The point on X-axis which is equidistant from the points (2, -2) and (-4,2) is(a) (1,0) (b) (2,0) (c) (0,2) (d) (-1,0)

To find the point on the X-axis that is equidistant from the points (2, -2) and (-4, 2), we can use the distance formula.

The distance formula is given by: d = sqrt((x2 - x1)^2 + (y2 - y1)^2)

Let's assume the point on the X-axis is (x, 0).

The distance from (x, 0) to (2, -2) is: d1 = sqrt((2 - x)^2 + (-2 - 0)^2)

The distance from (x, 0) to (-4, 2) is: d2 = sqrt((-4 - x)^2 + (2 - 0)^2)

Since we want the point that is equidistant from both points, we can set d1 equal to d2 and solve for x:

sqrt((2 - x)^2 + (-2 - 0)^2) = sqrt((-4 - x)^2 + (2 - 0)^2)

Squaring both sides of the equation, we get:

(2 - x)^2 + (-2 - 0)^2 = (-4 - x)^2 + (2 - 0)^2

Expanding and simplifying the equation, we have:

4 - 4x + x^2 + 4 = 16 + 8x + x^2 + 4

Combining like terms, we get:

x^2 - 12x + 4 = 0

To solve this quadratic equation, we can use the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

In this case, a = 1, b = -12, and c = 4. Plugging these values into the quadratic formula, we get:

x = (-(-12) ± sqrt((-12)^2 - 4(1)(4))) / 2(1)

Simplifying further, we have:

x = (12 ± sqrt(144 - 16)) / 2

x = (12 ± sqrt(128)) / 2

x = (12 ± 8√2) / 2

Simplifying the expression, we have:

x = 6 ± 4√2

So, the possible values for x are 6 + 4√2 and 6 - 4√2.

However, since we are looking for a point on the X-axis, the y-coordinate must be 0. Therefore, the point on the X-axis that is equidistant from the given points is (6 + 4√2, 0).

Therefore, the correct answer is not listed among the options provided.