# Express this in an algebraic equation - the point (x, y) is equidistant from (0, 0) and (4, -2). What is the locus of the point (x, y)?

*print*Print*list*Cite

### 1 Answer

Consider the distance formula between two points. The formula is:

`d = sqrt((x_2-x1)^2 + (y_2 - y_1)^2 )`

Then, let `d_1` the distance between (0,0) and (x,y) and

`d_2` the distance between (4,-2) and (x,y).

Since `d_1` and `d_2` are equidistant , set `d_1` ` ` and `d_2` equal to each other.

`d_1 = d_2`

`sqrt((x-0)^2 + (y-0)^2) = sqrt((x-4)^2 + (y - (-2))^2 )`

`sqrt(x^2+y^2) = sqrt((x-4)^2 + (y+2)^2)`

Then, square both sides.

`x^2 + y^2 = (x-4)^2 + (y+2)^2`

Expand right side. Use FOIL method.

`x^2 + y^2 = x^2 - 4x - 4x + 16 + y^2 + 2y + 2y + 4`

`x^2 + y^2 = x^2 - 8x + 16 + y^2 + 4y + 4`

`x^2 + y^2 = x^2 - 8x + y^2 + 4y + 20`

Move all the terms with exponent 2 on one side of the equation.

`x^2 + y^2 - x^2 - y^2 = -8x + 4y + 20`

` 0 = -8x +4y + 20`

Since -8, 4 and 20 are all divisible by 4, then divide both sides by 4 to simplify.

`0 = -2x + y + 5`

Then, we may express the above expression in slope intercept form.

`y = 2x - 5`

**Hence, the locus of the point is y = 2x - 5.**