Question

A point on the X− axis which is equidistant from the points (7,6) and (3,4) is​

Answer 1

Answer:

Point is ( 3 , 0 )

Step-by-step explanation:

Let the point on the x axis be ( x , 0 )

Distance between ( x , 0 ) and ( 7 , 6 ) =

$- > \sqrt{ {(7 - x)}^{2} + {(6 - 0)}^{2} }$

$- > \sqrt{ {7}^{2} + {x}^{2} - 14x + 36 }$

$- > \sqrt{ {x}^{2} - 14x + 85}$

Distance between ( x , 0 ) and ( 3 , 4 ) =

$- > \sqrt{ {(3 - x)}^{2} + {(4 - 0)}^{2} }$

$- > \sqrt{ {3}^{2} + {x}^{2} + 6x + 16 }$

$- > \sqrt{ {x}^{2} + 6x + 25}$

As the point ( x , 0 ) is eqidistant from the two points , both the distances claculated are equal .

$- > \sqrt{ {x}^{2} - 14x + 85 } = \sqrt{ {x}^{2} + 6x + 25 }$

${x}^{2} - 14x + 85 = {x}^{2} + 6x + 25$

$- > 85 - 25 = 6 x + 14x$

$- > 60 = 20x$

$- > x = 3$

Thus point is ( 3 , 0 )

I hope it helps u dear friend ^_^♡♡