A(4,7) and B(-3,2) are points on a coordinate plane.Find the coordinates of a point C on the x-axis such that AC=BC .
If you make a sketch it should be easy to see that point C must be on the right-bisector of AB
midpoint of AB = ( (4-3)/2 , (7+2)/2) = (1/2 , 9/2)
slope o AB = (2-7)/(-3-4) = 5/7
so the slope of the right-bisector is -7/5
equation of right-bisector:
y - 9/2 = (-7/5)(x - 1/2)
for the x-intercept, let y = 0
-9/2 = (-7/5)(x-1/2)
-9/2 = -7x/5 + 7/10
-45 = -14x + 7
14x = 52
x = 52/14 = 26/7
point C is (26/7, 0)
let the point be C(x,0)
AC = BC
√( (x-4)^2 + 7^2) = √( (x+3)^2 + 4^2)
square both sides
(x-4)^2 + 49 = (x+3)^2 + 4
x^2 - 8x + 16 + 49 = x^2 + 6x + 9 + 4
-14x = -52
x = -52/-14 = 26/7 , just like before
THANKS A LOT !!:)
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