Аналитичка геометрија у равни – решени задаци 6

Текст задатака објашњених у видео лекцији.

Пр.1)   Одредити угао пресека између правих $p: - 3x + 2y - 5 = 0$ и

            $q: - x + 5y + 7 = 0$.

Пр.2)   Одредити угао пресека између правих $p:5x + y - 9 = 0$ и

            $q:x - 5y - 7 = 0$.

Пр.3)   Одредити једначину праве која пролази кроз тачку $A\left( {2, - 4} \right)$

            и папаралелна је са правом $p:2x - 3y + 6 = 0$.

Пр.1)  

\[\begin{gathered}
\begin{array}{*{20}{c}}
{p:{\text{ }} - {\text{ }}3x{\text{ }} + {\text{ }}2y{\text{ }} - {\text{ }}5{\text{ }} = {\text{ }}0}&{}&{q:{\text{ }} - {\text{ }}x{\text{ }} + {\text{ }}5y{\text{ }} + {\text{ }}7{\text{ }} = {\text{ }}0} \\
{p:{\text{ }}y{\text{ = }}\frac{3}{2}x{\text{ }} - {\text{ }}\frac{5}{2}{\text{ }}}&{}&{q:{\text{ }}y{\text{ = }}\frac{1}{5}x{\text{ }} - {\text{ }}\frac{7}{5}} \\
{{k_p} = \frac{3}{2}}&{}&{{k_q} = \frac{1}{5}}
\end{array} \hfill \\
\hfill \\
tg\varphi = \left| {\frac{{{k_2} - {k_1}}}{{1 + {k_1}{k_2}}}} \right| \hfill \\
tg\varphi = \left| {\frac{{\frac{3}{2} - \frac{1}{5}}}{{1 + \frac{1}{5} \cdot \frac{3}{2}}}} \right| = \left| {\frac{{\frac{{15 - 2}}{{10}}}}{{1 + \frac{3}{{10}}}}} \right| = 1 \hfill \\
\varphi = {45^ \circ } \hfill \\
\end{gathered} \]

Пр.2) 

\[\begin{gathered}
\begin{array}{*{20}{c}}
{p:5x{\text{ }} + {\text{ }}y{\text{ }} - {\text{ }}9{\text{ }} = {\text{ }}0}&{}&{q:x{\text{ }} - {\text{ }}5y{\text{ }} - {\text{ }}7{\text{ }} = {\text{ }}0} \\
{p:{\text{ }}y{\text{ = }} - 5x{\text{ }} + 9{\text{ }}}&{}&{q:{\text{ }}y{\text{ = }}\frac{1}{5}x{\text{ }} - {\text{ }}\frac{7}{5}} \\
{{k_p} = - 5}&{}&{{k_q} = \frac{1}{5}}
\end{array} \hfill \\
\hfill \\
{k_p} = - \frac{1}{{{k_q}}} \hfill \\
- 5 = - \frac{1}{{\frac{1}{5}}} = - 5 \hfill \\
\measuredangle \left( {p,q} \right) = {90^ \circ } \hfill \\
\end{gathered} \]

Пр.3) 

\[\begin{gathered}
p:2x{\text{ }} - {\text{ }}3y{\text{ }} + {\text{ }}6{\text{ }} = {\text{ }}0 \hfill \\
p: - 3y = - 2x + 2 \hfill \\
p:{\text{ }}y{\text{ = }}\frac{2}{3}x{\text{ }} + 2{\text{ }} \hfill \\
{k_p} = \frac{2}{3} \hfill \\
q:{\text{ }}y{\text{ = }}{k_q}x{\text{ }} + n \hfill \\
q\parallel p \Leftrightarrow {k_q} = {k_p} \Rightarrow {k_q} = \frac{2}{3} \hfill \\
q:{\text{ }}y{\text{ = }}\frac{2}{3}x{\text{ }} + n \hfill \\
A\left( {2, - 4} \right) \in q \Rightarrow - 4{\text{ = }}\frac{2}{3} \cdot 2{\text{ }} + n \Rightarrow n = - \frac{{16}}{3} \hfill \\
q:{\text{ }}y{\text{ = }}\frac{2}{3}x{\text{ }} - \frac{{16}}{3} \hfill \\
\end{gathered} \]