12.find the point on the x-axis which is equidistant from(2,-5)and(-2,9).

(- 7, 0 )

Step-by-step explanation:

let the point on the x- axis be (x, 0 )

Then the distance from (x, 0) to (2, - 5) and (- 2, 9) is equal

Use the distance formula to calculate the distances

d = √ (x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = (2, - 5) and (x₂, y₂ ) = (x, 0)

d =

  = =

Repeat with

(x₁, y₁ ) = (- 2, 9) and (x₂, y₂ ) = (x, 0)

d =

  = =

Equating both distances

=

Squaring both sides

(x - 2)² + 25 = (x + 2)² + 81 ← expand parenthesis on both sides

x² - 4x + 4 + 25 = x² + 4x + 4 + 81, that is

x² - 4x + 29 = x² + 4x + 85 ← subtract x² + 4x from both sides

- 8x + 29 = 85 ( subtract 29 from both sides )

- 8x = 56 ( divide both sides by - 8 )

x = - 7

Thus the point on the x- axis is (- 7, 0 )

d

step-by-step explanation:

d

. + λrh() e 1 1 2 4 4 4 4 4 3 4 4 4 4 4 + th−1(− s(x)2 dx ω + λrh−1() e 2 1 2 4 4

step-by-step explanation: