First things first it will definately be useful to make a quick sketch of the x-y axes, the two points, and roughtly what the perpendicular bisector will look like.

This kind of question can be made much more painless if you make the following observation: for any point (x,y) on the perpendicular bisector (lets call this P) the distance from (x,y) to (0,2) and the distance from (x,y) to (1,0) are equal.

Let (x,y) lie on P. Using Pythagoras' Theorem the distance (x,y) to (0,2) is sqrt[(x - 0)² + (y - 2)²]. Similairly the distance from (x,y) to (1,0) is sqrt[(x - 1)² + (y - 0)²]. These two distances must then be equal so:

sqrt[(x - 0)² + (y - 2)²] = sqrt[(x - 1)² + (y - 0)²] , now squaring gives:

(x - 0)² + (y - 2)² = (x - 1)² + (y - 0)² , then simplifying gives:

x² + (y - 2)² = (x - 1)² + y² , expanding the brackets gives:

x² + y² - 4y +4 = x²- 2x +1 + y² , then cancelling the x² and y² terms:

-4y + 4 = -2x + 1 , which rearranges to give the line:

y = x/2 + 3/4

There was nothing special here about finding the perpendicular bisector of (0,2) and (1,0), the exact some method can be used for any two points in the plane. The only thing that must be altered is our two expressions for the distances from (x,y) to the first point and the distance form (x,y) to the second.

Factorise and solve x^2 -8x=15=0

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