Question

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

Answer 1

Option D = (3, 0)

Step-by-step explanation:

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

Then the distance of (p, 0) from the point (- 4, 0) is

= √{(p + 4)² + (0 - 0)²} units

= √{(p + 4)²} units

= √(p² + 8p + 16) units

and the distance of (p, 0) from the point (10, 0) is

= √{(p - 10)² + (0 - 0)²} units

= √{(p - 10)²} units

= √(p² - 20p + 100) units

By the given condition,

√(p² + 8p + 16) = √(p² - 20p + 100)

or, p² + 8p + 16 = p² - 20p + 100

or, 8p + 16 = - 20p + 100

or, 28p = 84

or, p = 3

∴ the required point on the x-axis is (3, 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

Answer 2

Given: Two points : (-4,0) and (10,0)

To find: A point which is equidistant from both the given points.

Solution:

• Now, we know that the point is on x axis (given in the question),

• So the y co-ordinate will be 0, y = 0

• So, Let the point be (x,0)

• The distance formula is $\sqrt{(x-x1)^2 + (y-y1)^2}$

• Here x1 is -4 and y1 is 0

• Then the distance of (x, 0) from the point (- 4, 0) will be,

           = $\sqrt{(x+4)^2 + (0-0)^2}$

           = √(x+4)²

• Now, the distance of (x, 0) from the point (10, 0) will be

             $\sqrt{(x-10)^2 + (0-0)^2}$

           = √(x - 10)²

• By the given condition, equating the both distance, we get

           √(x+4)² = √(x - 10)²

• squaring both sides, we get

          (x+4)² = (x - 10)²

          x² + 8x +16 = x² - 20x + 100

          28x = 84

          x = 84/28

          x = 3

Answer:

          So the point is (D) (3,0).