Question

find the coordinates for the point equidistant from (-1,5) (8,2) (6,-2)

Answer 1

Answer:

Step-by-step explanation:

Answer 2

The given question is we have to find the coordinates for the point equidistant from (-1,5) (8,2) (6,-2).

let the points be

P(x, y) A(-1,5), B(8,2), C(6,-2).

The P is A, B, and C.

let us find the square value of PA, PB and PC

${pa}^{2} = {(x + 1)}^{2} + {(y - 5)}^{2} \\ = {x}^{2} + {y}^{2} + 2x - 10y + 26$

The value of PB is

${pb}^{2} = {(x - 8)}^{2} + {(y - 2)}^{2} \\ = {x}^{2} + {y}^{2} - 16x - 4y + 68$

The value of a PC is

${(x - 6)}^{2} + {(y + 2)}^{2} \\ = {x}^{2} + {y}^{2} - 12x + 4y + 40$

The P is equidistant from points A, B, and C.

PA=PB=PC.

${pa}^{2} = {pb}^{2} = {pc}^{2}$

equalise the value of the square of PA and square of PB, we get

${ pa}^{2} = {pb}^{2} \\ {x}^{2} + {y}^{2} + 2x - 10y + 26 \\ = {x}^{2} + {y}^{2} - 16x - 4y + 68$

Equalise the value of the square of PA and the square of PC, and we get

${pa}^{2} = {pc}^{2} \\ {x}^{2} + {y}^{2} + 2x - 10y + 26 = {x}^{2} + {y}^{2} - 12x + 4y + 40$

On simplifying the above expressions we get the below value as

$18x - 6y = 42 \\ 14x - 14y = 14$

on further simplification of the above equation we get,

above equation we get, $3x - y = 7 \\ x - y = 1$

above equation we get, $3x - y = 7 \\ x - y = 1$on subtracting these two equations by the elimination method we get,

above equation we get, $3x - y = 7 \\ x - y = 1$on subtracting these two equations by the elimination method we get,$2x = 6 \\ 2y = 4$

above equation we get, $3x - y = 7 \\ x - y = 1$on subtracting these two equations by the elimination method we get,$2x = 6 \\ 2y = 4$Therefore, the value of x and y will be

above equation we get, $3x - y = 7 \\ x - y = 1$on subtracting these two equations by the elimination method we get,$2x = 6 \\ 2y = 4$Therefore, the value of x and y will be $x = 3 \: \: and \: \: y = 2$

above equation we get, $3x - y = 7 \\ x - y = 1$on subtracting these two equations by the elimination method we get,$2x = 6 \\ 2y = 4$Therefore, the value of x and y will be $x = 3 \: \: and \: \: y = 2$Hence, point p is (3,2).

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