Question
Find the locus of a point which is equidistant from the origin and the point (-2,5)
Answers
Steve
the locus is the perpendicular bisector of the line from (0,0) to (-2,5).
It's easy to find the midpoint, and the slope of the line.
Now just find the line with perpendicular slope, through the midpoint.
It's easy to find the midpoint, and the slope of the line.
Now just find the line with perpendicular slope, through the midpoint.
Reiny
Let a point on the locus be P (x,y)
then
√(x^2 + y^2) = √((x+2)^2 + (y-5)^2)
square both sides:
x^2 + y^2 = (x+2)^2 + (y-5)^2
x^2 + y^2 = x^2 + 4x + 4 + y^2 - 10y + 25
4x - 10y = -29
which is the perpendicular bisector of the line joining (0,0) and (-2,5) , as analysed by Steve above.
then
√(x^2 + y^2) = √((x+2)^2 + (y-5)^2)
square both sides:
x^2 + y^2 = (x+2)^2 + (y-5)^2
x^2 + y^2 = x^2 + 4x + 4 + y^2 - 10y + 25
4x - 10y = -29
which is the perpendicular bisector of the line joining (0,0) and (-2,5) , as analysed by Steve above.
Anonymous
draw the grid