(-1;-1) is equidistant from (0;2) and (x;-2)

Answer 1

To find the value of x, we can use the distance formula.

The distance between two points (x1, y1) and (x2, y2) is given by:

√[(x2 - x1)^2 + (y2 - y1)^2]

In this case, we have the points (0, 2) and (-1, -1). The distance between (-1, -1) and (0, 2) is equal to the distance between (-1, -1) and (x, -2).

So, we can set up the equation:

√[(-1 - 0)^2 + (-1 - 2)^2] = √[(-1 - x)^2 + (-1 - (-2))^2]

Simplifying both sides:

√[1 + 9] = √[(-1 - x)^2 + 1]

√[10] = √[(1 + x)^2 + 1]

Squaring both sides:

10 = (1 + x)^2 + 1

9 = (1 + x)^2

Taking the square root of both sides:

±3 = 1 + x

x = -1 ± 3

So, x can be either -4 or 2.

Therefore, the point (-1, -1) is equidistant from (0, 2) and either (-4, -2) or (2, -2).