Question

find the coordinates on x axis which is equidistant from point - 3, 4 and 2, 5​

Answer 1

Answer:

( 0.4, 0 )

Step-by-step explanation:

As the point it is on the x-axis, its y-coordinate is 0.  Its coordinates are then (x, 0) for some value of x.

The distance from (x, 0) to (-3, 4) is

         √( (diff in x coords)² + (diff in y coords)² )

      = √( ( x - (-3) )² + 4² )

      = √( ( x + 3 )² + 16 )

The distance from (x, 0) to (2, 5) is likewise

         √( ( x - 2 )² + 5² )  =  √( ( x - 2 )² + 25 ).

As (x, 0) is equidistant from (-3, 4) and from (2, 5), these two distances must be equal, so

   √( ( x + 3 )² + 16 )  =  √( ( x - 2 )² + 25 )

⇒ ( x + 3 )² + 16  =  ( x - 2 )² + 25

⇒ ( x + 3 )² - ( x - 2 )²  =  25 - 16

⇒  ( x² + 6x + 9 ) - ( x² - 4x + 4 ) = 9

⇒  10x + 5 = 9

⇒  10x = 4

⇒  x = 0.4

Hope this helps.

Answer 2

AnsWer :

P( 4/10 , 0 )

Solution :

We have, Point.

Let point be,

• A( -3 , 4 )
• B( 2 , 5 )
• P( x , 0 ) which equidistant from A and B
• When, any point equidistant of any point from x - Axis then, y should be 0.

A/Q,

$\tt\dagger \: \: \: \: \: PA = PB.$

$\tt : \implies \sqrt{ { \big(- 3 - x \big)}^{2} + { \big(4 - 0 \big)}^{2} } = \sqrt{ { \big(2 - x \big) + \big(5 - 0 \big)}^{2} }$

$\tt : \implies \sqrt{9 + {x}^{2} - 6x + 16} = \sqrt{4 + {x}^{2} - 8x + 16 }$

$\tt : \implies \sqrt{9 + {x}^{2} + 6x + 16 } = \sqrt{4 + {x}^{2} - 4x + 25 }$

Remove both sides Roots.

$\tt : \implies {x}^{2} + 6x + 25 = {x}^{2} - 4x + 29.$

$\tt : \implies 6x + 4x= 29 - 25.$

$\tt : \implies 10x = 4.$

$\tt : \implies x = \dfrac{4}{10}$

Therefore, the required answer is 4/10.