## College Algebra (10th Edition)

$a. \quad(2.65,1.6)$ $b. \quad \approx 1.285$ units
$a.$ The shortest side is between $(2.6,1.5)$ and $(2.7,1.7)$. Midpoint: $(\displaystyle \frac{x_{1}+x_{2}}{2},\displaystyle \frac{y_{1}+y_{2}}{2})=(\frac{2.6+2.7}{2},\frac{1.5+1.7}{2})=(\frac{5.3}{2},\frac{3.2}{2})$ $=(2.65,1.6)$ $b.$ Distance between $(1.4,1.3)$ and $(2.65,1.6)$: $d=\sqrt{(2.65-1.4)^{2}+(1.6-1.3)^{2}}$ $=\sqrt{(1.25)^{2}+(0.3)^{2}}$ $=\sqrt{1.5625+0.09}$ $=\sqrt{1.6525}$ $\approx 1.285$ units