Question

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

Answer 1

Step-by-step explanation:

According to the question we have A(2,−5) and B(−2,9) Let the points be P(x,0). So, AP=PB and AP2=PB2 ⇒(x−2)2+(0+5)2 = (x+2)2+(0−9)2 ⇒x2+4−4x+25=x2+4+4x+81 ⇒x2+29−4x=x2+85+4x ⇒−4x−4x=85−29 ⇒−8x=56 ⇒x=−7 Hence, point on the x-axis which is equidistant from (2,−5) and (−2,9) is (−7,0).Read more on Sarthaks.com - https://www.sarthaks.com/2428/find-the-point-on-the-x-axis-which-is-equidistant-from-2-5-and-2-9

Answer 2

Given,

The points are (2,-5) and (-2,9).

To find,

The point on the x-axis which is equidistant from the two given points.

Solution,

Now,the point is on x-axis,so the value 'y' of the given point will be zero.

Let,the value of the 'x' value of the given point = x

[Assume,x as a variable to do the further mathematical calculations.]

So,the point is = (x,0)

As mentioned in the question,the two points are equidistant from (x,0).

So,

Distance between (x,0) and (2,-5)

= ✓(x-2)²+(0+5)²

Distance between (x,0) and (-2,9).

=✓(x+2)²+(0-9)²

Now,the points are equidistant.

So,

✓(x-2)²+(0+5)² = ✓(x+2)²+(0-9)²

(x-2)²+(0+5)² = (x+2)²+(0-9)²

x²-4x+4+25 = x²+4x+4+81

x²-4x-x²-4x = 4+81-4-25

-8x = 56

x = -7

Hence,the point on x-axis is (-7,0).

Hope it will helps you ❤️