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Аналитичка геометрија у равни - решени задаци 9

Међусобни положај правих. Угао између две праве, услов паралелености и услов нормалности. Примена на троугао. Решени задаци.

Задаци

Текст задатака објашњених у видео лекцији.

Пр.7)   Дата су темена троугла $A\left( { - 3, - 2} \right)$, $B\left( {2, - 1} \right)$ и $C\left( {1,4} \right)$.

           Одредити: а) једначину странице $b$

                             б) једначину висине ${h_b}$

                             в) дужину тежишне дужи ${t_a}$

                             г) угао $\alpha $

                             д) једначину симетрале странице $b$

                             ђ) дужину висине ${h_c}$

                             е) координате центра описане кружнице $O$

                             ф) тачку $D$ која дели дуж $AB$ у размери 1:5

Пр.7)  

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a) 

\[\begin{gathered}
b = p\left( {AC} \right):y - {y_1} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\left( {x - {x_1}} \right) \hfill \\
y - 2 = \frac{{4 + 2}}{{1 + 3}}\left( {x - 3} \right) \hfill \\
y - 2 = \frac{6}{4}\left( {x + 3} \right) \hfill \\
y = \frac{3}{2}x + \frac{9}{2} - 2 \hfill \\
y = \frac{3}{2}x + \frac{5}{2} \hfill \\
\end{gathered} \]

б)

\[\begin{gathered}
\begin{array}{*{20}{c}}
{{h_b}:{h_b} \bot b}& \cap &{B \in {h_b}} \\
{{k_{{h_b}}} \cdot {k_b} = - 1}&{}&{y = - \frac{2}{3}x + n} \\
{{k_{{h_b}}} \cdot \frac{3}{2} = - 1}&{}&{ - 1 = - \frac{2}{3} \cdot 2 + n} \\
{{k_{{h_b}}} = - \frac{2}{3}}&{}&{n = - 1 + \frac{4}{3}} \\
{}&{}&{n = \frac{1}{3}}
\end{array} \hfill \\
{h_b}:y = - \frac{2}{3}x + \frac{1}{3} \hfill \\
\end{gathered} \]

в) 

\[\begin{gathered}
{A_1}\left( {\frac{{{x_B} + {x_C}}}{2},\frac{{{y_B} + {y_C}}}{2}} \right) \hfill \\
{A_1}\left( {\frac{3}{2},\frac{3}{2}} \right) \hfill \\
\left| {{t_a}} \right| = d\left( {A{A_1}} \right) = \sqrt {{{\left( {{x_A} - {x_{{A_1}}}} \right)}^2} + {{\left( {{y_A} - {y_{{A_1}}}} \right)}^2}} = \hfill \\
= \sqrt {{{\left( { - 3 - \frac{3}{2}} \right)}^2} + {{\left( { - 2 - \frac{3}{2}} \right)}^2}} = \frac{{\sqrt {130} }}{2} \hfill \\
\end{gathered} \]

г)

\[\begin{gathered}
\alpha = \measuredangle \left( {b,c} \right) \hfill \\
c = p\left( {AB} \right):y - {y_1} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\left( {x - {x_1}} \right) \hfill \\
y + 2 = \frac{{ - 1 + 2}}{{2 + 3}}\left( {x + 3} \right) \hfill \\
y + 2 = \frac{1}{5}\left( {x + 3} \right) \hfill \\
y = \frac{1}{5}x + \frac{3}{5} - 2 \hfill \\
c:y = \frac{3}{2}x + \frac{5}{2} \hfill \\
tg\alpha = \left| {\frac{{{k_2} - {k_1}}}{{1 + {k_1} \cdot {k_2}}}} \right| = \left| {\frac{{\frac{3}{2} - \frac{1}{5}}}{{1 + \frac{3}{2} \cdot \frac{1}{5}}}} \right| = \left| {\frac{{\frac{{13}}{{10}}}}{{\frac{{13}}{{10}}}}} \right| = 1 \hfill \\
\alpha = {45^ \circ } \hfill \\
\end{gathered} \]

д)

\[\begin{gathered}
\begin{array}{*{20}{c}}
{{s_b}:{s_b} \bot b}& \cap &{{B_1} \in {s_b}} \\
{{k_{{s_b}}} \cdot {k_{_b}} = - 1}&{}&{{B_1}\left( {\frac{{ - 3 + 1}}{2},\frac{{ - 2 + 4}}{2}} \right)} \\
{{k_{{s_b}}} \cdot \frac{3}{2} = - 1}&{}&{{B_1}\left( { - 1,1} \right)} \\
{{k_{{s_b}}} = - \frac{2}{3}}&{}&{}
\end{array} \hfill \\
{s_b}:y = - \frac{2}{3}x + n \hfill \\
{B_1} \in {s_b}:1 = - \frac{2}{3} \cdot \left( { - 1} \right) + n \hfill \\
n = 1 - \frac{2}{3} = \frac{1}{3} \hfill \\
{s_b}:y = - \frac{2}{3}x + \frac{1}{3} \hfill \\
\end{gathered} \]

ђ)

\[\begin{gathered}
\left| {{h_c}} \right| = d\left( {C,c} \right) = \left| {\frac{{A{x_0} + B{y_0} + C}}{{\sqrt {{A^2} + {B^2}} }}} \right| \hfill \\
c:y = \frac{1}{5}x - \frac{7}{5} \hfill \\
5y = x - 7 \hfill \\
x - 5y - 7 = 0 \hfill \\
\left| {{h_c}} \right| = \left| {\frac{{1 \cdot 1 + \left( { - 5} \right) \cdot 4 - 7}}{{\sqrt {1 + {{\left( { - 5} \right)}^2}} }}} \right| = \left| {\frac{{1 - 20 - 7}}{{\sqrt {26} }}} \right| = \frac{{26}}{{\sqrt {26} }} = \sqrt {26} \hfill \\
\end{gathered} \]

е) 

\[\begin{gathered}
\begin{array}{*{20}{c}}
{{s_c}:{s_c} \bot c}& \cap &{{C_1} \in {s_c}} \\
{{k_{{s_b}}} \cdot {k_{_b}} = - 1}&{}&{{C_1}\left( { - \frac{1}{2}, - \frac{3}{2}} \right)} \\
{{k_{{s_b}}} \cdot \frac{3}{2} = - 1}&{}&{} \\
{{k_{{s_b}}} = - \frac{2}{3}}&{}&{}
\end{array} \hfill \\
{s_c}:y = - 5x + n \hfill \\
{C_1} \in {s_c}: - \frac{3}{2} = - 5 \cdot \left( { - \frac{1}{2}} \right) + n \hfill \\
n = - \frac{3}{2} - \frac{5}{2} \hfill \\
{s_c}:y = - 5x - 4 \hfill \\
\left\{ O \right\} = {s_b} \cap {s_c} \hfill \\
{s_b}:y = - \frac{2}{3}x + \frac{1}{3} \hfill \\
\underline {{s_c}:y = - 5x - 4} \hfill \\
- \frac{2}{3}x + \frac{1}{3} = - 5x - 4 \hfill \\
- 2x + 1 = - 15x - 12 \hfill \\
13x = - 13 \hfill \\
x = - 1 \hfill \\
y = - 5 \cdot \left( { - 1} \right) - 4 = 1 \hfill \\
O\left( { - 1,1} \right) \hfill \\
\end{gathered} \]

ф)

\[\begin{gathered}
D\left( {\frac{{ - 3 \cdot 5 + 2}}{{5 + 1}},\frac{{ - 2 \cdot 5 + \left( { - 1} \right) \cdot 1}}{{5 + 1}}} \right) \hfill \\
D\left( { - \frac{{13}}{6}, - \frac{{11}}{6}} \right) \hfill \\
\end{gathered} \]