Question

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

Answer 1

AB²=(2-5)²+(0-4)²=(-3)²+(-4)²=9+16=25

AC²=(2-(-2))²+(0-3)²=(2+2)²+9==16+9=25

BC²=(5-(-2))²+(4-3)²=(5+2)²+1=49+1=50

25+25=50

BC²=AB²+BC²

Triunghiul    este     dreptunghic

Answer 2

$A(2,0)$

$B(5,4)$

$C(-2,3)$

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

$AB=\sqrt{{(5-2)}^{2}+{(4-0)}^{2}}$

$AB=\sqrt{{3}^{2}+{4}^{2}}$

$AB=\sqrt{9+16}$

$AB=\sqrt{25}$

$AB=5$

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

$AC=\sqrt{{( - 2-2)}^{2}+{(3-0)}^{2}}$

$AC=\sqrt{{( - 4)}^{2}+{3}^{2}}$

$AC=\sqrt{16+9}$

$AC= \sqrt{25}$

$AC=5$

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

$BC=\sqrt{{( - 2-5)}^{2}+{(3-4)}^{2}}$

$BC=\sqrt{{( - 7)}^{2}+{( - 1)}^{2}}$

$BC=\sqrt{49+1}$

$BC = \sqrt{50}$

$BC = 5 \sqrt{2}$

${AB}^{2}+{AC}^{2}={BC}^{2}$

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

$25 + 25 = 50$

$50 = 50 \: (A)$

$= > \Delta \: ABC \: dreptunghic$

$AB = AC = 5 = > \Delta \: ABC \: dreptunghic \: isoscel$