Question

find the point on x axis which is equidistant from the points(7, 6) &(-3, 4).

Answer 1

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

Assumption

A(7, 6) , B(-3, 4) are equidistant from P(x,0)

So,

AP = BP

Also here

A(7,6) , P(x,0) , B(-3,4)

As we know that,

Distance Formula = √(x2 - x1)² + (y2 - y1)²

So,

AP =√(x - 7)² + (0 - 6)²

AP = √( x - 7)² + (6)²

As we know that,

(a - b)² = a² + b² - 2ab

So,

= √{x² + 7² - 2 × x × x 7 + (6)²}

AP = √x² + 49 - 14x + 36

AP = √x² - 14 x + 85 ....... (1)

Now,

BP = √[{x - (-3)}² + (0 - 4)²]

BP = √(x + 3 )² + (-4)²

BP = √x² + 9 + 6 x + 16

BP = √x² + 6x + 25 ...... (2)

Here

AP = BP (Equidistant)

√x² - 14x + 85 = √x² + 6x + 25

Now

Using (1) and (2)

Squaring both sides we get,

x² - 14 x + 85 = x² + 6x + 25

x² - x² - 14 x - 6x = 25 - 85

-20x = -60

x = 60/20

x = 3

Therefore ,

Point on x-axis = (3 , 0)