Аналитичка геометрија у равни – решени задаци 1

Задаци

Текст задатака објашњених у видео лекцији.

Пр.1)   Одреди дужину тежишних дужи троугла $ABC$ ако је $A\left( { - 8, - 2} \right)$,

            $B\left( {0, - 4} \right)$ и $C\left( {8,6} \right)$.

Пр.2)   Нека су $A\left( { - 4,3} \right)$ и $B\left( {4, - 3} \right)$ два темана једнакостра-

            ничног троугла. Одредити координате трећег темена.

Пр.1) 

$$\begin{aligned}&{A_1}\left( {\frac{{{x_B} - {x_C}}}{2},\frac{{{y_B} - {y_C}}}{2}} \right) \\ &{A_1}\left( {\frac{{0 + 8}}{2},\frac{{ - 4 + 6}}{2}} \right) \\ &{A_1}\left( {4,1} \right) \\ &\left| {{t_a}} \right| = d\left( {A,{A_1}} \right) = \sqrt {{{\left( {{x_A} - {x_{{A_1}}}} \right)}^2} + {{\left( {{y_A} - {y_{{A_1}}}} \right)}^2}} \\ &\left| {{t_a}} \right| = \sqrt {{{\left( { - 8 - 4} \right)}^2} + {{\left( { - 2 - 1} \right)}^2}} = \sqrt {144 + 9} \\ &\left| {{t_a}} \right| = 3\sqrt {17}\end{aligned}$$

$$\begin{aligned}&{B_1}\left( {\frac{{{x_A} - {x_C}}}{2},\frac{{{y_A} - {y_C}}}{2}} \right) \\ &{B_1}\left( {\frac{{ - 8 + 8}}{2},\frac{{ - 2 + 6}}{2}} \right) \\ &{B_1}\left( {0,2} \right) \\ &\left| {{t_b}} \right| = d\left( {B,{B_1}} \right) = \sqrt {{{\left( {{x_B} - {x_{{B_1}}}} \right)}^2} + {{\left( {{y_B} - {y_{{B_1}}}} \right)}^2}} \\ &\left| {{t_b}} \right| = \sqrt {{{\left( {0 - 0} \right)}^2} + {{\left( { - 4 - 2} \right)}^2}} = \sqrt {36} \\ &\left| {{t_b}} \right| = 6\end{aligned}$$

$$\begin{aligned}&{C_1}\left( {\frac{{{x_A} - {x_B}}}{2},\frac{{{y_A} - {y_B}}}{2}} \right) \\ &{C_1}\left( {\frac{{ - 8 + 0}}{2},\frac{{ - 2 - 4}}{2}} \right) \\ &{C_1}\left( { - 4, - 3} \right) \\ &\left| {{t_c}} \right| = d\left( {C,{C_1}} \right) = \sqrt {{{\left( {{x_C} - {x_{{C_1}}}} \right)}^2} + {{\left( {{y_C} - {y_{{C_1}}}} \right)}^2}} \\ &\left| {{t_c}} \right| = \sqrt {{{\left( {8 + 4} \right)}^2} + {{\left( {6 + 3} \right)}^2}} = \sqrt {225} \\ &\left| {{t_c}} \right| = 15\end{aligned}$$

Пр.2)

$$\begin{aligned}&\left| {AB} \right| = \sqrt {{{\left( {{x_A} - {x_B}} \right)}^2} + {{\left( {{y_A} - {y_B}} \right)}^2}} \\ &\left| {AB} \right| = \sqrt {{{\left( { - 4 - 4} \right)}^2} + {{\left( {3 + 3} \right)}^2}} = \sqrt {100} \\ &\left| {AB} \right| = 10, \\ &\begin{array}{cccccccccccccccccccc} {\left| {AC} \right| = 10}&{}&{\left| {CB} \right| = 10} \\ {\sqrt {{{\left( {{x_A} - {x_C}} \right)}^2} + {{\left( {{y_A} - {y_C}} \right)}^2}} = 10}&{}&{\sqrt {{{\left( {{x_C} - {x_B}} \right)}^2} + {{\left( {{y_C} - {y_B}} \right)}^2}} = 10} \\ \begin{aligned}&\sqrt {{{\left( { - 4 - x} \right)}^2} + {{\left( {3 - y} \right)}^2}} = 10 \\ &{\left( { - 4 - x} \right)^2} + {\left( {3 - y} \right)^2} = 100\end{aligned} &{}&\begin{aligned}&\sqrt {{{\left( {4 - x} \right)}^2} + {{\left( { - 3 - y} \right)}^2}} = 10 \\ &{\left( {4 - x} \right)^2} + {\left( { - 3 - y} \right)^2} = 100\end{aligned} \end{array} \\ &{\left( { - 4 - x} \right)^2} + {\left( {3 - y} \right)^2} = 100 \\ &\underline {{{\left( {4 - x} \right)}^2} + {{\left( { - 3 - y} \right)}^2} = 100} \\ &16 + 8x + {x^2} + 9 - 6y + {y^2} = 100 \\ &\underline {16 - 8x + {x^2} + 9 - 6y + {y^2} = 100} \\ &- 16x + 12y = 0 \\ &\underline {16 + 8x + {x^2} + 9 - 6y + {y^2} = 100} \\ & \\ &12y = 16x \Rightarrow y = \frac{{16x}}{{12}} = \frac{{4x}}{3} \\ & \\ &16 + 8x + {x^2} + 9 - 6 \cdot \frac{{4x}}{3} + {\left( {\frac{{4x}}{3}} \right)^2} = 100 \\ &\underline {y = \frac{{4x}}{3}} \\ &225 + 9{x^2} + 16{x^2} = 900 \\ &\underline {y = \frac{{4x}}{3}} \\ &25{x^2} = 675 \Rightarrow {x^2} = 27 \\ &\underline {y = \frac{{4x}}{3}} \\ &\begin{array}{cccccccccccccccccccc} {x = 3\sqrt 3 }&{}&{x = - 3\sqrt 3 } \\ {y = \frac{{4 \cdot 3\sqrt 3 }}{3}}&{}&{y = \frac{{4 \cdot \left( { - 3\sqrt 3 } \right)}}{3}} \\ {y = 4\sqrt 3 }&{}&{y = - 4\sqrt 3 } \end{array} \\ & \\ &{C_1} = \left( {3\sqrt 3 ,4\sqrt 3 } \right);{C_2} = \left( { - 3\sqrt 3 , - 4\sqrt 3 } \right)\end{aligned}$$