Calcule a distância entre os pontos A(26,36) e B(10,28)
Soluções para a tarefa
Resposta:
\begin{gathered}d_{AB}=\sqrt{(x_{A}-x_{B})^{2}+(y_{A}-y_{B})^{2}}\\ \\ \text{substituimos na f\'{ormula}}:\\ \\ \text{*A(1, 9) e B(2, 8)}\\ \\ d_{AB}=\sqrt{(1-2)^{2}+(9-8)^{2}}\\ \\ d_{AB}=\sqrt{1+1}\\ \\ \boxed{d_{AB}=\sqrt{2}}\\ \\ \text{*C(-3, 5) e D(-3, 12)}\\ \\ d_{CD}=\sqrt{(-3-(-3))^{2}+(5-12)^{2}}\\ \\ d_{CD}=\sqrt{0+49}\\ \\ \boxed{d_{CD}=7}\\ \\ \text{*P(-5, 4) e Q(-2, 7)}\\ \\ d_{PQ}=\sqrt{(-5-(-2))^{2}+(4-7)^{2}}\\ \\ d_{PQ}=\sqrt{9+9}\\ \\ d_{PQ}=\sqrt{18}\\ \\ \boxed{d_{PQ}=3\sqrt{2}}\end{gathered}
d
AB
=
(x
A
−x
B
)
2
+(y
A
−y
B
)
2
substituimos na f
ormula
ˊ
:
*A(1, 9) e B(2, 8)
d
AB
=
(1−2)
2
+(9−8)
2
d
AB
=
1+1
d
AB
=
2
*C(-3, 5) e D(-3, 12)
d
CD
=
(−3−(−3))
2
+(5−12)
2
d
CD
=
0+49
d
CD
=7
*P(-5, 4) e Q(-2, 7)
d
PQ
=
(−5−(−2))
2
+(4−7)
2
d
PQ
=
9+9
d
PQ
=
18
d
PQ
=3
2
\begin{gathered}\\ \text{*M(0, 12) e N(9, 0)}\\ \\ d_{MN}=\sqrt{(0-9)^{2}+(12-0)^{2}}\\ \\ d_{MN}=\sqrt{81+144}\\ \\ d_{MN}=\sqrt{225}\\ \\ \boxed{d_{MN}=15}\end{gathered}
*M(0, 12) e N(9, 0)
d
MN
=
(0−9)
2
+(12−0)
2
d
MN
=
81+144
d
MN
=
225
d
MN
=15