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Calculus/finding the equation of the tangent line

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Question
My problem is:
The point P(5,2) lies on the curve y=sqrt(x-1).  Find an equation of the tangent line to the curve at P(5,2).  

I've been taught to solve this by finding the slope of the secant line, however, I'm really unsure of how to go about doing this.  

So far, I've graphed the equation and found a second point Q(10,3) and the slope of the line between those to points.  So, mPQ = 5.  I'm really not sure where to go from here though or even if I'm going in the right direction.  

Thanks for your help!  

Answer
I'm not sure if that's the right direction or not.  Here is what I do know.

To find a tangent line, we have a point and we know that it is parallel to the curve at (5,2).  Note that x^0.5 is the same as x^(1/2).  The slope of the line is the dervative at this point.  Since the function is (x-1)^0.5, the derivative is 0.5(x-1)^-0.5 = 1/(2(x-1)^0.5).  Put in the x value to find the slope and use (5,2) as (x0,y0).  That all fits into the point-slope form of a line.

Calculus

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