#### Answer

Distance $=\dfrac{5\sqrt{2}}{2}$

#### Work Step by Step

Point: $(-2,1)$ $;$ Line: $x-y-2=0$
The distance between the point $(x_{1},y_{1})$ and the line $Ax+By+C=0$ is given by $\dfrac{|Ax_{1}+By_{1}+C|}{\sqrt{A^{2}+B^{2}}}$.
In this case, $(x_{1},y_{1})=(-2,1)$ and $A=1$, $B=-1$ and $C=-2$
Substitute the known values into the formula and evaluate:
Distance $=\dfrac{|(1)(-2)+(-1)(1)-2|}{\sqrt{1^{2}+(-1)^{2}}}=...$
$...=\dfrac{|-2-1-2|}{\sqrt{1+1}}=\dfrac{|-5|}{\sqrt{2}}=\dfrac{5}{\sqrt{2}}=...$
Rationalize the denominator:
$...=\dfrac{5}{\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}=\dfrac{5\sqrt{2}}{(\sqrt{2})^{2}}=\dfrac{5\sqrt{2}}{2}$