Find the equation of the set of all points P(x,y) that is twice as far from (-8,8) as from (-2,2).
You clearly have a split personality and can't even tell that you keep posting the same problems again and again.
To find the equation of the set of all points P(x, y) that are twice as far from (-8, 8) as from (-2, 2), we can use the distance formula.
The distance between two points (x1, y1) and (x2, y2) is given by the formula:
d = √((x2 - x1)^2 + (y2 - y1)^2)
Let P(x, y) be the point we want to find. We know that the distance between P and (-8, 8) is twice the distance between P and (-2, 2).
Let's denote the distance between P and (-8, 8) as d1 and the distance between P and (-2, 2) as d2.
We can set up the following equation: d1 = 2 * d2
Using the distance formula, we have:
d1 = √((x - (-8))^2 + (y - 8)^2) = √((x + 8)^2 + (y - 8)^2)
d2 = √((x - (-2))^2 + (y - 2)^2) = √((x + 2)^2 + (y - 2)^2)
Now we can substitute these expressions into the equation d1 = 2 * d2:
√((x + 8)^2 + (y - 8)^2) = 2 * √((x + 2)^2 + (y - 2)^2)
To eliminate the square roots, we can square both sides of the equation:
((x + 8)^2 + (y - 8)^2) = 4 * ((x + 2)^2 + (y - 2)^2)
Expanding and simplifying the equation:
x^2 + 16x + 64 + y^2 - 16y + 64 = 4x^2 + 16x + 16 + 4y^2 - 16y + 16
Now, let's bring all terms to one side of the equation:
3x^2 + 3y^2 - 94 = 0
Therefore, the equation of the set of all points P(x, y) that are twice as far from (-8, 8) as from (-2, 2) is 3x^2 + 3y^2 - 94 = 0.
To find the equation of the set of all points P(x, y) that are twice as far from (-8, 8) as from (-2, 2), we can use the distance formula.
Let's denote the distance between P and (-8, 8) as d1 and the distance between P and (-2, 2) as d2.
According to the problem, the distance between P and (-8, 8) is twice the distance between P and (-2, 2). Mathematically, this can be expressed as:
d1 = 2 * d2
Using the distance formula for two points (x1, y1) and (x2, y2), the distance can be calculated as:
d = square root of [(x2 - x1)^2 + (y2 - y1)^2]
For the points (-8, 8) and (-2, 2), we have:
d1 = square root of [(-2 - (-8))^2 + (2 - 8)^2]
= square root of (6^2 + (-6)^2)
= square root of (36 + 36)
= square root of 72
Similarly, for the points (-2, 2) and (x, y), we have:
d2 = square root of [(x - (-2))^2 + (y - 2)^2]
= square root of (x^2 + 4 + (y - 2)^2)
Since d1 = 2 * d2, we substitute these values and solve for x and y:
square root of 72 = 2 * square root of (x^2 + 4 + (y - 2)^2)
Squaring both sides of the equation, we get:
72 = 4(x^2 + 4 + (y - 2)^2)
Expanding and simplifying, we have:
18 = x^2 + (y - 2)^2
Therefore, the equation of the set of all points P(x, y), which are twice as far from (-8, 8) as from (-2, 2), is:
x^2 + (y - 2)^2 = 18