## College Algebra 7th Edition

Point $Q(-1,3)$ is closer to point $R$ .
$d(P, R)=\sqrt{(-1-3)^{2}+(-1-1)^{2}}=\sqrt{(-4)^{2}+(-2)^{2}}$ $=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}$. $d(Q, R)=\sqrt{(-1-(-1))^{2}+(-1-3)^{2}}$ $=\sqrt{0+(-4)^{2}}=\sqrt{16}=4$. $4<2\sqrt{5}$, so point $Q(-1,3)$ is closer to point $R$ than point P.