In reperul cartezian xOy se considera punctele A(2,0) B(5,4) C(-2,3).
Aratati ca triunghiul ABC este drepunghic
Va rog sa ma ajutati

[tex]A(2,0)[/tex]

[tex]B(5,4)[/tex]

[tex]C(-2,3)[/tex]

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

[tex]AB=\sqrt{{(5-2)}^{2}+{(4-0)}^{2}}[/tex]

[tex]AB=\sqrt{{3}^{2}+{4}^{2}}[/tex]

[tex]AB=\sqrt{9+16}[/tex]

[tex]AB=\sqrt{25}[/tex]

[tex]AB=5[/tex]

[tex]AC=\sqrt{{(x_{C}-x_{A})}^{2}+{(y_{C}-y_{A})}^{2}}[/tex]

[tex]AC=\sqrt{{( - 2-2)}^{2}+{(3-0)}^{2}}[/tex]

[tex]AC=\sqrt{{( - 4)}^{2}+{3}^{2}}[/tex]

[tex]AC=\sqrt{16+9}[/tex]

[tex]AC= \sqrt{25} [/tex]

[tex]AC=5[/tex]

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

[tex]BC=\sqrt{{( - 2-5)}^{2}+{(3-4)}^{2}}[/tex]

[tex]BC=\sqrt{{( - 7)}^{2}+{( - 1)}^{2}}[/tex]

[tex]BC=\sqrt{49+1}[/tex]

[tex]BC = \sqrt{50} [/tex]

[tex]BC = 5 \sqrt{2} [/tex]

[tex] {AB}^{2}+{AC}^{2}={BC}^{2}[/tex]

[tex] {5}^{2} + {5}^{2} = {(5 \sqrt{2}) }^{2} [/tex]

[tex]25 + 25 = 50[/tex]

[tex]50 = 50 \: (A)[/tex]

[tex] = > \Delta \: ABC \: dreptunghic[/tex]

[tex] AB = AC = 5 = > \Delta \: ABC \: dreptunghic \: isoscel[/tex]