SOLUTION: Find the coordinates of point M algebraically: if it is equidistant from the points (-8, 8) and (6,1) and if it's y-coordinate is already shown to be -6. PLEASE HELP! THANKS!

Question 124881:
Find the coordinates of point M algebraically: if it is equidistant from the points (-8, 8) and (6,1) and if it's y-coordinate is already shown to be -6.
PLEASE HELP! THANKS!
Found 2 solutions by stanbon, checkley71:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Find the coordinates of point M algebraically: if it is equidistant from the points (-8, 8) and (6,1) and if it's y-coordinate is already shown to be -6.
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Let the missing point be (x,-6)
It's distance from each of the given points is the same.
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EQUATION:
sqrt((x+8)^2 + (-6-8)^2) = sqrt((x-6)^2 + (-6-1)^2)
x^2+16x+4 + 196 = x^2-12x+36 + 49
28x = -115
x = -4.107143..
=================
Cheers,
Stan H.
Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
FIRST YOU NEED TO FIND THE SLOPE THROUGH (-8,8) & (6,1).
SLOPE=(1-8)/(6+8)=-7/14=-1/2.
THE EQUATION FOR THIS IS:
1=-6/2+b
1=-3+b
b=1+3
b=4 WHICH IS THE Y INTERCEPT
THUS THE EQUATION IS:
Y=-X/2+4 (RED LINE)
USING THE LINE FORMULA Y=mX+b & USING THE Y VALUE=-6 & THE SLOPE OF 2 (THE NEGATIVE RECIPRICAL OF -1/2) WHICH IS THE PERPENDICULAR LINE WHICH BISECTS ( 2 THE MID POINT) THE LINE BETWEEN (-8,8) & (6,1)WHICH IS
(8-1)/2=7/2=3.5+1=4.5 FOR THE Y VALUE
(6+8)/2=14/2=7-8=-1 FOR THE X VALUE.
SO WE HAVE A SLOPE=2, A Y VALUE=4.5 & AN X VALUE=-1 WE SOLVE THE LINE FORMULA FOR b.
4.5=2*-1+b
4.5=-2+b
b=4.5+2
b=6.5
THUS THE LINE EQUATION IS:
Y=2X+6.5 (GREEN LINE)
NOW TO FIND THE X COORDINATE ASSOCIASTED WIT A Y VALUE=-6 WE SUBSTITUTE -6 FOR Y & SOLVE FOR X
-6=2X+6.5
2X=-6-6.5
2X=-12.5
X=-12.5/X
X=-6.25 IS THE X COORDINATE OR (-6.25,6) FOR THE POINT EQUIDISTANCE FROM THE END POINT A (-8,8) & (6,1)
PROOF: (graph 300x300 pixels, x from -10 to 10, y from -10 to 10, of TWO functions y = -x/2 +4 and y = 2x +6.5).

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