Question

find the point on x-axis which is equidistant from point (2,-2) (-4,2)​

Answer 1

Given

We have given two points A( 2,-2) & B(-4,2)

To Find

Point on x-axis which is equidistant from the point A and B

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

Let us assume that the point P (x,0) on x-axis which is equidistant from both both points.

If P is equidistant from two points then

PA = PB

Here,we use distance formula to find out the Distance between two points.

D=√(x₂- x₁)²+(y₂- y₁)²

Distance between P and A

P (x,0) & A( 2,-2)

let x be 'a'

x₁=a ;x₂ = 2 ; y₁= 0 & y₂ = -2

Distance (PA)=√(2-a)²+(-2-0)²

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$=>√4+a²-4a+4

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$=>√a²-4a+8

Distance between P and B

P(x,0) & B(-4,2)

x₁=a ;x₂ = -4 ; y₁= 0 & y₂ = 2

Distance (PB)=√(-4-a)²+(2-0)²

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$=>√16+a²+8a+4

=>√a²+8a+20

Now ,PA =PB

squaring both sides to remove root

=>(√a²+8a+20)²=(√a²-4a+8)²

=>a²+8a+20=a²-4a+8

=>8a+20= -4a+8

=>8a+4a=8-20

=>12a=-12

=>a=-1

a=x= -1

Therefore,our point P is (-1,0)

P(-1,0) is the point on x-axis which is equidistant from the point A(2,-2) & B(-4,2)

Answer 2

Let P(x,0) be a point on x axis

PA = PB  PA2

= PB2  (x - 2)2 + (0 + 2)2

= (x + 4)2 + (0 - 2)2  x2 + 4 - 4x + 4

= x2 + 16 + 8x + 4  -4x + 4

= 8x + 16  x

= -1  P(-1,0)