Question

A(1,2), B (2, -3) and C (-2, 3) are three points. A point P moves such that PA²+ PB² = 2PC². Show that the equation to the locus of P is 7x-7y+4=0.
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Answer 1

Let point P be (x,y)

Given A(1,2),B(2,−3),C(−2,3) and

PA

2

+PB

2

=2PC

2

⟹(

(x−1)

2

+(y−2)

2

)

2

+(

(x−2)

2

+(y+3)

2

)

2

=2(

(x+2)

2

+(y−3)

2

)

2

⟹x

2

+1−2x+y

2

+4−4y+x

2

+4−4x+y

2

+9+6y=2(x

2

+4+4x+y

2

+9−6y)

⟹2x

2

+2y

2

−6x+2y+18=2x

2

+2y

2

+8x−12y+26

⟹14x−14y+8=0

⟹7x−7y+4=0

Therefore, the locus of point P is 7x−7y+4=0