Okay, so I need to find the shortest distance from the point (6, 0) to the line y = 2x - 2. Hmm, how do I approach this? I remember there's a formula for the distance from a point to a line, but let me think if I can derive it or recall it properly. 

First, the line is given in slope-intercept form, y = 2x - 2. The general formula for the distance from a point (x0, y0) to the line Ax + By + C = 0 is |Ax0 + By0 + C| / sqrt(A^2 + B^2). Right? So maybe I can rewrite the given line into the standard form and then apply the formula.

Wait, let me check. The standard form is Ax + By + C = 0. The given line is y = 2x - 2. Let me rearrange that. Subtract 2x from both sides: -2x + y + 2 = 0. So A is -2, B is 1, and C is 2. Is that right? Let me verify. Original equation: y = 2x - 2. Subtract 2x: -2x + y = -2. Then add 2 to both sides: -2x + y + 2 = 0. Yes, so A = -2, B = 1, C = 2. 

So plugging into the distance formula: |A*x0 + B*y0 + C| / sqrt(A^2 + B^2). Here, x0 is 6, y0 is 0. So that becomes |-2*6 + 1*0 + 2| / sqrt((-2)^2 + 1^2). Let's compute numerator and denominator step by step.

Numerator: |-2*6 + 0 + 2| = |-12 + 2| = |-10| = 10.

Denominator: sqrt(4 + 1) = sqrt(5).

So the distance is 10 / sqrt(5). But they want it in simplest radical form. To simplify 10/sqrt(5), multiply numerator and denominator by sqrt(5): (10*sqrt(5)) / (sqrt(5)*sqrt(5)) = (10*sqrt(5))/5 = 2*sqrt(5). 

Wait, that seems straightforward. But maybe I should verify using another method to ensure I didn't make a mistake. Alternatively, I can use the formula for the distance from a point to a line by projecting the vector from the point to the line onto the direction perpendicular to the line. Let me try that.

Another way is to find the equation of the line perpendicular to y = 2x - 2 that passes through (6, 0). The slope of the given line is 2, so the slope of the perpendicular line is -1/2. The equation of the perpendicular line is y - 0 = -1/2(x - 6), which simplifies to y = -1/2 x + 3. 

Now, find the intersection point of the two lines: y = 2x - 2 and y = -1/2 x + 3. Set them equal:

2x - 2 = -1/2 x + 3

Multiply both sides by 2 to eliminate fractions:

4x - 4 = -x + 6

Bring variables to one side:

4x + x = 6 + 4

5x = 10

x = 2

Substitute back into one of the equations, say y = 2x - 2:

y = 2*2 - 2 = 4 - 2 = 2

So the intersection point is (2, 2). Now, the distance from (6, 0) to (2, 2) is sqrt[(6 - 2)^2 + (0 - 2)^2] = sqrt[16 + 4] = sqrt[20] = 2*sqrt(5). 

Okay, so both methods give the same answer. That makes me confident that 2*sqrt(5) is correct. But let me just check if I did everything correctly. 

In the first method, converting the line to standard form: y = 2x - 2 becomes -2x + y + 2 = 0. Then plugging into the formula: |-2*6 + 1*0 + 2| = |-12 + 2| = 10, denominator sqrt(4 + 1) = sqrt(5), so 10/sqrt(5) simplifies to 2*sqrt(5). Correct.

In the second method, finding the perpendicular line: slope -1/2, equation y = -1/2 x + 3. Solving for intersection with y = 2x -2. 2x -2 = -1/2 x +3. Multiply by 2: 4x -4 = -x +6. 5x =10, x=2, y=2. Distance from (6,0) to (2,2): sqrt[(6-2)^2 + (0-2)^2] = sqrt(16 +4) = sqrt(20) = 2*sqrt(5). 

Everything checks out. So the answer is 2*sqrt(5). 

Wait, but just to ensure I didn't mix up any signs in the first method. The standard form was -2x + y +2 =0. So A is -2, B is 1, C is 2. The formula is |A x0 + B y0 + C| / sqrt(A^2 + B^2). So plugging in x0=6, y0=0: |-2*6 +1*0 +2| = |-12 +2|=10. Then sqrt(4 +1)=sqrt(5). So 10/sqrt(5) is indeed 2*sqrt(5). Yep. 

Alternatively, if I had written the standard form as 2x - y -2 =0 instead, then A=2, B=-1, C=-2. Then the formula would be |2*6 + (-1)*0 + (-2)| = |12 -2|=10. Denominator sqrt(4 +1)=sqrt(5). Same result. So regardless of how I rearrange the line equation, the distance is the same. 

Therefore, the answer is 2√5. I think that's solid. Both methods confirm it. So I can be confident that's the correct answer.
Therefore, the shortest distance from the point $(6, 0)$ to the line $y = 2x - 2$ is $\boxed{2\sqrt{5}}$.
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