## Precalculus (6th Edition)

False, the distance is $4\sqrt{2}$.
$d(P, R)=\sqrt{(x_{2}-x_{\mathrm{I}})^{2}+(y_{2}-y_{1})^{2}}$ ----------- $P(0,0), R(4,4),$ $d(P, R)=\sqrt{(4-0)^{2}+(4-0)^{2}}$ $=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}$