Question

the coordinatea of point on x axis which is equidisantfrom points(5,4) and (-2,3)​

Answer 1

Answer:-

The required point is (2,0)

Explanation:-

Given:

• A = (5,4)
• B = k-2,3)

To Find:

Point on x- axis which is equidistant from A and A

Solution:

Let, P(x,y) be the point.

On x- axis Y = 0

Since, P is equidistant from A and B

PA = PB

By using the identity

$\boxed{\bold{\sqrt{{(x_1-x_2)}^{2}+ {(y_1 -y_2 )}^{2}}}}$

${\rightarrow}\sqrt{{(5-x)}^{2}+ {(4-y)}^{2} }= \sqrt{{(-2-x)}^{2}+ {(3-y)}^{2} }$

${\rightarrow}{(5-x)}^{2}+ {(4-0)}^{2} = {(-2-x)}^{2}+ {(3-0)}^{2}$

${\rightarrow}{x}^{2}+ 25-10x +16= {x}^{2}+4+4x+9$

${\rightarrow} 41-10x = 13+4x$

${\rightarrow} 10x+4x= 41-13$

${\rightarrow} 14x= 28$

$\bold{{\rightarrow} x = 2}$

Hence, required point is

P = (2,0)

Answer 2

Answer:

Step-by-step explanation: since point is on x axis y cordinate is zero

So let required coordinate be (x,y)

Or(x, 0)