Question

Qual a distancia entre os pontos p (3, -3) e Q (-6,2)?

Answer 1

d= √(x2-x1)^2+(y2-y1)^2

d= √(-6-3)+(2+3)^2

d= √(-9)^2+(5)^2

d= √81+25

d= √106

d= 10,30

Answer 2

Boa noite!

P(3, - 3)  Q(- 6, 2)

$dPQ = \sqrt{(xq - xp)^2+(yq - yp)^2} \\ \\ dPQ= \sqrt{(-6 - 3)^2+(2 - (-3))^2} \\ \\ dPQ= \sqrt{(-9)^2 + (5)^2} \\ \\ dPQ= \sqrt{81+25} \\ \\ dPQ = \sqrt{106} \\ \\ Bons\; estudos$