# Please explain the following in easy to understand steps.Point P lies on the x-axis. The coordinates of A and B are (3,4) and (5,8). Determine the possible coordinates of P if it is equidistant...

Please explain the following in easy to understand steps.

Point P lies on the x-axis. The coordinates of A and B are (3,4) and (5,8). Determine the possible coordinates of P if it is equidistant from A and B. The answers are P (16,0)

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### 2 Answers

Since the point P lies on the x-axis, then the coordinate is ( x, 0).

Now given A(3,4) and B(5,8) such that:

The distance AP = the distance BP

==> l AP l = l BP l

Now we will use the distance formula:

l AP l = sqrt[(xA- xP)^2 + (yA-yP)^2]

= sqrt(3 - x)^2 +(4 -0)^2

= sqrt(9-6x+x^2 + 16)

= sqrt(x^2 -6x + 25)...............(1)

l BP l = sqrt[(xB - xP)^2 + (yB - yP)^2]

= sqrt[(5-x)^2 + (8-0)^2]

= sqrt(25-10x+x^2 + 64)

= sqrt(x^2 -10x + 89)..............(2)

We know that:

l AP l = lBP l

==> eq. (1) = eq. (2)

==>sqrt(x^2 -6x + 25) = sqrt(x^2 - 10x + 89)

Square both sides:

==>x^2 - 6x +25 = x^2 -10x + 89

==>-6x + 25 = -10x + 89

==> 4x = 89-25

==> 4x = 64

==>x= 16

**Then the point P is (16, 0)**

The coordinates of any point P on the x axis is given by ( p , 0).

Therefore by distance formula, the distance formula , we get:

AP^2 = (3-p)^2+(4-0)^2 = 9-6p+p^2+16

BP^2 = (5-p)^2+(8-0)^2 = 25-10p+p^2+64.

Since AP = BP, 9-6p+p^2 +16= 25-10p+p^2+64. Subtracting 9-6p+p^2+16 fro both sides, we get:

0 = 25-10p+p^2-(9-6p+p^2).

0 = 25+64-9-16-10p+6p +p^2-p^2.

0 = 64 -4p.

-64 = -4p.

4p = 64.

p = 64/4 = 16.

Threfore the point (16,0) is equidistant from A(3,4) ) and B( 5,8).