Question

If A =(1, 0)B=(-1,0)and C=(2, 0)then the locus of the point P such that PA ^2+PB ^2=2PC^2 is a

If you know then only answer or else it will be reported

Please answer soon. ​

Answer 1

Step-by-step explanation:

Given :-

A =(1, 0)

B=(-1,0)

and C=(2, 0)

To find :-

Find the locus of the point P such that

PA^2+PB^2=2PC^2 ?

Solution :-

Let the point of the locus be P(x,y)

Given points are A (1,0) B(-1,0) and C (2,0)

Finding PA^2:-

Let (x1, y1) = P(x,y)=>x1 = x and y1 = y

Let (x2, y2) = A(1,0)=>x2 = 1 and y2 = 0

We know that

The distance between two points (x1, y1) and

( x2, y2) is √[(x2-x1)^2+(y2-y1)^2] units

=> PA =√[(1-x)^2+(0-y)^2]

=>PA =√[(1-x)^2+y^2]

=> PA =√(1+x^2-2x+y^2)

On squaring both sides then

=> PA^2 =[ √(1+x^2-2x+y^2) ]^2

Since (a-b)^2=a^2-2ab+b^2

PA^2 = (1+x^2-2x+y^2) ----------------(1)

Finding PB^2:-

Let (x1, y1) = P(x,y)=>x1 = x and y1 = y

Let (x2, y2) = B(-1,0)=>x2 = -1 and y2 = 0

We know that

The distance between two points (x1, y1) and

( x2, y2) is √[(x2-x1)^2+(y2-y1)^2] units

=> PB =√[(-1-x)^2+(0-y)^2]

=>PB =√[(1+x)^2+y^2]

=> PB =√(1+x^2+2x+y^2)

Since (a+b)^2=a^2+2ab+b^2

On squaring both sides then

=> PB ^2 =[ √(1+x^2+2x+y^2) ]^2

PB^2 = (1+x^2+2x+y^2) ----------------(2)

Finding PC^2:-

Let (x1, y1) = P(x,y)=>x1 = x and y1 = y

Let (x2, y2) = C(2,0)=>x2 = 2 and y2 = 0

We know that

The distance between two points (x1, y1) and

( x2, y2) is √[(x2-x1)^2+(y2-y1)^2] units

=> PC =√[(2-x)^2+(0-y)^2]

=>PC =√[(2-x)^2+y^2]

=> PC=√(4+x^2-4x+y^2)

Since (a-b)^2=a^2-2ab+b^2

On squaring both sides then

=> PC ^2 =[ √(4+x^2-4x+y^2) ]^2

PC^2 = (4+x^2-4x+y^2) ----------------(3)

Given that

PA^2 + PB^2 = 2PC^2

From (1),(2)&(3)

(1+x^2-2x+y^2) + (1+x^2+2x+y^2) = 2(4+x^2-4x+y^2)

=> 1+x^2-2x+y^2+1+x^2+2x+y^2=8+2x^2-8x+2y^2

=> 2+2x^2+2y^2 = 8+2x^2-8x+2y^2

=> 2+2x^2+2y^2 -8-2x^2+8x-2y^2 = 0

=> (2-8)+(2x^2-2x^2)+(2y^2-2y^2)+8x = 0

=> (-6)+0+0+8x = 0

=> -6+8x = 0

=> 8x = 6

=> x = 6/8

=> x = 3/4

The value of x = 3/4

Since the given points are A ,B and C lies on X-axis ,because the coordinate of y is zero.

So, the point of locus = (x,0)

=> P(x,0) = (3/4,0)

Answer:-

The required locus of the point is P(3/4,0)

Used formulae:-

Distance formula:-

The distance between two points (x1, y1) and( x2, y2) is √[(x2-x1)^2+(y2-y1)^2] units

• (a-b)^2=a^2-2ab+b^2

• (a+b)^2=a^2+2ab+b^2