<Question>
What is the shortest distance from the point $(6, 0)$ to the line $y = 2x-2$? Express your answer in simplest radical form.
</Question>

<think>
<Plan_1>
Okay, so I need to find the shortest distance from the point (6, 0) to the line y = 2x - 2. Hmm, let me think. I remember that the shortest distance from a point to a line is the perpendicular distance. Right? So, if I can find the equation of the line that's perpendicular to y = 2x - 2 and passes through (6, 0), then the point where these two lines intersect would be the closest point on the original line to (6, 0). Then, I can calculate the distance between (6, 0) and that intersection point. That should give me the shortest distance.

<prompt_1.1> Plan how to find the shortest distance from the point (6, 0) to the line y = 2x - 2 using the perpendicular distance method. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
First, let me confirm the details. The given line is y = 2x - 2. The point is (6, 0). The slope of the given line is 2, so the slope of the perpendicular line should be the negative reciprocal, which is -1/2. Yeah, because perpendicular lines have slopes that multiply to -1. So 2 * (-1/2) = -1. That checks out.

So, the equation of the perpendicular line passing through (6, 0) would be y - 0 = -1/2(x - 6). Simplifying that, it's y = (-1/2)x + 3.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Now, I need to find the intersection point of y = 2x - 2 and y = (-1/2)x + 3. Let me set them equal to each other:

<prompt_2.1> Find the intersection point of the lines y = 2x - 2 and y = (-1/2)x + 3. </prompt_2.1>

Wait, let me verify that. Let me check if the slopes and intercepts make sense. The original line has a slope of 2, and the perpendicular line has -1/2, which is correct.

Alternatively, maybe I can use the distance formula. The formula for the distance from a point (x0, y0) to the line ax + by + c = 0 is |ax0 + by0 + c| / sqrt(a² + b²). Let me recall that.

<prompt_2.2> Use the distance formula to verify the shortest distance from the point (6, 0) to the line y = 2x - 2. </prompt_2.2>

Wait, but just to make sure I didn't make any mistakes in my first method. Let me recheck that.

<prompt_2.3> Recheck the calculations for finding the intersection point and distance in the first method. </prompt_2.3>
</Plan_2>

<Execution_2>
<execution_2.1>
2x - 2 = (-1/2)x + 3

Let me solve for x. Multiply both sides by 2 to eliminate the fraction:

4x - 4 = -x + 6

Now, add x to both sides:

5x - 4 = 6

Add 4 to both sides:

5x = 10

Divide by 5:

x = 2
</execution_2.1>

<execution_2.2>
So first, I need to write the line y = 2x - 2 in the standard form ax + by + c = 0. Let's rearrange that:

2x - y - 2 = 0. So, a = 2, b = -1, c = -2.

Then, plugging into the distance formula: |2*6 + (-1)*0 -2| / sqrt(2² + (-1)²) = |12 - 0 -2| / sqrt(4 + 1) = |10| / sqrt(5) = 10 / sqrt(5) = 2*sqrt(5). Same result. So that's reassuring.

But the problem says to express in simplest radical form, which sqrt(5) is already simplified, so 2√5.
</execution_2.2>

<execution_2.3>
So solving 2x - 2 = (-1/2)x + 3. Adding (1/2)x to both sides gives (2 + 1/2)x - 2 = 3. 2 + 1/2 is 2.5 or 5/2. So (5/2)x = 5. Then x = 5 * (2/5) = 2. Then y = 2*2 - 2 = 2. So intersection at (2, 2).
</execution_2.3>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Then plug back into one of the equations to find y. Let's use the original line y = 2x - 2:
y = 2(2) - 2 = 4 - 2 = 2. So the intersection point is (2, 2).

Now, the distance between (6, 0) and (2, 2) is sqrt[(6 - 2)^2 + (0 - 2)^2] = sqrt[16 + 4] = sqrt[20] = 2*sqrt(5).
So that's the distance. Let me check if that's correct. Alternatively, maybe I can use another method to cross-verify. For example, parametric equations.

<prompt_3.1> Cross-verify the intersection point and distance using parametric equations. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
Let me see. The line y = 2x - 2 can be parametrized. Let x = t, then y = 2t - 2. So any point on the line is (t, 2t - 2). The distance from (6, 0) to this point is sqrt[(t - 6)^2 + (2t - 2 - 0)^2]. To minimize this distance, we can minimize the square of the distance, which is (t - 6)^2 + (2t - 2)^2. Let's compute that:

(t - 6)^2 + (2t - 2)^2 = (t² -12t + 36) + (4t² -8t +4) = 5t² -20t +40. To find the minimum, take derivative with respect to t and set to zero. The derivative is 10t -20. Setting to zero: 10t -20 =0 => t=2. So t=2. Then the point is (2, 2), same as before. So distance squared is 5*(2)^2 -20*(2) +40= 20 -40 +40=20. So sqrt(20)=2*sqrt(5). Same result.
</execution_3.1>
</Execution_3>

<Final_answer>
Therefore, the shortest distance from the point $(6, 0)$ to the line $y = 2x-2$ is $\boxed{2\sqrt{5}}$.
</Final_answer>
</think>