What is the shortest distance from the point $(6, 0)$ to the line $y = 2x-2$? Express your answer in simplest radical form.
To find the shortest distance from a point to a line, I need to draw a perpendicular line segment from the point to the line. The length of this segment will be the distance I'm looking for. The slope of the given line is $2$, so the slope of the perpendicular line must be the negative reciprocal, which is $-\frac{1}{2}$. Using the point-slope form of a line, I can write the equation of the perpendicular line as $y - 0 = -\frac{1}{2}(x - 6)$, or $y = -\frac{1}{2}x + 3$. To find where the perpendicular line intersects the given line, I need to solve the system of equations $y = 2x-2$ and $y = -\frac{1}{2}x + 3$. I can do this by substituting $y$ from one equation into the other and solving for $x$. Substituting $y = -\frac{1}{2}x + 3$ into $y = 2x-2$, I get $-\frac{1}{2}x + 3 = 2x - 2$. Simplifying, I get $-\frac{5}{2}x = -5$, or $x = 2$. To find the corresponding $y$-value, I can plug in $x = 2$ into either equation. I'll use the simpler one, $y = -\frac{1}{2}x + 3$. I get $y = -\frac{1}{2}(2) + 3$, or $y = 2$. So the point of intersection is $(2, 2)$. To find the distance from $(6, 0)$ to $(2, 2)$, I can use the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. Plugging in the coordinates, I get $d = \sqrt{(2 - 6)^2 + (2 - 0)^2}$, or $d = \sqrt{(-4)^2 + 2^2}$, or $d = \sqrt{16 + 4}$, or $d = \sqrt{20}$. I can simplify $\sqrt{20}$ by factoring out the largest perfect square that divides it, which is $4$. I get $d = \sqrt{4 \cdot 5}$, or $d = \sqrt{4} \cdot \sqrt{5}$, or $d = 2\sqrt{5}$. 