrt {52} = 2\sqrt {13} \end{array}\)

 شعاع دایره  \( \Rightarrow r = \sqrt {13} \)

\(\left\{ \begin{array}{l}{x_O} = \frac{{{x_A} + {x_B}}}{2} = \frac{{2 + 6}}{2} = 4\\\\{y_O} = \frac{{{y_A} + {y_B}}}{2} = \frac{{ - 2 + 4}}{2} = 1\end{array} \right.\)

 ب برای اینکه بفهمیم این نقطه بر روی محیط دایره قرار دارد، بایستی فاصله آن را نسبت به مرکز دایره بدست آوریم. اگر برابر با شعاع دایره باشد، این نقطه روی محیط دایره قرار دارد:

\(\begin{array}{l}\overline {CO} = \sqrt {{{\left( {{x_O} - {x_C}} \right)}^2} + {{\left( {{y_O} - {y_C}} \right)}^2}} = \\\\\sqrt {{{\left( {4 - 7} \right)}^2} + {{\left( {1 - 3} \right)}^2}} = \\\\\sqrt {{{\left( { - 3} \right)}^2} + {{\left( { - 2} \right)}^2}} = \\\\\sqrt {9 + 4} = \sqrt {13} \end{array}\)

چون فاصله نقطه C از مرکز دایره یعنی O برابر با شعاع دایره هست، بنابراین بر روی محیط دایره واقع شده است.

 \(\begin{array}{l}MA = MC \Rightarrow \left\{ \begin{array}{l}{x_M} = \frac{{{x_A} + {x_C}}}{2} = \frac{{2 + 1}}{2} = \frac{3}{2}\\{y_M} = \frac{{{y_A} + {y_C}}}{2} = \frac{{3 + \left( { - 2} \right)}}{2} = \frac{1}{2}\end{array} \right. \Rightarrow M\left( {\frac{3}{2}\;,\;\frac{1}{2}} \right)\\MB = MD \Rightarrow \left\{ \begin{array}{l}{x_M} = \frac{{{x_B} + {x_D}}}{2} \Rightarrow {x_D} = 2{x_M} - {x_B} = 4\\{y_M} = \frac{{{y_B} + {y_D}}}{2} \Rightarrow {y_D} = 2{y_M} - {y_B} = 1\end{array} \right. \Rightarrow D\left( {4\;,\;1} \right)\\\\\end{array}\)

راه حل کوتاه تر :

\(AD = BC \Rightarrow \left\{ \begin{array}{l}{x_D} - {x_A} = {x_C} - {x_B} \Rightarrow {x_D} - 2 = 2 \Rightarrow {x_D} = 4\\{y_D} - {y_A} = {y_C} - {y_B} \Rightarrow {y_D} - 3 = - 2 \Rightarrow {y_D} = 1\end{array} \right. \Rightarrow D\left( {4\;,\;1} \right)\)

\(\begin{array}{l}MA = MB \Rightarrow \left\{ \begin{array}{l}{x_M} = \frac{{{x_A} + {x_B}}}{2} = \frac{{ - 3 + 5}}{2} = 1\\{y_M} = \frac{{{y_A} + {y_B}}}{2} = \frac{{ - 1 + 3}}{2} = 1\end{array} \right. \Rightarrow M\left( {1\;,\;1} \right)\\MC = MD \Rightarrow \left\{ \begin{array}{l}{x_M} = \frac{{{x_C} + {x_D}}}{2} \Rightarrow {x_D} = 2{x_M} - {x_C} = 3\\{y_M} = \frac{{{y_C} + {y_D}}}{2} \Rightarrow {y_D} = 2{y_M} - {y_C} = 1\end{array} \right. \Rightarrow D\left( {3\;,\;3} \right)\end{array}\)

نقطه A روی خط قرار ندارد، بنابراین از نقطه A بر خط L عمود می کنیم. فاصله این نقطه از خط، طول ضلع مربع است :

\(\left\{ \begin{array}{l}A\left( {3\;,\;0} \right) = \left( {{x_ \circ }\;,\;{y_ \circ }} \right)\\y = 2x - 1 \Rightarrow - 2x + y + 1 = 0\end{array} \right. \Rightarrow AB = \frac{{\left| {a{x_ \circ } + b{y_ \circ } + c} \right|}}{{\sqrt {{a^2} + {b^2}} }} = \frac{{\left| { - 2\left( 3 \right) + 1\left( 0 \right) + 1} \right|}}{{\sqrt {{{\left( { - 2} \right)}^2} + {{\left( 1 \right)}^2}} }} = \frac{5}{{\sqrt 5 }} = \sqrt 5 \)

\( \Rightarrow S = A{B^2} = {\left( {\sqrt 5 } \right)^2} = 5\)

الف

\(\left\{ \begin{array}{l}L:\quad 5x - 12y + 8 = 0 \Rightarrow y = \frac{5}{{12}}x + \frac{2}{3} \Rightarrow {m_L} = \frac{5}{{12}}\\L':\quad - 10x + 24y + 10 = 0 \Rightarrow y = \frac{5}{{12}}x - \frac{5}{{12}} \Rightarrow {m_{L'}} = \frac{5}{{12}}\end{array} \right. \Rightarrow {m_L} = {m_{L'}} \Rightarrow L\parallel L'\)

ب

\(\left\{ \begin{array}{l}L:\quad 5x - 12y + 8 = 0 \Rightarrow A\left( {8,4} \right)\\L':\quad - 10x + 24y + 10 = 0\end{array} \right. \Rightarrow d = \frac{{\left| {a'{x_A} + b'{y_A} + c'} \right|}}{{\sqrt {{{a'}^2} + {{b'}^2}} }} = \frac{{\left| { - 10\left( 8 \right) + 24\left( 4 \right) + 10} \right|}}{{\sqrt {{{\left( { - 10} \right)}^2} + {{24}^2}} }} = \frac{{26}}{{26}} = 1\)

\(\begin{array}{l}\left. \begin{array}{l}A\left( {{{46}^ \circ }\;,\;{{38}^ \circ }} \right)\\B\left( {{{61}^ \circ }\;,\;{{25}^ \circ }} \right)\end{array} \right\} \Rightarrow AB = \sqrt {{{\left( {{x_A} - {x_B}} \right)}^2} + {{\left( {{y_A} - {y_B}} \right)}^2}} \\\begin{array}{*{20}{c}}{}&{}&{}&{}\end{array}\begin{array}{*{20}{c}}{}&{}&{}\end{array}\quad \; = \sqrt {{{\left( {46 - 61} \right)}^2} + {{\left( {38 - 25} \right)}^2}} = \sqrt {{{\left( { - 15} \right)}^2} + {{\left( {13} \right)}^2}} \simeq 19/{8^ \circ } \times 110 = 2178\end{array}\)