To find the point on the x-axis that is equidistant from the points (-1, 0) and (5, 0), let the required point be (x, 0), since it lies on the x-axis (its y-coordinate is 0).

 Step 1: Use the distance formula
The distance between two points (x₁, y₁) and (x₂, y₂) is given by:

d = √((x₂ – x₁)² + (y₂ – y₁)²)

The distances from (x, 0) to (-1, 0) and (5, 0) must be equal. Therefore:

Distance to (-1, 0) = Distance to (5, 0)

√((x – (-1))² + (0 – 0)²) = √((x – 5)² + (0 – 0)²)

Simplify both sides:

√((x + 1)²) = √((x – 5)²)

Step 2: Eliminate the square roots
Square both sides to remove the square roots:

(x + 1)² = (x – 5)²

Expand both sides:

x² + 2x + 1 = x² – 10x + 25

 Step 3: Simplify the equation
Cancel out x² from both sides:

2x + 1 = -10x + 25

Combine like terms:

2x + 10x = 25 – 1
12x = 24

Solve for x:

x = 24 / 12
x = 2

 Step 4: Verify the solution
The point on the x-axis is (2, 0). To verify, calculate the distances from (2, 0) to (-1, 0) and (5, 0):

1. Distance to (-1, 0):
√((2 – (-1))² + (0 – 0)²) = √((2 + 1)²) = √(3²) = 3

2. Distance to (5, 0):
√((2 – 5)² + (0 – 0)²) = √((-3)²) = √(3²) = 3

Since both distances are equal, the point (2, 0) is indeed equidistant from (-1, 0) and (5, 0).

The correct answer is:
b) (2, 0)
This question related to Chapter 7 Mathematics Class 10th NCERT. From the Chapter 7 Coordinate Geometry. Give answer according to your understanding.

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