Answer:
[tex] (0, 8\sqrt{3}) [/tex] and [tex] (0, -8\sqrt{3}) [/tex] are both 14 units from points (-2, 0) and (2, 0).
Step-by-step explanation:
distance formula
[tex] d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} [/tex]
We want the distance, d, from points (-2, 0) and (2, 0) to be 14.
Point (-2, 0):
[tex] 14 = \sqrt{(x - (-2))^2 + (y - 0)^2} [/tex]
[tex] \sqrt{(x + 2)^2 + y^2} = 14 [/tex]
Point (2, 0):
[tex] 14 = \sqrt{(x - 2)^2 + (y - 0)^2} [/tex]
[tex] \sqrt{(x - 2)^2 + y^2} = 14 [/tex]
We have a system of equations:
[tex] \sqrt{(x + 2)^2 + y^2} = 14 [/tex]
[tex] \sqrt{(x - 2)^2 + y^2} = 14 [/tex]
Since the right sides of both equations are equal, we set the left sides equal.
[tex] \sqrt{(x + 2)^2 + y^2} = \sqrt{(x - 2)^2 + y^2} [/tex]
Square both sides:
[tex] (x + 2)^2 + y^2 = (x - 2)^2 + y^2 [/tex]
Square the binomials and combine like terms.
[tex] x^2 + 4x + 4 + y^2 = x^2 - 4x + 4 + y^2 [/tex]
[tex] 4x = -4x [/tex]
[tex] 8x = 0 [/tex]
[tex] x = 0 [/tex]
Now we substitute x = 0 in the first equation of the system of equations:
[tex] \sqrt{(x + 2)^2 + y^2} = 14 [/tex]
[tex] \sqrt{(0 + 2)^2 + y^2} = 14 [/tex]
[tex] \sqrt{4 + y^2} = 14 [/tex]
Square both sides.
[tex] y^2 + 4 = 196 [/tex]
[tex] y^2 = 192 [/tex]
[tex] y = \pm \sqrt{192} [/tex]
[tex] y = \pm \sqrt{64 \times 3} [/tex]
[tex] y = \pm 8\sqrt{3} [/tex]
The points are:
[tex] (0, 8\sqrt{3}) [/tex] and [tex] (0, -8\sqrt{3}) [/tex]
