Question

Find a pointon x-axis which is
equidistant from A (2,-5) and B(-2,9)

Answer 1

Answer

Required point is (-7,0)

Solution

Let assume that the point R(x,0) is equidistant from A(2,-5) and B(-2,9).

So , it means that

AR = BR

Firstly finding distance of AR -

By distance formula :-

$ar \: = \sqrt{( {x2 - x1)}^{2} + ( {y2 - y1)}^{2} }$

Here x1 = x ; x2 = 2 ; y1 = 0 ; y2 = -5 .

$= > \sqrt{( {2 - x)}^{2} + ( { - 5 - 0) }^{2} }$

As we know that ( a-b)² → a²+b²-2ab

$= > \sqrt{4 + {x}^{2} - 4x + 25}$

Now finding distance of BR :-

x1 = x ; x2= -2 ; y1= 0 ; y2 = 9

$\sqrt{ {( - 2 - x)}^{2} + {9 - 0)}^{2} }$

Now (-a-b)² = (a+b)² → a²+b²+2ab

$= > \sqrt{ {x}^{2} - 4x + 29}$

Now finding the distance of BR

Here x1 = x ; x2 = -2 , y1 = 0 , y2 = 9

$\sqrt{ {( - 2 - x)}^{2} + ( {9 - 0)}^{2} }$

As we know that(-a-b)² =( a+b)²

$= > \sqrt{4 + {x}^{2} + 4x + 81 }$

$= > \sqrt{ {x}^{2} + 4x + 85 }$

So if AR = BR then

AR² = BR ² , So

x² -4x +29 = x² +4x +85

-8x. = 85-29

x. = - (56/8)

x. = -7

So the coordinates of P are (-7,0)