Answer:
Step-by-step explanation:
[tex]\text{Let}\\\\Ax+By=C-\text{a line}\\\\(x_0,\ y_0)-\text{a point}\\\\\text{The formula of a distance between a point and a line}:\\\\d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}[/tex]
[tex]\text{We have}\\\\y=\dfrac{1}{2}x+1,\ A(-2,\ 3)\\\\\text{Convert the equation of a line to the standard form}:\\\\y=\dfrac{1}{2}x+1\qquad\text{multiply both sides by 2}\\\\2y=x+2\qquad\text{subtract}\ x\ \text{from both sides}\\\\-x+2y=2\qquad\text{change the signs}\\\\x-2y=-2\\\\\text{Substitute}\ A=1,\ B=-2,\ C=-2,\ x_0=-2,\ y_0=3,\ \text{to the formula:}\\\\d=\dfrac{|(1)(-2)+(-2)(3)+(-2)|}{\sqrt{1^2+(-2)^2}}=\dfrac{|-2-6-2|}{\sqrt{1+4}}=\dfrac{|-10|}{\sqrt5}[/tex]
[tex]=\dfrac{10}{\sqrt5}\cdot\dfrac{\sqrt5}{\sqrt5}=\dfrac{10\sqrt5}{5}=2\sqrt5[/tex]
