Question

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

Answer 1

☆Given

A point on X axis where A(2,-5) and B(-2,9)

To find

the point on X axis

Solution

Let the point on X axis be p(X,0)

Here,

y is 0 because the point is only on X axis

From section formula

x=$\frac{mx_2+nx_2}{m+n}$

y=$\frac{my_2+ny_2}{m+n}$

Applying on coordinate y of p

x=$\frac{mx_2+nx_2}{m+n}$

0=$\frac{9m-5n}{m+n}$

O=9m-5n

5n=9m

$\frac{m}{n}$=$\frac{5}{9}$

Now to getting coordinate X we have to put value of m:n on X

X=$\frac{mx_2+nx_2}{m+n}$

x=$\frac{-10+18}{14}$

X=$\frac{4}{7}$

Hence,the point p is ($\frac{4}{7}$,0)