Find the locus of a point which is equidistant from the origin and the point (-2,5)
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.
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.
draw the grid
To find the locus of a point that is equidistant from the origin and the point (-2,5), we need to find the set of all points that have the same distance from both.
Let's call the unknown point (x, y). The distance between (x, y) and the origin (0, 0) is given by the distance formula:
d₁ = √((x - 0)² + (y - 0)²)
= √(x² + y²)
Similarly, the distance between (x, y) and the point (-2, 5) is given by:
d₂ = √((x - (-2))² + (y - 5)²)
= √((x + 2)² + (y - 5)²)
Since we want the distance from (x, y) to the origin and (x, y) to (-2, 5) to be equal, we can set up the following equation:
d₁ = d₂
√(x² + y²) = √((x + 2)² + (y - 5)²)
To simplify the equation, we can square both sides:
(x² + y²) = (x + 2)² + (y - 5)²
Expanding and simplifying further:
x² + y² = x² + 4x + 4 + y² - 10y + 25
Rearranging terms:
4x - 10y + 29 = 0
This is the simplified equation of the locus. It represents a straight line in the coordinate plane. Hence, the locus of points equidistant from the origin and (-2, 5) is a straight line with the equation 4x - 10y + 29 = 0.