Question

the point P on x axis equidistant from the points A(-1,0) and B(5,0) is​

Answer 1

Required point P on the x-axis is (2, 0).

Step-by-step explanation:

Let the point on the x-axis be (p, 0).

∴ the distance of (p, 0) from the point A (- 1, 0) is

= √{(p + 1)² + (0 - 0)²} units

= √{(p + 1)²} units

= √(p² + 2p + 1) units

and the distance of (p, 0) from the point B (5, 0) is

= √{(p - 5)² + (0 - 0)²} units

= √{(p - 5)²} units

= √(p² - 10p + 25) units

By the given condition,

√(p² + 2p + 1) = √(p² - 10p + 25)

or, p² + 2p + 1 = p² - 10p + 25

or, 2p + 1 = - 10p + 25

or, 12p = 24

or, p = 2

∴ the required point on the x-axis is (2, 0).

Equidistance related problem:

• If the point R(x,y) is equidistant from two points P (-3, 4) and Q (2, -1), prove that y = x + 2.

- https://brainly.in/question/13073351

• The point on the x-axis which is equidistant from (- 4, 0) and (10, 0) is

A. (7, 0)

B. (5, 0)

C. (0, 0)

D. (3, 0)

- https://brainly.in/question/15921962

Answer 2

the point P on x axis equidistant from the points A(-1,0) and B(5,0) is​ (2, 0)

Given,

since the point P is on x axis, therefore, we have (x, 0)

distance  formula is given by, d = √ [ (x2-x1)^2 + (y2-y1)^2]

AP = BP

√ [ (-1-x)^2 + (0-0)^2] = √ [ (5-x)^2 + (0-0)^2]

(-1-x)^2 + (0-y)^2 = (5-x)^2 + (0-y)^2

1 + x^2 + 2x = x^2 + 25 - 10x

12x = 24

x = 2

(2, 0) is the coordinates of P.