Question

find the coordinates of the point equidistant from three given points A (5,1) , B(-3,-7) and C(7,-1).

Answer 1

if points are equidistant then the length is same

Answer 2

Answer:

The coordinates of the point equidistant from three given points A (5,1) , B(-3,-7) and C(7,-1) is (2,-4)

Step-by-step explanation:

Let the point which is equidistant from A , B and C is P

Let the coordinates of P be (x,y)

A (5,1)

B(-3,-7)

C(7,-1)

Find the distance between Point P and A

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

$A=(x_1,y_1)= (5,1)$

$P=(x_2,y_2)= (x,y)$

$PA=\sqrt{(x-5)^2+(y-1)^2}$

Find the distance between Point P and B

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

$B=(x_1,y_1)= (-3,-7)$

$P=(x_2,y_2)= (x,y)$

$PB=\sqrt{(x+3)^2+(y+7)^2}$

Find the distance between Point P and C

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

$C=(x_1,y_1)= (7,-1)$

$P=(x_2,y_2)= (x,y)$

$PC=\sqrt{(x-7)^2+(y+1)^2}$

Since we are given that P  is equidistant from A , B and C

So. PA = PB =PC

$\sqrt{(x-5)^2+(y-1)^2}=\sqrt{(x+3)^2+(y+7)^2}=\sqrt{(x-7)^2+(y+1)^2}$

PA = PB

$\sqrt{(x-5)^2+(y-1)^2}=\sqrt{(x+3)^2+(y+7)^2}$

$(x-5)^2+(y-1)^2=(x+3)^2+(y+7)^2$

$x^2+25-10x+y^2+1-2y=x^2+9+6x+y^2+49+14y$

$26-10x-2y=58+6x+14y$

$16x+16y=-32$

$x+y=-2$ ---1

PB = PC

$\sqrt{(x+3)^2+(y+7)^2}=\sqrt{(x-7)^2+(y+1)^2}$

$(x+3)^2+(y+7)^2=(x-7)^2+(y+1)^2$

$x^2+9+6x+y^2+49+14y=x^2+49-14x+y^2+1+2y$

$6x+58+14y=-14x+50+2y$

$20x+12y=-8$

$5x+3y=-2$ ---2

Solve 1 and 2

Substitute value of x from 1 in 2

$5(-2-y)+3y=-2$

$-10-5y+3y=-2$

$-10-2y=-2$

$-2y=8$

$y=-4$

Substitute the value of y in 1

$x+(-4)=-2$

$x=2$

So, the coordinates of P is (2,-4)

Hence the coordinates of the point equidistant from three given points A (5,1) , B(-3,-7) and C(7,-1) is (2,-4)