To determine the equation of the locus of the point [tex]\( P(x, y) \)[/tex] such that the distances [tex]\( |PS| \)[/tex] and [tex]\( |PR| \)[/tex] are equal, we can start by using the distance formula for points in a Cartesian plane.
Given:
- Point [tex]\( S \)[/tex] has coordinates [tex]\( (4, 6) \)[/tex].
- Point [tex]\( R \)[/tex] has coordinates [tex]\( (12, 6) \)[/tex].
The distance from [tex]\( P(x, y) \)[/tex] to [tex]\( S(4, 6) \)[/tex] is given by:
[tex]\[
|PS| = \sqrt{(x - 4)^2 + (y - 6)^2}
\][/tex]
The distance from [tex]\( P(x, y) \)[/tex] to [tex]\( R(12, 6) \)[/tex] is given by:
[tex]\[
|PR| = \sqrt{(x - 12)^2 + (y - 6)^2}
\][/tex]
Since the point [tex]\( P \)[/tex] moves such that [tex]\( |PS| = |PR| \)[/tex], we set the two distance expressions equal to each other:
[tex]\[
\sqrt{(x - 4)^2 + (y - 6)^2} = \sqrt{(x - 12)^2 + (y - 6)^2}
\][/tex]
To remove the square roots, we square both sides of the equation:
[tex]\[
(\sqrt{(x - 4)^2 + (y - 6)^2})^2 = (\sqrt{(x - 12)^2 + (y - 6)^2})^2
\][/tex]
This simplifies to:
[tex]\[
(x - 4)^2 + (y - 6)^2 = (x - 12)^2 + (y - 6)^2
\][/tex]
We see that the [tex]\( (y - 6)^2 \)[/tex] terms are common on both sides of the equation and cancel each other out, leaving:
[tex]\[
(x - 4)^2 = (x - 12)^2
\][/tex]
Expanding both squares, we get:
[tex]\[
x^2 - 8x + 16 = x^2 - 24x + 144
\][/tex]
We can subtract [tex]\( x^2 \)[/tex] from both sides of the equation:
[tex]\[
-8x + 16 = -24x + 144
\][/tex]
Solving for [tex]\( x \)[/tex], we first add [tex]\( 24x \)[/tex] to both sides:
[tex]\[
16x + 16 = 144
\][/tex]
Then, subtract 16 from both sides:
[tex]\[
16x = 128
\][/tex]
Finally, divide both sides by 16:
[tex]\[
x = 8
\][/tex]
Thus, the equation of the locus of [tex]\( P \)[/tex] is:
[tex]\[
x = 8
\][/tex]
This means that the point [tex]\( P(x, y) \)[/tex] moves along the vertical line [tex]\( x = 8 \)[/tex] in the Cartesian plane.
