Answer
The point $C$ is closer to the point $E$
Work Step by Step
$C(-6,3);$ $D(3,0);$ $E(-2,1)$
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 points $C$ and $E$. For these two points, $x_{1}=-6$, $y_{1}=3$, $x_{2}=-2$ and $y_{2}=1$
Substitute these values into the formula:
$d_{CE}=\sqrt{(-2+6)^{2}+(1-3)^{2}}=\sqrt{4^{2}+(-2)^{2}}=...$
$...=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}\approx4.4721$
Find the distance between the points $D$ and $E$. For these two points, $x_{1}=3$, $y_{1}=0$, $x_{2}=-2$ and $y_{2}=1$
Substitute these values into the formula:
$d_{DE}=\sqrt{(-2-3)^{2}+(1-0)^{2}}=\sqrt{(-5)^{2}+1^{2}}=...$
$...=\sqrt{25+1}=\sqrt{26}\approx5.099$
Since $d_{CE}\lt d_{DE}$, the point $C$ is closer to the point $E$