Alright! Let's go through the problem step-by-step to find the point [tex]\( P(x, y) \)[/tex] such that the distances from [tex]\( P \)[/tex] to [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are equal, and [tex]\( P \)[/tex] lies on the given line [tex]\( 10x - 12y - 15 = 0 \)[/tex].
Given points:
- [tex]\( A = (4, 3) \)[/tex]
- [tex]\( B = (-1, -3) \)[/tex]
- The equation of the line: [tex]\( 10x - 12y - 15 = 0 \)[/tex]
We are asked to find point [tex]\( P(x, y) \)[/tex] such that:
1. [tex]\( PA = PB \)[/tex]
2. Point [tex]\( P \)[/tex] satisfies the line equation [tex]\( 10x - 12y - 15 = 0 \)[/tex]
### Step 1: Establish the distance equality
Firstly, we'll use the distance formula to express [tex]\( PA \)[/tex] and [tex]\( PB \)[/tex]:
[tex]\[
PA = \sqrt{(x - 4)^2 + (y - 3)^2}
\][/tex]
[tex]\[
PB = \sqrt{(x + 1)^2 + (y + 3)^2}
\][/tex]
### Step 2: Equate the distances
Given [tex]\( PA = PB \)[/tex], we can set up the following equation:
[tex]\[
\sqrt{(x - 4)^2 + (y - 3)^2} = \sqrt{(x + 1)^2 + (y + 3)^2}
\][/tex]
Square both sides to eliminate the square roots:
[tex]\[
(x - 4)^2 + (y - 3)^2 = (x + 1)^2 + (y + 3)^2
\][/tex]
### Step 3: Expand both sides
Expand the expressions:
[tex]\[
(x^2 - 8x + 16) + (y^2 - 6y + 9) = (x^2 + 2x + 1) + (y^2 + 6y + 9)
\][/tex]
Simplify by combining like terms:
[tex]\[
x^2 - 8x + 16 + y^2 - 6y + 9 = x^2 + 2x + 1 + y^2 + 6y + 9
\][/tex]
Remove [tex]\( x^2 \)[/tex] and [tex]\( y^2 \)[/tex] from both sides, since they are equal:
[tex]\[
-8x + 16 - 6y + 9 = 2x + 1 + 6y + 9
\][/tex]
Combine all constant terms:
[tex]\[
-8x - 6y + 25 = 2x + 6y + 10
\][/tex]
Combine like terms involving [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[
-8x - 2x - 6y - 6y = 10 - 25
\][/tex]
Simplify further:
[tex]\[
-10x - 12y = -15
\][/tex]
Which simplifies to:
[tex]\[
10x + 12y = 15
\][/tex]
### Step 4: Incorporate the given line equation
Given point [tex]\( P \)[/tex] lies on the line [tex]\( 10x - 12y - 15 = 0 \)[/tex], this is already our second equation.
Thus, the solution for [tex]\( P \)[/tex] should satisfy:
1. [tex]\( 10x + 12y = 15 \)[/tex]
2. [tex]\( 10x - 12y - 15 = 0 \)[/tex]
### Step 5: Solve the system of equations
Add the two equations:
[tex]\[
(10x + 12y) + (10x - 12y) = 15 + 15
\][/tex]
[tex]\[
20x = 30
\][/tex]
[tex]\[
x = \frac{30}{20} = \frac{3}{2}
\][/tex]
Substitute [tex]\( x = \frac{3}{2} \)[/tex] back into the line equation [tex]\( 10x - 12y - 15 = 0 \)[/tex]:
[tex]\[
10 \left(\frac{3}{2}\right) - 12y - 15 = 0
\][/tex]
[tex]\[
15 - 12y - 15 = 0
\][/tex]
Solve for [tex]\( y \)[/tex]:
[tex]\[
-12y = 0
\][/tex]
[tex]\[
y = 0
\][/tex]
Therefore, the point [tex]\( P \)[/tex] is:
[tex]\[
\left( \frac{3}{2}, 0 \right)
\][/tex]
