Question

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

Answer 1

We have to find a point on x-axis. Therefore, its y-coordinate will be 0.

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

Distance between (x, 0) and (2, -5) = `sqrt((x-2)^2+(0-(-5))^2)`

= `sqrt((x-2)^2+(5)^2)`

Distance between (x, 0) and (-2, -9) = `sqrt((x-(-2))^2+(0-(-9))^2)`

= `sqrt((x+2)^2+(9)^2)`

By the given condition, these distances are equal in measure.

`sqrt((x-2)^2 +(5)^2)`

= `sqrt((x+2)^2+(9)^2)`

= (x - 2)² + 25 = (x + 2)² + 81

= x² + 4 - 4x + 25

= x² + 4 + 4x + 81

8x = 25 - 81

8x = -56

x = -7

Therefore, the point is (−7, 0).

Answer 2

Let (x, 0) be the point on the x axis. Then as per the question, we have

⇒ `sqrt((x-2)^2 +(0+5)^2)`

⇒ `sqrt((x+2)^2 + (0-9)^2)`

⇒ `sqrt((x-2)^2 +(5)^2)=sqrt((x+2)^2 + (9)^2)`

⇒ (x - 2)² + (5)² = (x + 2)² + (-9)²          ...(Squaring both sides) 

⇒ x² - 4x + 4 + 25 = x² + 4x + 4 + 81 

8x = 25 - 81

8x = -56

x = -7

Therefore, the point is (−7, 0).