Triangle Problem: Calculate Perpendicular Bisector

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To calculate the perpendicular bisector of triangle side AB with vertices A(6,-15) and B(22,17), first determine the midpoint, which is (14,1). The slope of line AB is 2, leading to a perpendicular slope of -1/2. Using the point-slope form of the equation, the perpendicular bisector can be expressed as y - 1 = -1/2(x - 14). This equation represents the line that bisects side AB at a right angle.
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assignment question URGENT!

I having trouble and its urgent some one help
Triangle wit the vertices A(6,-15)B(22,17)and C(24,11)

-midpoint of side A = (14,1)

Question

1.Calculate the equation of the line that is perpindicular bisector of the side triangle.
 
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Use the fact that m_{2} = -\frac{1}{m_{1}} for a perpendicular line.
 
i Still do not get it
 
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but how do i figure out the equation of the line
 
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The slope of AB is 2. So the slope of the perpendicular bisector is -\frac{1}{2}.

So use the equation y-y_{1} = m(x-x_{1}).

So y - 1 = -\frac{1}{2}( x-14)
 
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I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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