Find the equation of set of all points P(x,y) that is twice as far from (-8,8) as from (-2,2).
Name your points A(-8,8) and B(-2,2)
So you want PA = 2PB
Use you distance between points formula:
√( (x+8)^2 + (y-8)^2 ) = 2√( (x+2)^2 + y-2)^2 )
square both sides
(x+8)^2 + (y-8)^2 = 4( (x+2)^2 + y-2)^2 )
I will leave it up to you to expand and simplify the equation
Ah, finding equations, huh? Don't worry, I've got just the right amount of humor for you. Let's solve this step by step!
First, let's find the distance between the point P(x, y) and (-8, 8). We'll call this distance d1. And since P(x, y) is twice as far from (-8, 8) as from (-2, 2), we can say:
d1 = 2 * (distance between P and (-2, 2))
Now, let's calculate the distance between two points using the distance formula:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Let's apply this to our situation. For d1, we have:
d1 = 2 * sqrt((x - (-8))^2 + (y - 8)^2)
Now, let's find the distance between P(x, y) and (-2, 2). We'll call this distance d2:
d2 = sqrt((x - (-2))^2 + (y - 2)^2)
Now, we know that d1 = 2 * d2. So we can write the equation:
2 * sqrt((x - (-8))^2 + (y - 8)^2) = sqrt((x - (-2))^2 + (y - 2)^2)
And there you have it! The equation that represents all the points P(x, y) that are twice as far from (-8, 8) as from (-2, 2). Now, if this equation doesn't leave you in splits, I don't know what will!
To find the equation of the set of all points P(x, y) that is twice as far from (-8, 8) as from (-2, 2), we can use the distance formula.
Step 1: Find the distance from P(x, y) to (-8, 8).
The distance formula is:
d1 = √((x2 - x1)² + (y2 - y1)²)
Using (-8, 8) as (x1, y1) and (x, y) as (x2, y2), we have:
d1 = √((x - (-8))² + (y - 8)²)
= √((x + 8)² + (y - 8)²)
Step 2: Find the distance from P(x, y) to (-2, 2).
Using (-2, 2) as (x1, y1) and (x, y) as (x2, y2), we have:
d2 = √((x - (-2))² + (y - 2)²)
= √((x + 2)² + (y - 2)²)
Step 3: Set up the equation.
Since the distance from P(x, y) to (-8, 8) is twice the distance from P(x, y) to (-2, 2), we can write:
d1 = 2d2
Substituting the expressions we found in steps 1 and 2:
√((x + 8)² + (y - 8)²) = 2√((x + 2)² + (y - 2)²)
Squaring both sides to eliminate the square root:
(x + 8)² + (y - 8)² = 4((x + 2)² + (y - 2)²)
Expanding and simplifying:
x² + 16x + 64 + y² - 16y + 64 = 4(x² + 4x + 4 + y² - 4y + 4)
Simplifying further:
x² + 16x + y² - 16y + 128 = 4x² + 16x + 16 + 4y² - 16y + 16
Combining like terms:
3x² + 3y² - 116 = 0
Therefore, the equation of the set of all points P(x, y) that is twice as far from (-8, 8) as from (-2, 2) is 3x² + 3y² - 116 = 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 need to apply the concept of distance formula.
Let the distance from P(x, y) to (-8, 8) be d1, and the distance from P(x, y) to (-2, 2) be d2.
According to the given condition, we have:
d1 = 2 * d2
The distance formula between two points (x1, y1) and (x2, y2) is given by:
distance = √((x2 - x1)^2 + (y2 - y1)^2)
Using this formula, we can calculate the distances d1 and d2.
For d1, the coordinates are (-8, 8) and (x, y):
d1 = √((x - (-8))^2 + (y - 8)^2)
= √((x + 8)^2 + (y - 8)^2)
For d2, the coordinates are (-2, 2) and (x, y):
d2 = √((x - (-2))^2 + (y - 2)^2)
= √((x + 2)^2 + (y - 2)^2)
Now, we can substitute the values of d1 and d2 into the equation d1 = 2 * d2:
√((x + 8)^2 + (y - 8)^2) = 2 * √((x + 2)^2 + (y - 2)^2)
To simplify the equation further, we can square both sides:
(x + 8)^2 + (y - 8)^2 = 4 * ((x + 2)^2 + (y - 2)^2)
Expanding and simplifying, we get:
x^2 + 16x + 64 + y^2 - 16y + 64 = 4 * (x^2 + 4x + 4 + y^2 - 4y + 4)
Rearranging the terms, we have the equation of the set of all points P(x, y):
3x^2 - 4x + 3y^2 + 4y - 7 = 0