Question

1) Calcule a distância entre os pontos p (2,5) e Q ( -1,1) se a distância (d) de "P" a "Q" é obtida por: d2 pq= ( x2 -p) 2 + (y2 -yp).

Answer 1

$\large\begin{array}{l} \textsf{Considere dois pontos }P(x_{_P},\,y_{_P})\textsf{ e }Q(x_{_Q},\,y_{_Q})\textsf{ de modo que }\\d_{_{PQ}}\textsf{ \'e a dist\^ancia entre eles. Nestas condi\c{c}\~oes vale que}\\\\ d_{_{PQ}}^2=(x_{_P}-x_{_Q})^2+(y_{_P}-y_{_Q})^2\\\\\textsf{(aqui temos o quadrado da dist\^ancia entre }P\textsf{ e }Q\textsf{).}\\\\\\\textsf{Para esta quest\~ao, temos}\\\\ P(2,\,5)\textsf{ e }Q(-1,\,1)\\\\\\\textsf{Portanto,}\\\\ d_{_{PQ}}^2=(2-(-1))^2+(5-1)^2\\\\ d_{_{PQ}}^2=(2+1)^2+(5-1)^2\\\\ d_{_{PQ}}^2=3^2+4^2\\\\ d_{_{PQ}}^2=9+16 \end{array}$

$\large\begin{array}{l} d_{_{PQ}}^2=25\\\\ d_{_{PQ}}=\sqrt{25}\\\\ \boxed{\begin{array}{c}d_{_{PQ}}=5\mathrm{~u.c.} \end{array}} \end{array}$


$\large\begin{array}{l} \textsf{D\'uvidas? Comente.}\\\\\\ \textsf{Bons estudos!} \end{array}$