Question 7

Equidistant: A point is equidistant from two other points if its distance from both is exactly equal. Any point on the x-axis has coordinates $(x, 0)$ since its y-coordinate is always 0.

Distance Formula: The distance between two points $\text{A}(x_1, y_1)$ and $\text{B}(x_2, y_2)$:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

We use the distance formula to find the point on the x-axis that is equally far from the two given points.

Step 1 — Set up the problem

Let's call the two given points $\text{A}(\textbf{2}, \textbf{-5})$ and $\text{B}(\textbf{-2}, \textbf{9})$. We need to find a point on the x-axis. Any point on the x-axis has its y-coordinate as $\textbf{0}$. So, let's call this point $\text{P}(x, \textbf{0})$. The problem states that point $\text{P}$ is equidistant from $\text{A}$ and $\text{B}$. This means the distance $\text{PA}$ must be equal to the distance $\text{PB}$. We will use the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

$PA = \sqrt{(2 - x)^2 + (-5 - 0)^2}$

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

Since $\text{PA} = \text{PB}$, we can write:

$\sqrt{(2 - x)^2 + (-5)^2} = \sqrt{(-2 - x)^2 + (9)^2}$

Step 2 — Solve for x

To get rid of the square roots, we can square both sides of the equation.

$(2 - x)^2 + (-5)^2 = (-2 - x)^2 + (9)^2$

Let's expand the squared terms.

$(4 - 4x + x^2) + 25 = (4 + 4x + x^2) + 81$

Now, we can simplify both sides of the equation.

$x^2 - 4x + 29 = x^2 + 4x + 85$

We can subtract $x^2$ from both sides.

$-4x + 29 = 4x + 85$

Let's gather the $x$ terms on one side and constant terms on the other.

$29 - 85 = 4x + 4x$

$-56 = 8x$

Finally, we solve for $x$.

$x = \frac{-56}{8}$

$x = -7$

So, the x-coordinate of point $\text{P}$ is $\textbf{-7}$. The point on the x-axis is $\text{P}(\textbf{-7}, \textbf{0})$.

$\boxed{P(-7, 0)}$

Answer

The point on the x-axis equidistant from $(2, -5)$ and $(-2, 9)$ is $(-7, 0)$.