Answer
The point $Q$ is closer to the point $R$
Work Step by Step
$P(3,1);$ $Q(-1,3);$ $R(-1,-1)$
The distance between two points is given by the formula $d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$
Find the distance between the points $P$ and $R$. For these two points, $x_{1}=3$, $y_{1}=1$, $x_{2}=-1$ and $y_{2}=-1$
Substitute these values into the formula:
$d_{PR}=\sqrt{(-1-3)^{2}+(-1-1)^{2}}=\sqrt{(-4)^{2}+(-2)^{2}}=...$
$...=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}\approx4.4721$
Find the distance between the points $Q$ and $R$. For these two points, $x_{1}=-1$, $y_{1}=3$, $x_{2}=-1$ and $y_{2}=-1$
Substitute these values into the formula:
$d_{QR}=\sqrt{(-1+1)^{2}+(-1-3)^{2}}=\sqrt{0^{2}+(-4)^{2}}=...$
$...=\sqrt{0+16}=\sqrt{16}=4$
Since $d_{QR}\lt d_{PR}$, the point $Q$ is closer to the point $R$