Question

Find the point of x- axis
which is equidistance

from (2-5) and (-2, 9)​

Answer 1

Answer:

point equidistant from (2,-5) and (-2,9) is

(0,2)

hope it's helpful

Answer 2

Answer:

x=-7

Step-by-step explanation:

point of x-. axis =(X,0)

equidistant from (2,-5) and (-2,9)

$\sqrt{(x2- x1) {}^{2} + (y2 - y1) {}^{2} } = \sqrt{(x3 - x2) {}^{2} + (y3 - y2) {}^{2} } \\ \sqrt{(2 - x) {}^{2} + -( - 5 - 0) {}^{2} } = \sqrt{ ( - 2 - x) {}^{2} + (9 - 0) {}^{2} } \\ \sqrt{(2 - x) {}^{2} + ( - 5) {}^{2} } = \sqrt{( - 2 - x) {}^{2} + (9) {}^{2} } \\ \sqrt{4 + {x}^{2} - 4x + 25} = \sqrt{4 + {x}^{2} + 4x + 81} \\ \sqrt{ {x}^{2} - 4x + 29} = \sqrt{ {x}^{2} + 4x + 85} \\ {x}^{2} - 4x + 29 = {x}^{2} + 4x + 85 \\ - 4x - 4x = 85 - 29 \\ - 8x = 56 \\ x = \frac{ - 56}{8} \\ x = - 7 \: our \: answer$