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12. Fie A(-2,1), B(2:2) şi C(6;a). Să se afle a dacă [AB]= [BC]. ​

Răspuns :

Hei! :)

  • A(-2, 1)
  • B(2, 2)
  • C(6, a)

AB=BC

[tex]AB=\sqrt{(x_{B}-x_{A})^{2} + (y_{B}-y_{A})^{2} } \\AB=\sqrt{(2+2)^{2} +(2-1)^{2} }\\AB=\sqrt{16+1} =\sqrt{17}[/tex]

[tex]BC=\sqrt{(x_{C}-x_{B} )^{2}+(y_{C}-y_{B} )^{2}} } \\BC=\sqrt{(6-2)^{2}+(a-2)^{2} } \\BC=\sqrt{16+a^{2}-4a+ 4} \\BC=\sqrt{a^{2}-4a+20 }[/tex]

[tex]AB=BC=> \sqrt{a^{2}-4a+20 }=\sqrt{17} /^{2} \\a^{2}-4a+20 =17\\a^{2}-4a+3=0\\delta=16-12=4\\=> a_{1} =\frac{4+2}{2} =3\\=> a_{2}=\frac{4-2}{2} =1[/tex]