find an equation in x and y such that the distance between (x,y) and (-2,0) is twice the distance between (x,y) and (3,1)
Translating your condition gives me
√((x+2)^2 + y^2) = 2√(x-3)^2 + (y-1)^2)
square both sides
(x+2)^2 + y^2 = 4((x-3)^2 + (y-1)^2)
x^2 + 4x + 4 + y^2 = 4(x^2 - 6x + 9) + y^2 - 2y + 1)
expanding, collecting like terms and simplifying I got
3x^2 + 3y^2 + 20x - 8y = -36
which has the form of a circle, so I completed the square to end up with
(x + 10/3)^2 + (y - 4/3)^2 = 224/9
To find an equation in x and y such that the distance between (x, y) and (-2, 0) is twice the distance between (x, y) and (3, 1), we can use the distance formula.
The distance formula between two points (x1, y1) and (x2, y2) is:
√((x2 - x1)² + (y2 - y1)²)
Let the coordinates of (x, y) be (x, y). The distance between (x, y) and (-2, 0) is:
√((x - (-2))² + (y - 0)²)
The distance between (x, y) and (3, 1) is:
√((x - 3)² + (y - 1)²)
According to the given condition, the distance between (x, y) and (-2, 0) is twice the distance between (x, y) and (3, 1):
√((x - (-2))² + (y - 0)²) = 2 * √((x - 3)² + (y - 1)²)
Simplifying this equation will give us the desired equation in x and y.
To find the equation in terms of x and y that satisfies the given condition, we need to use the distance formula.
The distance formula between two points (x1, y1) and (x2, y2) is given by:
d = √((x2 - x1)^2 + (y2 - y1)^2)
Let's apply this formula to find the distance between (x, y) and (-2, 0) as well as the distance between (x, y) and (3, 1) and set up the equation.
Distance between (x, y) and (-2, 0):
d1 = √((x - (-2))^2 + (y - 0)^2)
= √((x + 2)^2 + y^2)
Distance between (x, y) and (3, 1):
d2 = √((x - 3)^2 + (y - 1)^2)
According to the problem, the distance between (x, y) and (-2, 0) is twice the distance between (x, y) and (3, 1). Mathematically, we can express this as:
2d2 = d1
Replacing the expressions for d1 and d2, we get:
2√((x - 3)^2 + (y - 1)^2) = √((x + 2)^2 + y^2)
Now, we can square both sides of the equation to eliminate the square roots:
4((x - 3)^2 + (y - 1)^2) = (x + 2)^2 + y^2
Expanding and simplifying:
4(x^2 - 6x + 9 + y^2 - 2y + 1) = x^2 + 4x + 4 + y^2
Simplifying further:
4x^2 - 24x + 36 + 4y^2 - 8y + 4 = x^2 + 4x + 4 + y^2
Rearranging terms:
3x^2 - 28x + 4y^2 - 8y + 36 = 0
Hence, the equation in terms of x and y is 3x^2 - 28x + 4y^2 - 8y + 36 = 0.
I assume you want to find a point (x, y) that fits the conditions.
Find the distance between (-2,0) and (3,1) but keep it in x,y form.
(3-(-2),1-0) = (5,1)
It may be helpful now for you to draw a graph to see whether to add (5,1) to (3,1) or subtract (5,1) from (-2,0).
However, because the distance between (x,y) and (-2,0) is twice the distance between (x,y) and (3,1), you add it to (3,1). (This is why a graph is helpful.)
So, add the distance (5,1) to (3,1).
(3+5, 1+1) = (8,2)