Explicație pas cu pas:
a) A(4, 6), B(-2,-2), C(12, 0) este isoscel
[tex]AB = \sqrt{ {(4 + 2)}^{2} + {(6 + 2)}^{2} } = \sqrt{36 + 64} = \sqrt{100} = 10 \\ [/tex]
[tex]BC = \sqrt{ {( - 2 - 12)}^{2} + {( - 2 - 0)}^{2} } = \sqrt{196 + 4} = \sqrt{200} = 2 \sqrt{10} \\ [/tex]
[tex]AC = \sqrt{ {(4 - 12)}^{2} + {(6 - 0)}^{2} } = \sqrt{64 + 36} = \sqrt{100} = 10 \\ [/tex]
AB = AC → triunghi isoscel
b) A(1, 8), B(3, 2), C(9, 4) este isoscel
[tex]AB = \sqrt{ {(1 - 3)}^{2} + {(8 - 2)}^{2} } = \sqrt{4 + 36} = \sqrt{40} = 2 \sqrt{10} \\ [/tex]
[tex]BC = \sqrt{ {(3 - 9)}^{2} + {(2 - 4)}^{2} } = \sqrt{36 + 4} = \sqrt{40} = 2 \sqrt{10} \\ [/tex]
[tex]AC = \sqrt{ {(1 - 9)}^{2} + {(8 - 4)}^{2} } = \sqrt{64 + 16} = \sqrt{80} =4 \sqrt{5} \\ [/tex]
AB = BC → triunghi isoscel
c) A (10, 4), B(12, -30), C(0, 0) este dreptunghic
[tex]AB = \sqrt{ {(10 - 12)}^{2} + {(4 + 30)}^{2} } = \sqrt{4 + 1156} \\ = \sqrt{1160} = 2 \sqrt{290} [/tex]
[tex]BC = \sqrt{ {(12 - 0)}^{2} + {( - 30 - 0)}^{2} } = \sqrt{144 + 900} \\ = \sqrt{1044} = 6 \sqrt{29} [/tex]
[tex]AC = \sqrt{ {(10 - 0)}^{2} + {(4 - 0)}^{2} } = \sqrt{100 + 16} \\ = \sqrt{116} = 2 \sqrt{29} [/tex]
AC² + BC² = AB² → triunghi dreptunghic
d) A(0, -2), B(0, –4), C(-√3, -3) este echilateral
[tex]AB = \sqrt{ {(0 - 0)}^{2} + {( - 2 + 4)}^{2} } = \sqrt{4} = 2 \\ [/tex]
[tex]BC = \sqrt{ {(0 + \sqrt{3} )}^{2} + {( - 4 + 3)}^{2} } = \sqrt{3 + 1} = \sqrt{4} = 2 \\ [/tex]
[tex]AC = \sqrt{ {(0 + \sqrt{3} )}^{2} + {( - 2 + 3)}^{2} } = \sqrt{3 + 1} = \sqrt{4} = 2 \\[/tex]
AB = BC = AC → triunghi echilateral
