Question

The point on y axis equidistant from (-3,4) and (7,6) is...plz anybody do it.
And its answer is (15,0)​

Answer 1

Answer:

Let the required point be (x,0)

Since, (x,0) is equidistant from the points (−3,4) and (2,5)

∴

(−3−x)

2

+(4−0)

2

=

(2−x)

2

+(5−0)

2

[ Using distance formula ]

⇒sqrt9+x

2

+6x+16=

4+x

2

−4x+25

⇒x

2

+6x+25=x

2

−4x+29

⇒10x=4

⇒x=

10

4

=

5

2

∴ Required point is (

5

2

,0)