Question

in reperul cartezian xOy se considera punctele A(1,5), B(1,1) si C(5,5). Aratati ca triunghiul ABC este isoscel​

Answer 1

$AB=\sqrt{{(x_{B}-x_{A})}^{2}+{(y_{B}-y_{A})}^{2}} </p><p>$

$AB = \sqrt{ {(1 - 1)}^{2} + {(1 - 5)}^{2} }$

$AB = \sqrt{ {0}^{2} + {( - 4)}^{2} }$

$AB = \sqrt{0 + 16}$

$AB = \sqrt{16}$

$AB = 4$

$AC=\sqrt{{(x_{C}-x_{A})}^{2}+{(y_{C}-y_{A})}^{2}}</p><p>$

$AC= \sqrt{ {(5 - 1)}^{2} + {(5 - 5)}^{2} } </p><p>$

$AC = \sqrt{ {4}^{2} + {0}^{2} }$

$AC = \sqrt{16 + 0}$

$AC = \sqrt{16}$

$AC = 4$

$BC=\sqrt{{(x_{C}-x_{B})}^{2}+{(y_{C}-y_{B})}^{2}}</p><p>$

$BC = \sqrt{ {(5 - 1)}^{2} + {(5 - 1)}^{2} }$

$BC = \sqrt{ {4}^{2} + {4}^{2} }$

$BC = \sqrt{16 + 16}$

$BC = \sqrt{32}$

$BC = 4 \sqrt{2}$

$AB=AC=4=>\Delta ABC \: isoscel$