Question

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

Answer 1

Given:  

Points $(2,-5)$ and $(-2,9)$.

To find:

Point on the x -axis which is equidistant from both the given points.

Explanation:

We know that any point on x axis is of the form $x,0$.  So its y coordinate will be zero. Now the distance between $(x,0)$ and $(2,-5)$ is given by the following

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

Now distance between $(x,0)$ and $(-2,9)$ is given by the following

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

Now according to the question these distances are equal. Then

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

$(x-2)^2+25$=$(x-2)^2+81$

$8x=25-81$

$8x=-56$

$x=-7$

Therefore the required point is $(-7,0)$