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Аналитичка геометрија у равни - решени задаци 7

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.4)   Наћи једначину праве која пролази кроз пресечну тачку правих

           $l:2x + 3y - 7 = 0$ и $m:x + 2y - 5 = 0$ и нормална је на праву

           $q:x - y + 7 = 0$.

Пр.5)   Одредити једначине висина и координате ортоцентра троугла

            ако су дате једначине страница: 

            $a:3x - y - 18 = 0$     $b:x - y - 2 = 0$    $c:x + 2y + 1 = 0$.

Пр.4)

412 png

\[\begin{gathered}
\left\{ {N\left( {x,y} \right)} \right\} = l \cap m \hfill \\
\hfill \\
l:2x + 3y = 7 \hfill \\
\underline {m:x + 2y = 5} \hfill \\
- y = - 3 \hfill \\
\underline {x + 2y = 5} \hfill \\
y = 3 \hfill \\
\underline {x + 6 = 5} \hfill \\
y = 3 \hfill \\
x = - 1 \hfill \\
N\left( { - 1,3} \right) \hfill \\
\hfill \\
n:y = {k_n}x + n \hfill \\
n \bot q \Leftrightarrow {k_n} = - \frac{1}{{{k_q}}} \hfill \\
\hfill \\
q: - y = - x - 7 \hfill \\
q:y = x + 7 \Rightarrow {k_q} = 1 \hfill \\
\hfill \\
{k_n} = - \frac{1}{{{k_q}}} = - \frac{1}{1} = - 1 \hfill \\
n:y = - x + n \hfill \\
N \in n:3 = - \left( { - 1} \right) + n \hfill \\
n = 2 \hfill \\
n:y = - x + 2 \hfill \\
\end{gathered} \]

Пр.5)

414 png

\[\begin{gathered}
\left\{ C \right\} = a \cap b \hfill \\
a:3x{\text{ }} - {\text{ }}y{\text{ }} - {\text{ }}18{\text{ }} = {\text{ }}0 \hfill \\
\underline {b:x{\text{ }} - {\text{ }}y{\text{ }} - {\text{ }}2{\text{ }} = {\text{ }}0} \hfill \\
2x = 16x - 8 \hfill \\
\underline {x - y = 2} \hfill \\
x = 8 \hfill \\
\underline {8 - y = 2} \hfill \\
x = 8 \hfill \\
y = 6 \hfill \\
C\left( {8,6} \right) \hfill \\
\hfill \\
{h_c}:y = kx + n \hfill \\
{h_c} \bot c \Leftrightarrow {k_{{h_c}}} = - \frac{1}{{{k_c}}} \hfill \\
c:2y = - x - 1 \hfill \\
c:y = - \frac{1}{2}x - \frac{1}{2};{k_c} = - \frac{1}{2} \hfill \\
{k_{{h_c}}} = - \frac{1}{{ - \frac{1}{2}}} = 2 \hfill \\
{h_c}:y = 2x + n \hfill \\
C \in {h_c} \Rightarrow 6 = 2 \cdot 8 + n \Rightarrow n = - 10 \hfill \\
{h_c}:y = 2x - 10 \hfill \\
\hfill \\
\left\{ A \right\} = b \cap c \hfill \\
b:x{\text{ }} - {\text{ }}y{\text{ }} - {\text{ }}2{\text{ }} = {\text{ }}0 \hfill \\
\underline {c:x{\text{ }} + {\text{ }}2y{\text{ }} + {\text{ }}1{\text{ }} = {\text{ }}0} \hfill \\
3y = - 3 \hfill \\
\underline {x - y = 2} \hfill \\
y = - 1 \hfill \\
\underline {x + 1 = 2} \hfill \\
x = 1 \hfill \\
y = - 1 \hfill \\
A\left( {1, - 1} \right) \hfill \\
{h_a}:y = kx + n \hfill \\
{h_a} \bot a \Leftrightarrow {k_{{h_a}}} = - \frac{1}{{{k_a}}} \hfill \\
a: - y = - 3x + 18 \hfill \\
a:y = 3x - 18;{k_a} = 3 \hfill \\
{k_{{h_a}}} = - \frac{1}{3} \hfill \\
{h_a}:y = - \frac{1}{3}x + n \hfill \\
A \in {h_a} \Rightarrow - 1 = - \frac{1}{3}x + n \Rightarrow n = - \frac{2}{3} \hfill \\
{h_a}:y = - \frac{1}{3}x - \frac{2}{3} \hfill \\
\hfill \\
\left\{ H \right\} = {h_a} \cap {h_c} \hfill \\
{h_a}:y = - \frac{1}{3}x - \frac{2}{3} \hfill \\
\underline {{h_c}:y = 2x - 10} \hfill \\
- \frac{1}{3}x - \frac{2}{3}y = 2x - 10 \hfill \\
- x - 2 = 6x - 30 \hfill \\
- 7x = - 28 \hfill \\
x = 4 \hfill \\
y = 2 \cdot 4 - 10 = - 2 \hfill \\
H\left( {4, - 2} \right) \hfill \\
\end{gathered} \]