SOLUTION: Find the standard form of the equation of the perpendicular bisector of the line defined by the points (-2,5) and (6,-1).

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Question 102180: Find the standard form of the equation of the perpendicular bisector of the line defined by the points (-2,5) and (6,-1).
Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
Y=mX+b IS THE LINE EQUATION. m=THE SLOPE & b=THE Y INTTERCEPT.
FIRST WE FIND THE SLOPE OF THE LINE THROUGH (-2,5) & (6,-1):
SLOPE=(-1-5)/(6+2)=-6/8=-3/4.
THE Y INTERCEPT IS:
5=-2*-3/4+b
5=6/4+b
b=5-6/4
b=5-1.5
b=3.5
Y=-3X/4+3.5 IS THE LINE EQUATION THROUGH THESE 2 POINTS(RED LINE)
THUS THE SLOPE OF A PERPENDICULAR LINE IS 4/3.
NOW WE NEED TO FIND THE MIDPOINT OF THE ORIGINAL LINE.
(6+2)/2=8/2=4 UNITS FROM EITHER END OR 6-4=2 FOR THE X COORDINATE OF THE MIDPOINT.
(5+1)/2=6/2=3 UNITS FROM EITHER END OR 5-3=2 FOR THE Y COORDINATE OF THE MIDPOINT.
(2,2) WITH A SLOPE OF 4/3 IS:
2=(4/3)2+b WHERE b IS THE Y INTERCEPT.
2=8/3+b
b=2-8/3
b=(6-8)/3
b=-2/3
THUS THIS LINE EQUATION IS:
Y=4X/3-2/3 (green line)
THE GRAPH OF THESE 2 LINES FOLLOWS:
+graph%28+300%2C+300%2C+-6%2C+5%2C+-6%2C+5%2C+y+=+-3x%2F4+%2B3.5%2C+y+=+4x%2F3+-2%2F3%29+ (graph 300x300 pixels, x from -6 to 5, y from -6 to 5, of TWO functions y = -3x/4 +3.5 and y = 4x/3 -2/3).