گام به گام تمرین صفحه 35 درس 1 حسابان (1) (جبر و معادله)

1)

الف)

ب)

\(\begin{array}{l}\left\{ \begin{array}{l}AB = \sqrt {{{\left( {{x_A} - {x_B}} \right)}^2} + {{\left( {{y_A} - {y_B}} \right)}^2}} \\BC = \sqrt {{{\left( {{x_B} - {x_C}} \right)}^2} + {{\left( {{y_B} - {y_C}} \right)}^2}} \end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}AB = \sqrt {{{\left( {\left( { - 1} \right) - \left( { - 6} \right)} \right)}^2} + {{\left( {7 - \left( { - 2} \right)} \right)}^2}} \\BC = \sqrt {{{\left( {\left( { - 6} \right) - 3} \right)}^2} + {{\left( {\left( { - 2} \right) - 3} \right)}^2}} \end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}AB = \sqrt {{5^2} + {9^2}}  = \sqrt {106} \\BC = \sqrt {{{\left( { - 9} \right)}^2} + {{\left( { - 5} \right)}^2}}  = \sqrt {106} \end{array} \right.\\\\ \Rightarrow AB = BC\end{array}\)

پ)

\(\begin{array}{l}\left\{ \begin{array}{l}{x_M} = \frac{{{x_B} + {x_C}}}{2} = \frac{{\left( { - 6} \right) + 3}}{2} = \frac{{ - 3}}{2}\\{y_M} = \frac{{{y_B} + {y_C}}}{2} = \frac{{\left( { - 2} \right) + 3}}{2} = \frac{1}{2}\end{array} \right.\\ \Rightarrow M\left| \begin{array}{l}\frac{{ - 3}}{2}\\\frac{1}{2}\end{array} \right.\\{m_{BC}} = \frac{{{y_B} - {y_C}}}{{{x_B} - {x_C}}} = \frac{{\left( { - 2} \right) - 3}}{{\left( { - 6} \right) - 3}} = \frac{{ - 5}}{{ - 9}} = \frac{5}{9}\\ \Rightarrow {{m'}_{BC}} = \frac{{ - 1}}{{{m_{BC}}}} =  - \frac{9}{5}\\ \Rightarrow y - {y_M} = {{m'}_{BC}}\left( {x - {x_M}} \right)\\ \Rightarrow y - \frac{1}{2} =  - \frac{9}{5}\left( {x - \frac{3}{2}} \right)\\ \Rightarrow y =  - \frac{9}{5}x + \frac{{16}}{5}\end{array}\)

ت)

\(\begin{array}{l}y - {y_C} = {m_{BC}}\left( {x - {x_C}} \right)\\ \Rightarrow y - 3 = \frac{5}{9}\left( {x - 3} \right)\\ \Rightarrow  - 5x + 9y - 12 = 0\\AH = \frac{{\left| { - 5{x_A} + 9{y_A} - 12} \right|}}{{\sqrt {{{\left( { - 5} \right)}^2} + {9^2}} }}\\ = \frac{{\left| { - 5\left( { - 1} \right) + 9\left( 7 \right) - 12} \right|}}{{\sqrt {106} }}\\ = \frac{{56}}{{\sqrt {106} }} = \frac{{28\sqrt {106} }}{{53}}\end{array}\)

2)

\(\begin{array}{l}\left\{ \begin{array}{l}{x_O} = \frac{{{x_A} + {x_B}}}{2} = \frac{{0 + 8}}{2} = 4\\{y_O} = \frac{{{y_A} + {y_B}}}{2} = \frac{{6 + \left( { - 8} \right)}}{2} =  - 1\end{array} \right.\\ \Rightarrow O\left| \begin{array}{l}4\\ - 1\end{array} \right.\\\\AB = \sqrt {{{\left( {{x_A} - {x_B}} \right)}^2} + {{\left( {{y_A} - {y_B}} \right)}^2}} \\ = \sqrt {{{\left( {0 - 8} \right)}^2} + {{\left( {6 - \left( { - 8} \right)} \right)}^2}} \\ = \sqrt {{{\left( { - 8} \right)}^2} + {{\left( {14} \right)}^2}}  = 2\sqrt {65} \\\\R = \frac{{AB}}{2} = \sqrt {65} \end{array}\)

3)

الف)

\(\begin{array}{l}{x^2} - 8x - 20 = 0\\ \Rightarrow \Delta  = {b^2} - 4ac = {\left( { - 8} \right)^2} - 4\left( 1 \right)\left( { - 20} \right) = 144\\x = \frac{{ - b \pm \sqrt \Delta  }}{{2a}} = \frac{{8 \pm 12}}{2}\\ \Rightarrow \left\{ \begin{array}{l}{x_A} = \frac{{8 - 12}}{2} =  - 2 \Rightarrow A\left| \begin{array}{l} - 2\\0\end{array} \right.\\{x_B} = \frac{{8 + 12}}{2} = 10 \Rightarrow B\left| \begin{array}{l}10\\0\end{array} \right.\end{array} \right.\end{array}\)

ب)

\(AB = \left| {{x_A} - {x_B}} \right| = \left| { - 2 - 10} \right| = 12\;cm\)

پ)

\(\begin{array}{l}{x_S} = \frac{{ - b}}{{2a}} = \frac{8}{2} = 4\\ \Rightarrow {y_{\min }} = {4^2} - 8 \times 4 - 20 =  - 16\\ \Rightarrow {D_{\max }} = \left| {{y_{\min }}} \right| = 16\;mm\end{array}\)

4)

\(\begin{array}{l}{L_1}:ax + by + c = 0\\{L_2}:ax + by + c' = 0\\ \Rightarrow A\left| \begin{array}{l}{x_A}\\{y_A}\end{array} \right. \in {L_1}:a{x_A} + b{y_A} + c = 0\\ \Rightarrow a{x_A} + b{y_A} =  - c\\\\ \Rightarrow AH = \frac{{\left| {a{x_A} + b{y_A} + c'} \right|}}{{\sqrt {{a^2} + {b^2}} }}\\ = \frac{{\left| { - c + c'} \right|}}{{\sqrt {{a^2} + {b^2}} }} = \frac{{\left| {c - c'} \right|}}{{\sqrt {{a^2} + {b^2}} }}\end{array}\)

5)

\(OT = \frac{{\left| {4{x_O} + 3{y_O} - 5} \right|}}{{\sqrt {{4^2} + {3^2}} }} = \frac{{\left| {4\left( { - 1} \right) + 3\left( 2 \right) - 5} \right|}}{5} = \frac{3}{5}\)

6)

الف)

\(\begin{array}{l}10 = \sqrt {x_S^2 + y_S^2}  = \sqrt {{x^2} + {8^2}} \\ \Rightarrow {x^2} + 64 = 100 \Rightarrow {x^2} = 36\\ \Rightarrow x = 6\end{array}\)

ب)

\(\begin{array}{l}P\left| \begin{array}{l} - 10\\0\end{array} \right.\quad ,\quad Q\left| \begin{array}{l}10\\0\end{array} \right.\\\left\{ \begin{array}{l}{m_{PS}} = \frac{{{y_P} - {y_S}}}{{{x_P} - {x_S}}} = \frac{{0 - 8}}{{ - 10 - 6}} = \frac{{ - 8}}{{ - 16}} = \frac{1}{2}\\{m_{SQ}} = \frac{{{y_S} - {y_Q}}}{{{x_S} - {x_Q}}} = \frac{{8 - 0}}{{6 - 10}} = \frac{8}{{ - 4}} =  - 2\end{array} \right.\end{array}\)

پ)

\({m_{PS}} \times {m_{SQ}} = \frac{1}{2} \times \left( { - 2} \right) =  - 1 \Rightarrow PS \bot SQ \Rightarrow P\hat SQ = {90^ \circ }\)

7)

\(\begin{array}{l}2 = \frac{{\left| {a\left( 1 \right) + 4\left( 2 \right) - 1} \right|}}{{\sqrt {{a^2} + {4^2}} }} = \frac{{\left| {a + 7} \right|}}{{\sqrt {{a^2} + {4^2}} }}\\ \Rightarrow 2\sqrt {{a^2} + {4^2}}  = \left| {a + 7} \right|\\\mathop  \Rightarrow \limits^{{{\left( {} \right)}^2}} 4\left( {{a^2} + {4^2}} \right) = {\left( {a + 7} \right)^2}\\ \Rightarrow 4{a^2} + 64 = {a^2} + 14a + 49\\ \Rightarrow 3{a^2} - 14a + 15 = 0\\ \Rightarrow \Delta  = {\left( { - 14} \right)^2} - 4\left( 3 \right)\left( {15} \right) = 16\\a = \frac{{14 \pm \sqrt {16} }}{6} = \frac{{14 \pm 4}}{6}\\ \Rightarrow \left\{ \begin{array}{l}{a_1} = \frac{{14 + 4}}{6} = 3\\{a_2} = \frac{{14 - 4}}{6} = \frac{5}{3}\end{array} \right.\end{array}\)

8)

الف)

\(\begin{array}{l}\left\{ \begin{array}{l}{x_M} = \frac{{{x_A} + {x_B}}}{2} = \frac{{\left( { - 11} \right) + \left( { - 3} \right)}}{2} =  - 7\\{y_M} = \frac{{{y_A} + {y_B}}}{2} = \frac{{\left( { - 13} \right) + 3}}{2} =  - 5\end{array} \right. \Rightarrow M\left| \begin{array}{l} - 7\\ - 5\end{array} \right.\\{m_{CM}} = \frac{{{y_C} - {y_M}}}{{{x_C} - {x_M}}} = \frac{{1 - \left( { - 5} \right)}}{{3 - \left( { - 7} \right)}} = \frac{3}{5}\\ \Rightarrow {L_{CM}}:y - {y_C} = {m_{CM}}\left( {x - {x_C}} \right)\\\\ \Rightarrow  - 3x + 5y + 4 = 0\\BH = \frac{{\left| { - 3{x_B} + 5{y_B} + 4} \right|}}{{\sqrt {{{\left( { - 3} \right)}^2} + {{\left( 5 \right)}^2}} }}\\ = \frac{{\left| { - 3\left( { - 3} \right) + 5\left( 3 \right) + 4} \right|}}{{\sqrt {{{\left( { - 3} \right)}^2} + {{\left( 5 \right)}^2}} }}\\ = \frac{{28}}{{\sqrt {34} }} = \frac{{14\sqrt {34} }}{{17}}\end{array}\)

ب) می دانیم که در متوازی الاضلاع ضلع های رو به روی هم موازی هستند.در نتیجه اگر خطوط موازی با دو ضلع BC و AB را به ترتیب از نقاط A و C رسم کنیم، محل تقاطع این دو خط  جدید رأس دیگر متوازی الاضلاع یا نقطه D را معین می کند:

\(\begin{array}{l}{m_{BC}} = \frac{{{y_B} - {y_C}}}{{{x_B} - {x_C}}} = \frac{{3 - 1}}{{\left( { - 3} \right) - 3}} =  - \frac{1}{3}\\ \Rightarrow {L_{AD}}:y - {y_A} = {m_{BC}}\left( {x - {x_A}} \right)\\ \Rightarrow {L_{AD}}:x + 3y =  - 50\\\\{m_{AB}} = \frac{{{y_A} - {y_B}}}{{{x_A} - {x_B}}} = \frac{{\left( { - 13} \right) - 3}}{{\left( { - 11} \right) - \left( { - 3} \right)}} = 2\\ \Rightarrow {L_{CD}}:y - {y_C} = {m_{AB}}\left( {x - {x_C}} \right)\\ \Rightarrow {L_{CD}}: - 2x + y =  - 5\\\\\left\{ \begin{array}{l}x + 3y =  - 50\\ - 2x + y =  - 5\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}{x_D} =  - 5\\{y_D} =  - 15\end{array} \right. \Rightarrow D\left| \begin{array}{l} - 5\\ - 15\end{array} \right.\end{array}\)

9)

\(\begin{array}{l}L:y = 2x \Rightarrow B\left| \begin{array}{l}{x_B}\\{y_B}\end{array} \right.\quad ,B \in L \Rightarrow {y_B} = 2{x_B}\\y = 2x \Rightarrow y - 2x = 0\\\left| {AB} \right| + \left| {OB} \right| = 5 \Rightarrow \\\left\{ \begin{array}{l}\left| {AB} \right| = \sqrt {{{\left( {{x_A} - {x_B}} \right)}^2} + {{\left( {{y_A} - {y_B}} \right)}^2}}  = \sqrt {{{\left( {2 - {x_B}} \right)}^2} + {{\left( {4 - {y_B}} \right)}^2}} \\\left| {OB} \right| = \sqrt {x_B^2 + y_B^2} \end{array} \right.\\\sqrt {{{\left( {2 - {x_B}} \right)}^2} + {{\left( {4 - 2{x_B}} \right)}^2}}  + \sqrt {x_B^2 + 4x_B^2}  = 5\\ \Rightarrow \sqrt {5x_B^2 - 20{x_B} + 20}  + \sqrt {5x_B^2}  = 5\\ \Rightarrow \sqrt 5 \left| {{x_B} - 2} \right| + \sqrt 5 \left| {{x_B}} \right| = 5\\ \Rightarrow \left| {{x_B} - 2} \right| + \left| {{x_B}} \right| = \sqrt 5 \\{x_B} < 0 \Rightarrow \left| {{x_B} - 2} \right| + \left| {{x_B}} \right| = \left( {2 - {x_B}} \right) + \left( { - {x_B}} \right)\\ = \sqrt 5  \Rightarrow {x_B} = \frac{{2 - \sqrt 5 }}{2}\\ \Rightarrow {y_B} = 2 - \sqrt 5  \Rightarrow B\left| \begin{array}{l}\frac{{2 - \sqrt 5 }}{2}\\2 - \sqrt 5 \end{array} \right.\\\\0 \le {x_B} < 2 \Rightarrow \left| {{x_B} - 2} \right| + \left| {{x_B}} \right| = \left( {{x_B} - 2} \right) + \left( { - {x_B}} \right)\\ = \sqrt 5  \Rightarrow  - 2 = \sqrt 5 \;\; \otimes \\\\2 \le {x_B} \Rightarrow \left| {{x_B} - 2} \right| + \left| {{x_B}} \right| = \left( {{x_B} - 2} \right) + \left( {{x_B}} \right)\\ = \sqrt 5  \Rightarrow {x_B} = \frac{{2 + \sqrt 5 }}{2}\\ \Rightarrow {y_B} = 2 + \sqrt 5  \Rightarrow B\left| \begin{array}{l}\frac{{2 + \sqrt 5 }}{2}\\2 + \sqrt 5 \end{array} \right.\end{array}\)

10)

\(\begin{array}{l}{m_{BC}} = \frac{{{y_B} - {y_C}}}{{{x_B} - {x_C}}} = \frac{{\left( { - 1} \right) - \left( { - 2} \right)}}{{1 - 8}} =  - \frac{1}{7}\\ \Rightarrow y - {y_B} = {m_{BC}}\left( {x - {x_B}} \right)\\ \Rightarrow {L_{BC}}:y =  - \frac{1}{7}x + \frac{8}{7}\\ \Rightarrow x + 7y = 8\\\\{m_{AH}} = \frac{{ - 1}}{{{m_{BC}}}} = 7\\ \Rightarrow y - {y_A} = {m_{AH}}\left( {x - {x_A}} \right)\\ \Rightarrow {L_{AH}}:y = 7x - 26\\ \Rightarrow  - 7x + y =  - 26\\\left\{ \begin{array}{l}x + 7y = 8\\ - 7x + y =  - 26\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}{x_H} = \frac{{19}}{5}\\{y_H} = \frac{3}{5}\end{array} \right.\\ \Rightarrow H\left| \begin{array}{l}\frac{{19}}{5}\\\frac{3}{5}\end{array} \right.\\\left\{ \begin{array}{l}{x_M} = \frac{{{x_B} + {x_C}}}{2} = \frac{{1 + 8}}{2} = \frac{9}{2}\\{y_M} = \frac{{{y_B} + {y_C}}}{2} = \frac{{\left( { - 1} \right) + \left( { - 2} \right)}}{2} =  - \frac{3}{2}\end{array} \right.\\ \Rightarrow M\left| \begin{array}{l}\frac{9}{2}\\ - \frac{3}{2}\end{array} \right.\\MH = \sqrt {{{\left( {{x_M} - {x_H}} \right)}^2} + {{\left( {{y_M} - {y_H}} \right)}^2}} \\ = \sqrt {{{\left( {\frac{9}{2} - \frac{{19}}{5}} \right)}^2} + {{\left( {\left( { - \frac{3}{2}} \right) - \frac{3}{5}} \right)}^2}}  = \frac{{7\sqrt {10} }}{{10}}\end{array}\)