Help me with this math and explaining the question solution and quickly and explain it

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.