## Calculus: Early Transcendentals 8th Edition

(a) The midpoint is $(4,9)$ (b) The midpoint is $(\frac{7}{2},-3)$
(a) We can find the midpoint of the line segment joining the two points $(1,3)$ and $(7,15)$: $x = \frac{1+7}{2} = 4$ $y = \frac{3+15}{2} = 9$ The midpoint is $(4,9)$ (b) We can find the midpoint of the line segment joining the two points $(-1,6)$ and $(8,-12)$: $x = \frac{-1+8}{2} = \frac{7}{2}$ $y = \frac{6+(-12)}{2} = -3$ The midpoint is $(\frac{7}{2},-3)$