well, let's notice that the x-coordinate is the same for both points, the same -3, which means is a vertical line, running from y = 5 to y = -5/2.
so we can just find their distance from 0.
y=-5/2 is 5/2 units down below from 0.
y=5 is 5 units up above 0.
[tex]\bf \cfrac{5}{2}+5\implies \cfrac{5}{2}+\cfrac{5}{1}\implies \stackrel{\textit{using the LCD of 2}}{\cfrac{5+10}{2}}\implies \stackrel{improper}{\cfrac{15}{2}}\implies \stackrel{mixed}{7\frac{1}{2}}[/tex]
