Question

find the equation of locus of point p such
that AP²+ BP²= 50 if A=(3, 1) and B=(4,-5)​

Answer 1

Answer:

Step-by-step explanation:

Given,

Co-ordinates of :-

A = ( 3 , 1 )

B = ( 4 , - 5)

Let , the co-ordinates of 'P' be :-

P = (x , y)

AP²+ BP²= 50

to Find :-

Equation of the locus .

Solution :-

Distance formula :-

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

According to AP :-

(3 , 1 ) = (x_1 , y_1)

(x , y) = (x_2 , y_2)

$(AP)^2=\bigg(\sqrt{(x-3)^2+(y-1)^2}\bigg)^2$

$= x^2-6x+9+y^2-2y+1\\\\= x^2+y^2-6x-2y+10$

According to BP :-

(4 , -5) = (x_1 , y_1)

(x , y) = (x_2 , y_2)

$(BP)^2=\bigg(\sqrt{(x-4)^2+(y+5)^2}\bigg)^2$

$= x^2 - 8x + 16 + y^2 + 10y + 25\\\\= x^2+y^2-8x+10y+41$

Substituting in " AP²+ BP²= 50"

$x^2+y^2-6x-2y+10+ x^2+y^2-8x+10y+41=50$

$2x^2+2y^2-14x+8y+51=50$

$2x^2+2y^2-14x+8y+1 = 0$