To find the distance between V1 and the aquarium we can use the formula of the distance between two points in the plane, that is,
[tex]\begin{gathered} d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ \text{ Where }(x_1,y_1)\text{ and }(x_2,y_2)\text{ are the coordinates of the points} \end{gathered}[/tex]
So, in this case, we have
[tex]\begin{gathered} V1(-6,5) \\ AQ(5,5) \\ d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ d=\sqrt[]{(5-(-6))^2+(5-5)^2} \\ d=\sqrt[]{(5+6)^2+(5-5)^2} \\ d=\sqrt[]{(11)^2+(0)^2} \\ d=\sqrt[]{(11)^2} \\ d=11 \end{gathered}[/tex]
Therefore, the distance between v1 and the aquarium is 11 units.
