Question

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

Answer 1

Answer:

Let a point on X-axis A(a,0).

B(–2,9) and C(2,–5) are equidistant from point A.

therefore AB=AC.

The distance between (x1,y1) and (x2,y2)=√x2-x1)²+ (y2-y1)²

The distance between A(a,0) and B(-2,9)

AB=√(-2-a)²(9-0)²

=√(-2-a)²+(9)²

=√[–(2+a)]²+81

=√(2+a)²+81 units.

The distance between A(a,0) and C(2,-5)

AC=√(2-a)²+(-5-0)²

=√(2-a)²+(-5)²

=√(2-a)²+25 units.

By problem,

AB=AC

→ √(2+a)²+81=√(2-a)²+25

squaring on both sides

(2+a)²+81=(2-a)²+25

(2)²+(a)²+2(2)(a)+81=(2)²+(a)²-2(2)(a)

4+a²+4a+81=4 +a²-4a

4a+4a=25-81

8a=-56

a=-56/8

a=-7

Therefore the required point on X-axis is A(-7,0).

hope it works..........................................™✌️✌️