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find the value of x and y if the distance of the point (x,y) from (-2,0) and (2,0) are both 14 units.​

1 Answer

1 vote

Answer:


(0, 8√(3)) and
(0, -8√(3)) are both 14 units from points (-2, 0) and (2, 0).

Explanation:

distance formula


d = √((x_2 - x_1)^2 + (y_2 - y_1)^2)

We want the distance, d, from points (-2, 0) and (2, 0) to be 14.

Point (-2, 0):


14 = √((x - (-2))^2 + (y - 0)^2)


√((x + 2)^2 + y^2) = 14

Point (2, 0):


14 = √((x - 2)^2 + (y - 0)^2)


√((x - 2)^2 + y^2) = 14

We have a system of equations:


√((x + 2)^2 + y^2) = 14


√((x - 2)^2 + y^2) = 14

Since the right sides of both equations are equal, we set the left sides equal.


√((x + 2)^2 + y^2) = √((x - 2)^2 + y^2)

Square both sides:


(x + 2)^2 + y^2 = (x - 2)^2 + y^2

Square the binomials and combine like terms.


x^2 + 4x + 4 + y^2 = x^2 - 4x + 4 + y^2


4x = -4x


8x = 0


x = 0

Now we substitute x = 0 in the first equation of the system of equations:


√((x + 2)^2 + y^2) = 14


√((0 + 2)^2 + y^2) = 14


√(4 + y^2) = 14

Square both sides.


y^2 + 4 = 196


y^2 = 192


y = \pm √(192)


y = \pm √(64 * 3)


y = \pm 8√(3)

The points are:


(0, 8√(3)) and
(0, -8√(3))

User Carlos Rodrigez
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