Аналитичка геометрија у равни – решени задаци 8

Текст задатака објашњених у видео лекцији.

Пр.6)   Дата су темена троугла $A\left( {0, - 3} \right)$, $B\left( { - 2, - 1} \right)$, $C\left( {2,3} \right)$.

           Одредити: а) једначину странице $c$

                             б) једначину висине ${h_c}$

                             в) дужину тежишне дужи ${t_a}$

                             г) угао $\beta $

                             д) једначину симетрале странице $C$

                             е) координате центра описане кружнице $O$

Пр.6)

\[\begin{gathered}
c = p\left( {AB} \right):y - {y_1} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\left( {x - {x_1}} \right) \hfill \\
y + 3 = \frac{{ - 1 + 3}}{{ - 2 - 0}}\left( {x - 0} \right) \hfill \\
y + 3 = - x \hfill \\
c:y = - x - 3 \hfill \\
{k_c} = - 1 \hfill \\
\hfill \\
{h_c}:y = kx + n \hfill \\
{h_c} \bot c \Leftrightarrow {k_{{h_c}}} = - \frac{1}{{{k_c}}} = 1 \hfill \\
{h_c}:y = x + n \hfill \\
C \in {h_c}:3 = 2 + n \Rightarrow n = 1 \hfill \\
{h_c}:y = x + 1 \hfill \\
\hfill \\
{A_1}\left( {\frac{{{x_B} + {x_C}}}{2},\frac{{{y_B} + {y_C}}}{2}} \right) \hfill \\
{A_1}\left( {\frac{{ - 2 + 2}}{2},\frac{{ - 1 + 3}}{2}} \right) \hfill \\
{A_1}\left( {0,1} \right) \hfill \\
\hfill \\
\left| {{t_a}} \right| = \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( {0 - 0} \right)}^2} + {{\left( { - 3 - 1} \right)}^2}} = \sqrt {0 + 16} = 4 \hfill \\
\hfill \\
a = p\left( {CB} \right):y - {y_1} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\left( {x - {x_1}} \right) \hfill \\
y + 1 = \frac{{3 + 1}}{{2 + 2}}\left( {x - 0} \right) \hfill \\
y + 1 = x + 2 \hfill \\
a:y = x + 1 \hfill \\
{k_a} = 1 \hfill \\
\hfill \\
tg\beta = \left| {\frac{{{k_a} - {k_b}}}{{1 + {k_a} \cdot {k_b}}}} \right| = \left| {\frac{{1 + 1}}{{1 - 1}}} \right| = \left| {\frac{2}{0}} \right| = \infty \Rightarrow \beta = {90^ \circ } \hfill \\
\hfill \\
{C_1}\left( {\frac{{{x_A} + {x_B}}}{2},\frac{{{y_A} + {y_B}}}{2}} \right) \hfill \\
{C_1}\left( {\frac{{0 - 2}}{2},\frac{{ - 3 - 1}}{2}} \right) \hfill \\
{C_1}\left( { - 1, - 2} \right) \hfill \\
\hfill \\
{s_c}:y = kx + n \hfill \\
{s_c} \bot c \Leftrightarrow {k_{{s_c}}} = - \frac{1}{{{k_c}}} = 1 \hfill \\
{s_c}:y = x + n \hfill \\
{C_1} \in {s_c}: - 2 = - 1 + n \Rightarrow n = - 1 \hfill \\
{s_c}:y = x - 1 \hfill \\
\hfill \\
{s_a}:y = kx + n \hfill \\
{s_a} \bot a \Leftrightarrow {k_{{s_a}}} = - \frac{1}{{{k_a}}} = - 1 \hfill \\
{s_a}:y = - x + n \hfill \\
{A_1} \in {s_a}:1 = - 0 + n \Rightarrow n = 1 \hfill \\
{s_a}:y = - x + 1 \hfill \\
\hfill \\
\left\{ O \right\} = {s_a} \cap {s_c} \hfill \\
{s_c}:y = x - 1 \hfill \\
\underline {{s_a}:y = - x + 1} \hfill \\
x - 1 = - x + 1 \hfill \\
2x = 2 \hfill \\
x = 1 \hfill \\
y = - 1 + 1 = 0 \hfill \\
O\left( {1,0} \right) \hfill \\
\hfill \\
D\left( {\frac{{0 \cdot 3 + 2 \cdot 4}}{7},\frac{{ - 3 \cdot 3 + 3 \cdot 4}}{7}} \right) \hfill \\
D\left( {\frac{8}{7},\frac{3}{7}} \right) \hfill \\
\end{gathered} \]