College Algebra 7th Edition

$B=(10,13)$.
Let $B=(x,y)$. Since (6,8) is the midpoint of AB, $(6,8)=(\displaystyle \frac{2+x}{2}, \frac{3+y}{2})$ Equating the x-coordinates, $6=\displaystyle \frac{2+x}{2}$ $12=2+x$ $x=10$. Equating the y-coordinates, $8=\displaystyle \frac{3+y}{2}$ $16=3+y$ $y=13$. So, $B=(10,13)$.