# Deriving the equation

• Sep 9th 2007, 08:21 AM
Mr_Green
Deriving the equation
Derive the equation (general form) of all points that are equidistant from the point F(-6,3) and the line y = -5.

Thanks
• Sep 9th 2007, 08:23 AM
ThePerfectHacker
Quote:

Originally Posted by Mr_Green
Derive the equation (general form) of all points that are equidistant from the point F(-6,3) and the line y = -5.

Thanks

Let $(x,y)$ be such a point. Then what is that distance from this point to $(6,3)$? What is the distance from this point to $y=-5$? Now equate these two quantities and you have your equation.
• Sep 9th 2007, 08:30 AM
Mr_Green
sqrt[(x+6)^2 + (y-3)^2] = sqrt[(0^2) + (y+5)^2]

square both sides

x^2 + 12x + 36 + y^2 - 6y + 9 = y^2 + 10y +25

x^2 + 12x -16y -11 = 0

Is this correct?
• Sep 9th 2007, 09:07 AM
Jhevon
Quote:

Originally Posted by Mr_Green
sqrt[(x+6)^2 + (y-3)^2] = sqrt[(0^2) + (y+5)^2]

square both sides

x^2 + 12x + 36 + y^2 - 6y + 9 = y^2 + 10y +25

x^2 + 12x -16y -11 = 0

Is this correct?

the distance of (-6,3) to the line y = -5 is the distance between the points (-6,3) and (-6,-5), you don't really need the formula for this, but you can use it if you want to (we are looking for the shortest distance here, which is a vertical line, so the x's are the same)
• Sep 9th 2007, 09:27 AM
Mr_Green
its correct though?
• Sep 9th 2007, 09:31 AM
Jhevon
Quote:

Originally Posted by Mr_Green
its correct though?

yes, how you expanded before was correct. it's just that the right side should be a number. namely the square of the number i asked you to calculate