Question

find the point an the x_ axis which is equidistant fro (2,_5)and (_2,9)​

Answer 1

Answer:

Given ,

the coordinates of the point of P

= (x,0)

Now, according to the above question

the point P is equidistant from X -Axis from

(2,-5) & (-2,9)

Thus, the distance from (x,0) & (2,-5) =distance from (x,0) & (-2,9).

$\sqrt{ {(2 - x)}^{2} + ( - 5 - 0) {}^{2} } = \sqrt{( - 2 - x) {}^{2} +( 9 - 0) {}^{2} } \\ = > \sqrt{4 + {x}^{2} - 4x + 25 } = \sqrt{4 + {x}^{2} + 4x + 81 } \\ now \: squaring \: on \: both \: sides \: we \: get \: \\ 4 + {x}^{2} - 4x + 25 = 4 + {x}^{2} + 4x + 81 \\ = > 8x = - 56 \\ = > x = - 7$

Hence, the point on X-Axis is (-7,0)