Question

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.​

Answer 1

Answer:

$(0, 8\sqrt{3})$  and  $(0, -8\sqrt{3})$ are both 14 units from points (-2, 0) and (2, 0).

Step-by-step explanation:

distance formula

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

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

Point (-2, 0):

$14 = \sqrt{(x - (-2))^2 + (y - 0)^2}$

$\sqrt{(x + 2)^2 + y^2} = 14$

Point (2, 0):

$14 = \sqrt{(x - 2)^2 + (y - 0)^2}$

$\sqrt{(x - 2)^2 + y^2} = 14$

We have a system of equations:

$\sqrt{(x + 2)^2 + y^2} = 14$

$\sqrt{(x - 2)^2 + y^2} = 14$

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

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

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:

$\sqrt{(x + 2)^2 + y^2} = 14$

$\sqrt{(0 + 2)^2 + y^2} = 14$

$\sqrt{4 + y^2} = 14$

Square both sides.

$y^2 + 4 = 196$

$y^2 = 192$

$y = \pm \sqrt{192}$

$y = \pm \sqrt{64 \times 3}$

$y = \pm 8\sqrt{3}$

The points are:

$(0, 8\sqrt{3})$  and  $(0, -8\sqrt{3})$