Finding Equidistant Point on the X-axis

The question asks us to find a point on the x-axis that is the same distance away (equidistant) from two given points: (2, -5) and (-2, 9).

Understanding the Problem

• A point on the x-axis always has its y-coordinate equal to 0. Let the required point on the x-axis be P(x, 0).
• The two given points are A(2, -5) and B(-2, 9).
• The problem states that point P is equidistant from A and B. This means the distance from P to A (PA) is equal to the distance from P to B (PB).

Using the Distance Formula

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ in a coordinate plane is given by the distance formula:

$$\backslash text\{Distance\}\; =\; \backslash sqrt\{(x\_2\; -\; x\_1)^2\; +\; (y\_2\; -\; y\_1)^2\}$$

Since PA = PB, we can also say that $PA^2 = PB^2$ to avoid dealing with the square root.

Calculating $PA^2$:

Point P is (x, 0) and Point A is (2, -5).

$$PA^2\; =\; (x\; -\; 2)^2\; +\; (0\; -\; (-5))^2$$

$$PA^2\; =\; (x\; -\; 2)^2\; +\; (0\; +\; 5)^2$$

$$PA^2\; =\; (x\; -\; 2)^2\; +\; 5^2$$

Expanding $(x - 2)^2$:

$$(x\; -\; 2)^2\; =\; x^2\; -\; 2(x)(2)\; +\; 2^2\; =\; x^2\; -\; 4x\; +\; 4$$

So, $$PA^2\; =\; x^2\; -\; 4x\; +\; 4\; +\; 25$$

$$PA^2\; =\; x^2\; -\; 4x\; +\; 29$$

Calculating $PB^2$:

Point P is (x, 0) and Point B is (-2, 9).

$$PB^2\; =\; (x\; -\; (-2))^2\; +\; (0\; -\; 9)^2$$

$$PB^2\; =\; (x\; +\; 2)^2\; +\; (-9)^2$$

Expanding $(x + 2)^2$:

$$(x\; +\; 2)^2\; =\; x^2\; +\; 2(x)(2)\; +\; 2^2\; =\; x^2\; +\; 4x\; +\; 4$$

So, $$PB^2\; =\; x^2\; +\; 4x\; +\; 4\; +\; 81$$

$$PB^2\; =\; x^2\; +\; 4x\; +\; 85$$

Setting up the Equidistant Equation

Since PA = PB, we have $PA^2 = PB^2$:

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

Solving for x

Now, we solve this equation for x:

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

Subtract $x^2$ from both sides:

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

Subtract $4x$ from both sides:

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

$$-8x\; +\; 29\; =\; 85$$

Subtract 29 from both sides:

$$-8x\; =\; 85\; -\; 29$$

$$-8x\; =\; 56$$

Divide by -8:

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

$$x\; =\; -7$$

The Point on the X-axis

The required point on the x-axis is P(x, 0). Since we found x = -7, the point is (-7, 0).

Verification

Let's check if the distance from (-7, 0) to (2, -5) is the same as the distance from (-7, 0) to (-2, 9).

Distance from (-7, 0) to (2, -5):

$$\backslash sqrt\{(2\; -\; (-7))^2\; +\; (-5\; -\; 0)^2\}\; =\; \backslash sqrt\{(2\; +\; 7)^2\; +\; (-5)^2\}\; =\; \backslash sqrt\{9^2\; +\; 25\}\; =\; \backslash sqrt\{81\; +\; 25\}\; =\; \backslash sqrt\{106\}$$

Distance from (-7, 0) to (-2, 9):

$$\backslash sqrt\{(-2\; -\; (-7))^2\; +\; (9\; -\; 0)^2\}\; =\; \backslash sqrt\{(-2\; +\; 7)^2\; +\; 9^2\}\; =\; \backslash sqrt\{5^2\; +\; 81\}\; =\; \backslash sqrt\{25\; +\; 81\}\; =\; \backslash sqrt\{106\}$$

Since both distances are $\backslash sqrt\{106\}$, the point (-7, 0) is indeed equidistant from the two given points.

Conclusion

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

Revision Table: Key Concepts

Additional Information: Equidistant Points and Perpendicular Bisector

The set of all points that are equidistant from two distinct points forms a straight line called the perpendicular bisector of the line segment connecting the two points.

In this problem, we were looking specifically for the point where this perpendicular bisector intersects the x-axis. The point (-7, 0) lies on both the x-axis and the perpendicular bisector of the line segment connecting (2, -5) and (-2, 9).

The equation of the perpendicular bisector could be found by finding the midpoint of the segment AB, calculating the slope of AB, finding the negative reciprocal slope (perpendicular slope), and then using the point-slope form with the midpoint and the perpendicular slope. The x-intercept of this line would be the point we found.