Question

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

Answer 1

Answer:

Step-by-step explanation:

Distance formula ✓(x2-x1)^2+(y2-y1)^2

O - (x,0) ; P - (2,-5) ; Q - (-2,9)

Putting values

PO=QO

Equating

LHS PO=✓29+x^2-4

RHS QO=✓85+x^2+4x

Square both sides

And you'll get the answer

8x=-56

x= -7

Answer 2

Step-by-step explanation:

let the point be (a,0)

as it is equidistant from given points so,

$\sqrt{(0 + 5) {}^{2} +(a - 2) {}^{2} } = \sqrt{(0 - 9) {}^{2} +( a + 2 ) {}^{2} }$

$25 + a {}^{2} + 4 - 4a = 81 + a {}^{2} + 4 + 4a$

$8a = - 56$

$a = - 7$

so the points are (-7,0)

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