5 Considere os pontos A(−1, 3), B(2, 5) e C(6, −4), vertices de um triangulo.

Triângulo esboçado no Geogebra

a)

Encontraremos o ponto F e G:

\[\eqalign{ & F = \left( {\dfrac{{6 - \left( { - 1} \right)}}{3},\dfrac{{ - 4 - 3}}{3}} \right) \cr & F = \left( {\dfrac{7}{3},\dfrac{{ - 7}}{3}} \right) \cr & G = \left( {\dfrac{7}{3} + \dfrac{7}{3}, - \dfrac{7}{3} - \dfrac{7}{3}} \right) \cr & G = \left( {\dfrac{{14}}{3},\dfrac{{ - 14}}{3}} \right) }\]

b)

A mediana do vértice A será:

\[\eqalign{ & {M_{BC}} = \left( {\dfrac{{6 - 2}}{2},\dfrac{{ - 4 - 5}}{2}} \right) \cr & {M_{BC}} = \left( {2,\dfrac{{ - 9}}{2}} \right) }\]

Calculando a equação da reta temos:

\[\eqalign{ & m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} \cr & m = \dfrac{{ - 4,5 - 3}}{{2 - \left( { - 1} \right)}} \cr & m = \dfrac{{ - 7,5}}{3} \cr & m = - 2,5 \cr & \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y + 3 = - 2,5\left( {x + 1} \right) \cr & y = - 2,5x - 5x5 }\]

c)

\[\eqalign{ & H = \sqrt {{{\left( {x - {x_0}} \right)}^2} + {{\left( {y - {y_0}} \right)}^2}} \cr & H = \sqrt {{{\left( {2 + 1} \right)}^2} + {{\left( {5 - 3} \right)}^2}} \cr & H = \sqrt {9 + 4} \cr & H = \sqrt {13} }\]

d

\[\eqalign{ & \left| {3BA - 4CB} \right| \cr & \left| {3\left( {2 + 1,5 - 3} \right) - 4\left( {2 - 6,5 + 4} \right)} \right| \cr & \left| {3\left( {3,2} \right) - 4\left( { - 4,9} \right)} \right| \cr & \left| {\left( {9,6} \right) + \left( {16, - 36} \right)} \right| \cr & \left| {\left( {25, - 30} \right)} \right| \cr & \left| {\left( {25, - 30} \right)} \right| = \sqrt {{{25}^2} + {{\left( { - 30} \right)}^2}} \cr & \left| {\left( {25, - 30} \right)} \right| = \sqrt {625 + 900} \cr & \left| {\left( {25, - 30} \right)} \right| = \sqrt {1525} \cr & \left| {\left( {25, - 30} \right)} \right| = 39,05 }\]