Objective: Equidistant Point on X-axis

Find a point P located on the x-axis such that its distance to A(2,3) is the same as its distance to B(-4,5).

Key Concepts: X-axis & Distance Formula

• Point on X-axis: Any point on the x-axis has coordinates $(x, 0)$. Let P = $(x, 0)$.
• Equidistant Points: The distance PA must equal the distance PB. PA = PB.
• Distance Formula: The distance between $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.

Coordinate Calculation: Step-by-Step

1. Distance PA: Calculate the distance between P$(x, 0)$ and A$(2, 3)$.

   PA = $\sqrt{(x-2)^2 + (0-3)^2} = \sqrt{(x-2)^2 + 9}$

2. Distance PB: Calculate the distance between P$(x, 0)$ and B$(-4, 5)$.

   PB = $\sqrt{(x-(-4))^2 + (0-5)^2} = \sqrt{(x+4)^2 + 25}$

3. Equate Distances: Set PA = PB.

   $\sqrt{(x-2)^2 + 9} = \sqrt{(x+4)^2 + 25}$

4. Solve for x: Square both sides and simplify.

   $(x-2)^2 + 9 = (x+4)^2 + 25$

   $x^2 - 4x + 4 + 9 = x^2 + 8x + 16 + 25$

   $x^2 - 4x + 13 = x^2 + 8x + 41$

   Cancel $x^2$ terms:

   $-4x + 13 = 8x + 41$

   Group x terms and constants:

   $13 - 41 = 8x + 4x$

   $-28 = 12x$

   Calculate x:

   $x = \frac{-28}{12} = -\frac{7}{3}$

Final Coordinates of P

Since P lies on the x-axis, its coordinates are $(x, 0)$. Substituting the calculated value of x gives:

P = $(-\frac{7}{3}, 0)$