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 bcus at P is 7x-7y+4=0

Answer 1

Step-by-step explanation:

We have,

$A\equiv(1, 2) ;B\equiv(2, -3) ; C\equiv(-2, 3)$

$Let \: \:the\: \: coordinates \: \: of\:\:P\: \: be \equiv(h, k)$

By the given condition,

$(PA)^{2} + (PB)^{2} = 2(PC)^{2}$

$\implies(h - 1)^{2} + {(k - 2)}^{2} + {(h - 2)}^{2} + {(k + 3)}^{2} = 2 {(h + 2)}^{2} + 2 {(k - 3)}^{2}$

$\implies2 {h}^{2} + 2 {k}^{2} - 6h + 2k + 18 = 2 {h}^{2} + 2 {k}^{2} + 8h - 12k + 26$

$\implies - 6h + 2k + 18 = 8h - 12k + 26$

$\implies14h - 14k + 8 = 0$

$\implies7h - 7k + 4 = 0$

Hence, the required locus of the point is 7x - 7y +4 =0