SOLUTION: A curve is traced by a point (x,y) which moves such that its distance from the point A(-1,1) is three times its distance from the point B(2,-1). Determine the equation of the curve

Algebra ->  Graphs -> SOLUTION: A curve is traced by a point (x,y) which moves such that its distance from the point A(-1,1) is three times its distance from the point B(2,-1). Determine the equation of the curve      Log On


   

Question 1156566: A curve is traced by a point (x,y) which moves such that its distance from the point A(-1,1) is three times its distance from the point B(2,-1). Determine the equation of the curve.
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The distance from (x,y) to A(-1,1) is sqrt%28%28x%2B1%29%5E2%2B%28y-1%29%5E2%29.

The distance from (x,y) to B(2,-1) is sqrt%28%28x-2%29%5E2%2B%28y%2B1%29%5E2%29.

The distance from A is 3 times the distance from B:

sqrt%28%28x%2B1%29%5E2%2B%28y-1%29%5E2%29+=+3%28sqrt%28%28x-2%29%5E2%2B%28y%2B1%29%5E2%29%29

The square of the distance from A is 9 times the square of the distance from B:

%28x%2B1%29%5E2%2B%28y-1%29%5E2+=+9%28%28x-2%29%5E2%2B%28y%2B1%29%5E2%29

I started down that path, and the numbers got ugly; so I decided to try another path to the answer.

Without doing any detailed calculations with the above equation, we can see that the equation is going to be the equation of a circle (there are going to be x^2 and y^2 terms with the same coefficient).

There are two points on the line containing A and B that satisfy the condition that the distance from A is 3 times the distance from B. Those two points will be the endpoints of a diameter of the circle.

The point between A and B is 3/4 of the way from A to B; that is (5/4,-1/2).

The point to the right of B is such that its distance from B is twice the distance from A to B; that point is (7/2,-2).

The center of the circle is the midpoint of that diameter: (19/8,-5/4).

The equation is of the form

%28x-19%2F8%29%5E2%2B%28y%2B5%2F4%29%5E2+=+r%5E2

The square of the radius of the circle is the square of the distance from the center to either endpoint of the diameter: %289%2F8%29%5E2%2B%283%2F4%29%5E2+=+117%2F64

The equation is

%28x-19%2F8%29%5E2%2B%28y%2B5%2F4%29%5E2+=+117%2F64