Student Question

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)?

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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.

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial