Question

O triângulo POR, no plano cartesiano, de vértices P(0,0), O(6,0) e R(3,5), é:

a) Isósceles, mas não equilátero.
b) Equilátero.
c) Escaleno
d) Retângulo.
e) Obtusângulo

Answer 1

Resposta:

$\textsf{letra A}$

Explicação passo a passo:

$\mathsf{d_{PO} = \sqrt{(x_P - x_O)^2 + (y_P - y_O)^2}}$

$\mathsf{d_{PO} = \sqrt{(0 - 6)^2 + (0 - 0)^2}}$

$\mathsf{d_{PO} = \sqrt{(-6)^2 + (0)^2}}$

$\mathsf{d_{PO} = \sqrt{36 + 0}}$

$\mathsf{d_{PO} = \sqrt{36}}$

$\boxed{\boxed{\mathsf{d_{PO} = 6}}}$

$\mathsf{d_{PR} = \sqrt{(x_P - x_R)^2 + (y_P - y_R)^2}}$

$\mathsf{d_{PR} = \sqrt{(0 - 3)^2 + (0 - 5)^2}}$

$\mathsf{d_{PR} = \sqrt{(-3)^2 + (- 5)^2}}$

$\mathsf{d_{PR} = \sqrt{9 +25}}$

$\boxed{\boxed{\mathsf{d_{PR} = \sqrt{34}}}}$

$\mathsf{d_{RO} = \sqrt{(x_O - x_R)^2 + (y_O - y_R)^2}}$

$\mathsf{d_{RO} = \sqrt{(6 - 3)^2 + (0 - 5)^2}}$

$\mathsf{d_{RO} = \sqrt{(3)^2 + (-5)^2}}$

$\mathsf{d_{RO} = \sqrt{9 + 25}}$

$\boxed{\boxed{\mathsf{d_{RO} = \sqrt{34}}}}$

Answer 2

Resposta:

d²=(x1-x2)²+(y1-y2)²

P(0,0), O(6,0) e R(3,5)

dPO²=(0-6)²+(0-0)²=36  ==>dPO=6

dPR²=(0-3)²+(0-5)²=9+25 ==>dPR=√34

dOR²=(6-3)²+(0-5)²=9+25 ==>dPR=√34

doR=dPR≠dPO

a) Isósceles, mas não equilátero.