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Find the Point on The X-axis Which is Equidistant from (2, - 5) and (- 2, 9). - Mathematics

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

 

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Solution

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).

Concept: Distance Formula
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APPEARS IN

NCERT Class 10 Maths
Chapter 7 Coordinate Geometry
Exercise 7.1 | Q 7 | Page 161

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