Аналитичка геометрија у равни – решени задаци 9

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)  

a) 

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

б)

$$\begin{aligned}&\begin{array}{cccccccccccccccccccc} {{h_b}:{h_b} \bot b}& \cap &{B \in {h_b}} \\ {{k_{{h_b}}} \cdot {k_b} = - 1}&{}&{y = - \frac{2}{3}x + n} \\ {{k_{{h_b}}} \cdot \frac{3}{2} = - 1}&{}&{ - 1 = - \frac{2}{3} \cdot 2 + n} \\ {{k_{{h_b}}} = - \frac{2}{3}}&{}&{n = - 1 + \frac{4}{3}} \\ {}&{}&{n = \frac{1}{3}} \end{array} \\ &{h_b}:y = - \frac{2}{3}x + \frac{1}{3}\end{aligned}$$

в) 

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

г)

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

д)

$$\begin{aligned}&\begin{array}{cccccccccccccccccccc} {{s_b}:{s_b} \bot b}& \cap &{{B_1} \in {s_b}} \\ {{k_{{s_b}}} \cdot {k_{_b}} = - 1}&{}&{{B_1}\left( {\frac{{ - 3 + 1}}{2},\frac{{ - 2 + 4}}{2}} \right)} \\ {{k_{{s_b}}} \cdot \frac{3}{2} = - 1}&{}&{{B_1}\left( { - 1,1} \right)} \\ {{k_{{s_b}}} = - \frac{2}{3}}&{}&{} \end{array} \\ &{s_b}:y = - \frac{2}{3}x + n \\ &{B_1} \in {s_b}:1 = - \frac{2}{3} \cdot \left( { - 1} \right) + n \\ &n = 1 - \frac{2}{3} = \frac{1}{3} \\ &{s_b}:y = - \frac{2}{3}x + \frac{1}{3}\end{aligned}$$

ђ)

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

е) 

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

ф)

$$\begin{aligned}&D\left( {\frac{{ - 3 \cdot 5 + 2}}{{5 + 1}},\frac{{ - 2 \cdot 5 + \left( { - 1} \right) \cdot 1}}{{5 + 1}}} \right) \\ &D\left( { - \frac{{13}}{6}, - \frac{{11}}{6}} \right)\end{aligned}$$