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Аналитичка геометрија у равни - решени задаци 10

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.8)   Одредити теме $C$ троугла $ABC$ ако је $A\left( {1, - 1} \right)$ и $H\left( {0,5} \right)$

            где је $H$ ортоцентар троугла.

Пр.9)   Дата су  два суседна темена паралелограма $A\left( {2,5} \right)$ и $B\left( {5,3} \right)$

            и пресечна тачка дијагонала $S\left( { - 2,0} \right)$. Одреди дужину висине

            на страницу $AB$ и површину тог паралелограма.

Пр.8)

417 png

 

\[\begin{gathered}
{h_a} = p\left( {AH} \right):y - {y_1} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\left( {x - {x_1}} \right) \hfill \\
y + 1 = \frac{{5 + 1}}{{0 - 1}}\left( {x - 1} \right) \hfill \\
y + 1 = - 6\left( {x - 1} \right) \hfill \\
y = - 6x + 6 - 1 \hfill \\
\underline {{h_a}:y = - 6x + 5} \hfill \\
\hfill \\
a \bot {h_a} \Leftrightarrow {k_a} \cdot {k_{{h_a}}} = - 1 \Rightarrow {k_a} = \frac{1}{6} \hfill \\
a:y = \frac{1}{6}x + n \hfill \\
B \in a:1 = \frac{1}{6} \cdot 4 + n \hfill \\
n = 1 - \frac{4}{6} = \frac{1}{3} \hfill \\
\underline {a:y = \frac{1}{6}x + \frac{1}{3}} \hfill \\
\hfill \\
{h_b} = p\left( {BH} \right):y - {y_1} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\left( {x - {x_1}} \right) \hfill \\
y - 1 = \frac{{5 - 1}}{{0 - 4}}\left( {x - 4} \right) \hfill \\
y - 1 = - x + 4 \hfill \\
\underline {{h_b}:y = - x + 5} \hfill \\
\hfill \\
b \bot {h_b} \Leftrightarrow {k_b} \cdot {k_{{h_b}}} = - 1 \Rightarrow {k_b} = 1 \hfill \\
b:y = x + n \hfill \\
A \in b: - 1 = 1 + n \hfill \\
n = - 2 \hfill \\
\underline {b:y = x - 2} \hfill \\
\hfill \\
\left\{ C \right\} = a \cap b \hfill \\
y = \frac{1}{6}x + \frac{1}{3} \hfill \\
\underline {y = x - 2} \hfill \\
\frac{1}{6}x + \frac{1}{3}\underline {y = x - 2} \hfill \\
\underline {y = x - 2} \hfill \\
x + 2 = 6x - 12 \hfill \\
\underline {y = x - 2} \hfill \\
5x = 14;x = \frac{{14}}{5} \hfill \\
y = \frac{{14}}{5} - 2 = \frac{4}{5} \hfill \\
C\left( {\frac{{14}}{5},\frac{4}{5}} \right) \hfill \\
\end{gathered} \]

Пр.9)

418 png

\[\begin{gathered}
p\left( {AB} \right):y - {y_1} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\left( {x - {x_1}} \right) \hfill \\
y - 5 = \frac{{3 - 5}}{{5 - 2}}\left( {x - 2} \right) \hfill \\
y - 5 = - \frac{2}{3}x + \frac{4}{3} \hfill \\
y = - \frac{2}{3}x + \frac{{19}}{3} \hfill \\
\underline \begin{gathered}
3y = - 2x + 19 \hfill \\
p\left( {AB} \right):2x + 3y - 19 = 0 \hfill \\
\end{gathered} \hfill \\
\hfill \\
S\left( {\frac{{{x_B} + {x_D}}}{2},\frac{{{y_B} + {y_D}}}{2}} \right) \hfill \\
\left( { - 2,0} \right) = \left( {\frac{{5 + {x_0}}}{2},\frac{{3 + {y_0}}}{2}} \right) \hfill \\
\frac{{5 + {x_0}}}{2} = - 2,\frac{{3 + {y_0}}}{2} = 0 \hfill \\
{x_0} = - 9,{y_0} = - 3 \hfill \\
D\left( { - 9, - 3} \right) \hfill \\
\hfill \\
\left| {{h_a}} \right| = d\left( {D,p\left( {AB} \right)} \right) = \left| {\frac{{2 \cdot \left( { - 3} \right) + 3 \cdot \left( { - 3} \right) \cdot 19}}{{\sqrt {{2^2} + {3^2}} }}} \right| = \hfill \\
= \left| {\frac{{ - 46}}{{\sqrt {13} }}} \right| = \frac{{46\sqrt {13} }}{{13}} \hfill \\
\hfill \\
P = \left| a \right| \cdot \left| {{h_a}} \right| \hfill \\
\left| a \right| = \left| {AB} \right| = \sqrt {{{\left( {5 - 2} \right)}^2} + {{\left( {3 - 5} \right)}^2}} = \sqrt {13} \hfill \\
P = \sqrt {13} \cdot \frac{{46\sqrt {13} }}{{13}} = 46 \hfill \\
\end{gathered} \]