A moving point is always equidistant from (-3,2) and the line 2x-y+2=0. What is the equation of its locus?
how do you solve this?
Sounds like the definition of a parabola: The locus of points equidistant from a point (the focus) and a line (the directrix). It will be a tilted parabola.
Good luck. Good exercise for you.
To solve this problem, we first need to understand the concept of equidistance.
A point is said to be equidistant from two objects when its distance to each object is the same. In this case, the moving point is equidistant from the given point (-3,2) and the line 2x-y+2=0.
Let's break down the problem into two steps:
Step 1: Find the distance formula
The distance formula between two points (x1, y1) and (x2, y2) is given by:
d = √((x2 - x1)^2 + (y2 - y1)^2)
In this case, we need to find the distance between the moving point (x, y) and the given point (-3, 2). Therefore, the distance formula becomes:
d1 = √((x - (-3))^2 + (y - 2)^2)
Step 2: Find the distance between the moving point and the given line
The distance between a point (x1, y1) and a line Ax + By + C = 0 is given by:
d = |Ax1 + By1 + C| / √(A^2 + B^2)
In this case, the given line is 2x - y + 2 = 0. Thus, the distance between the moving point (x, y) and the line is:
d2 = |2x - y + 2| / √(2^2 + (-1)^2)
Since the moving point is equidistant from the given point and line, we can equate the two distances:
√((x - (-3))^2 + (y - 2)^2) = |2x - y + 2| / √5
Now let's simplify the equation further:
((x + 3)^2 + (y - 2)^2) = (2x - y + 2)^2 / 5
Expanding both sides of the equation:
(x^2 + 6x + 9 + y^2 - 4y + 4) = (4x^2 + y^2 + 4 + 4xy - 4x - 2y + 4) / 5
Simplifying:
5x^2 + 30x + 45 + 5y^2 - 20y + 20 = 4x^2 + y^2 + 4 + 4xy - 4x - 2y + 4
Bringing all terms to one side, we get the equation of the locus:
x^2 + 2xy - 6x - 3y + 6 = 0
Therefore, the equation of the locus of the moving point is x^2 + 2xy - 6x - 3y + 6 = 0.