Answer:
distance [tex]= \frac{\bf 1}{\sqrt{\bf 5}}[/tex] units
Step-by-step explanation:
distance [tex]= \frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}[/tex]
Given equation is [tex]y=2x-1[/tex]
it can be written as [tex]2x-y-1=0[/tex]
here [tex]a=2, y=-1, c=-1[/tex] and the point [tex](x_0, y_0)[/tex] is [tex](1, 2)[/tex]
distance [tex]= \frac{|2(1)+(-1)(2)-1|}{\sqrt{2^2+1^2}}[/tex]
distance [tex]= \frac{|2-2-1|}{\sqrt{4+1}}[/tex]
distance [tex]= \frac{|-1|}{\sqrt{5}}[/tex]
distance [tex]= \frac{1}{\sqrt{5}}[/tex] units
