The shortest line from the point $(6,0)$ to the given line will be perpendicular to it.  A line perpendicular to $y=2x-2$ will have a slope of $-1/2$.  This will give it a form of $y=-\frac{1}{2}x+b$.  Substituting the point $(6,0)$ that we know must lie on this line, we find: $$0=-\frac{1}{2}\cdot 6 +b$$ $$3=b$$ The equation of the perpendicular line is $y=-\frac{1}{2}x+3$.  Now, we can solve for the point where the two lines intersect: $$-\frac{1}{2}x+3=2x-2$$ $$5=\frac{5}{2}x$$ $$x=2$$ Plugging into either line, we find the point of intersection is $(2,2)$.  The coordinate plane now looks like: [asy]
size(150);
draw((-.5,0)--(7,0));
draw((0,-3)--(0,5));
draw((-.5,-3)--(4,6),linewidth(.7));
draw((6,0)--(0,3),linewidth(.7));
label("$(6,0)$",(6,0),S);
label("$(2,2)$",(2.3,2.1),E);
dot((2,2));
dot((6,0));
[/asy] The distance from the point $(6,0)$ to this point is: $$\sqrt{(6-2)^2+(0-2)^2}=\sqrt{16+4}=\boxed{2\sqrt{5}}$$
Question:
What is the shortest distance from the point $(6, 0)$ to the line $y = 2x-2$? Express your answer in simplest radical form.
Answer:
Let's think step by step.

user
2\sqrt{5}
What is the shortest distance from the point $(6, 0)$ to the line $y = 2x-2$? Express your answer in simplest radical form.
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