Răspuns:
Explicație pas cu pas:
3 puncte sunt coliniare daca suma a doua segmente este egala cu cea de-a treia
a. [tex]d(A,B)=\sqrt{(2-\frac{1}{3})^2+(4+1)^2 } =\sqrt{\frac{25}{9}+25 } =\frac{5\sqrt{10} }{3}[/tex]
[tex]d(C,A)=\sqrt{(\frac{1}{3} +\frac{2}{3})^2+(4-1)^2 } =\sqrt{1+9}=\sqrt{10}[/tex]
[tex]d(B,C)=\sqrt{(-\frac{2}{3}-2)^2+(-4-4)^2 } =\sqrt{\frac{64}{9} +64} =\frac{8\sqrt{10} }{3}[/tex]
Observam ca AB+AC=BC⇒A, B, C sunt coliniare
b. [tex]d(D,E)=\sqrt{(-3-6)^2+(-3-0)^2} =\sqrt{90} =3\sqrt{10} \\\\d(D,F)=\sqrt{(2-6)^2+(\frac{5}{3} -0)^2} =\sqrt{16+\frac{25}{9} } =\frac{13}{3} \\\\d(E,F)=\sqrt{(2+3)^2+(\frac{5}{3}+3)^2) } =\sqrt{25+\frac{196}{9} }=\frac{\sqrt{421} }{3}[/tex]
Nu sunt coliniare
c. [tex]d(M,N)=\sqrt{(-\frac{2}{3} +5)^2+(-3-16)^2} =\sqrt{\frac{169}{9}+361 } =\frac{39}{3} \\\\d(M,P)=\sqrt{(0+5)^2+(1-16)^2}=\sqrt{25+225} =5\sqrt{10} \\\\d(N,P)=\sqrt{(0+\frac{2}{3})^2+(1+3)^2 }=\sqrt{\frac{4}{9}+16 } =\frac{2\sqrt{37} }{3}[/tex]
Nu sunt coliniare
Mai verifica si tu calculele odata.
