A(-2,6) B (9,3)
We need to find C on x-axis ==> C (x,0)
such that :
AC = BC
AC= sqrt(0-6)^2 + (x+2)^2]= sqrt[(36 + (x+2)^2]
BC= sqrt[(0-3)^2+ (x-9)^2]= sqrt[(9 + (x-9)^2]
==> AC = BC
==> sqrt(36+(x+2)^2]= sqrt[(9+ (x-9)^2]
square both sides:
==> 36 + (x+2)^2 = 9 +...
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A(-2,6) B (9,3)
We need to find C on x-axis ==> C (x,0)
such that :
AC = BC
AC= sqrt(0-6)^2 + (x+2)^2]= sqrt[(36 + (x+2)^2]
BC= sqrt[(0-3)^2+ (x-9)^2]= sqrt[(9 + (x-9)^2]
==> AC = BC
==> sqrt(36+(x+2)^2]= sqrt[(9+ (x-9)^2]
square both sides:
==> 36 + (x+2)^2 = 9 + (x-9)^2
==> 36 + x^2 +4x + 4 = 9 + x^2 -18x + 81
Now group similars:
==> 22x -50 = 0
==> x= 50/22= 25/11
Then the point C is (25/11, 0)
Let the point C be at (x,y).
AC--> D^2 = (-2 - x)^2 + (6 - y)^2 = x^2 + 4x + 4 + y^2 - 12y + 36
BC--> D^2 = (x - 9)^2 + (y - 3)^2 = x^2 - 18x + 81 + y^2 - 6y + 9
AC = BC
x^2 + 4x + 4 + y^2 - 12y + 36 = x^2 - 18x + 81 + y^2 - 6y + 9
11x - 25 - 3y = 0
The point C lies on this line.
On the x-axis, y= 0. Therefore:
11x - 25 = 0
x = 25/11