# The coordinates of two points are A(-2,6) & B(9,3). Find the coordinates of the point C on the X-axis such that AC=BC.

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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**

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)

Any point C on X axis has the y coordinate zero. So we take P( h,0) as a point on X axis.

Since CA = CB, by distance formula,

CA^2 = ((h- -2)^2 +(0-6) ^2 = (h+2)^2+36 = h^2+4h+40....(1)

CB^2 = (h-9)^2+(0-3)^2 = = h^2-18h+ 90.....(2).

Right side of (1) and (2) are equal as CA = CB.

h^2+2h+40 = h^2-18h+90

2h+18h = 90 - 40 = 50

20 = 50 Or

h = 50/20 = 2.5

So C has the coordinates (2.5,0).