1. Find the equation of the set of all points P(x,y) that is equidistant from (-3,0) and (3,-5).
2. Find the equation of the set of all points P(x,y) that is twice as far from (-8,8) as from (-2,2).
3. Find the equation of the set of all points P(x,y) such that the sum of the distance from (-4,0) and (4,0) is equal to 12.
On the first, it is a line perpendiular to the line connecting the two ponts, which intersects the half way pont (0,-1.5).
On the last, would it be an eliplse?
On the second, look for a conic.
How do you get the equations of the last two?
Pick a point (x,y).
write the distance equations to the two given points.
for 2, set twice one distance equal to the distance of the other, then maniplate the equation into standard form for x and y.
for 3, add the distances, set them equal to 12.
1. To find the equation of the set of all points equidistant from (-3,0) and (3,-5), we can use the concept of the midpoint formula.
Let P(x, y) be a point that is equidistant from (-3,0) and (3,-5). The midpoint formula states that the coordinates of the midpoint between two points (x1, y1) and (x2, y2) is ( (x1 + x2)/2, (y1 + y2)/2 ).
Using this formula, we find the midpoint coordinates:
Midpoint(x, y) = ((-3 + 3)/2, (0 + -5)/2)
= (0, -2.5)
Since point P(x, y) is equidistant from (-3,0) and (3,-5), the distance from P to (-3,0) should be the same as the distance from P to (3,-5).
Using the distance formula, the distance from P(x, y) to (-3,0) is:
√((x - (-3))^2 + (y - 0)^2)
The distance from P(x, y) to (3,-5) is:
√((x - 3)^2 + (y - (-5))^2)
Since the distances are equal, we can set up the equation as follows:
√((x - (-3))^2 + (y - 0)^2) = √((x - 3)^2 + (y - (-5))^2)
Simplifying the equation:
(x + 3)^2 + y^2 = (x - 3)^2 + (y + 5)^2
Expanding and simplifying further:
x^2 + 6x + 9 + y^2 = x^2 - 6x + 9 + y^2 + 10y + 25
Simplifying:
12x + 10y = -16
Therefore, the equation of the set of all points equidistant from (-3,0) and (3,-5) is 12x + 10y = -16.
2. 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 again use the distance formula and the concept of proportions.
Let P(x, y) be a point that is twice as far from (-8,8) as from (-2,2).
Using the distance formula, the distance from P(x, y) to (-8,8) is:
√((x - (-8))^2 + (y - 8)^2)
The distance from P(x, y) to (-2,2) is:
√((x - (-2))^2 + (y - 2)^2)
Since P(x, y) is twice as far from (-8,8) as from (-2,2), we can set up the proportion:
√((x - (-8))^2 + (y - 8)^2) = 2 * √((x - (-2))^2 + (y - 2)^2)
Squaring both sides of the equation to eliminate the square root:
(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 + 4y^2 - 16y + 4
Rearranging the equation:
3x^2 + 3y^2 - 15 = 0
Simplifying further:
x^2 + y^2 - 5 = 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 x^2 + y^2 - 5 = 0.
3. To find the equation of the set of all points P(x,y) such that the sum of the distance from (-4,0) and (4,0) is equal to 12, we can again use the distance formula.
Let P(x, y) be a point that satisfies the given condition.
Using the distance formula, the distance from P(x, y) to (-4,0) is:
√((x - (-4))^2 + (y - 0)^2)
The distance from P(x, y) to (4,0) is:
√((x - 4)^2 + (y - 0)^2)
The sum of these distances should be equal to 12:
√((x - (-4))^2 + (y - 0)^2) + √((x - 4)^2 + (y - 0)^2) = 12
Squaring both sides of the equation to eliminate the square root:
((x + 4)^2 + y^2) + ((x - 4)^2 + y^2) + 2√((x + 4)^2 + y^2)√((x - 4)^2 + y^2) = 144
Expanding and simplifying the equation:
2x^2 + 32 + 2y^2 + 2√((x + 4)^2 + y^2)√((x - 4)^2 + y^2) = 144
Subtracting 2x^2 + 32 + 2y^2 from both sides:
2√((x + 4)^2 + y^2)√((x - 4)^2 + y^2) = 112 - 2x^2 - 2y^2
Squaring both sides of the equation to eliminate the square root:
4((x + 4)^2 + y^2)((x - 4)^2 + y^2) = (112 - 2x^2 - 2y^2)^2
Expanding and simplifying the equation:
4(x^2 + 8x + 16 + y^2)((x^2 - 8x + 16 + y^2) = (112 - 2x^2 - 2y^2)^2
Expanding further and rearranging the equation:
4x^4 + 128x^2 - 256 + 4y^4 = 256x^2 + 4x^4 + 256y^2 - 112x^2 - 224y^2 + 12544
Simplifying:
-128x^2 - 480y^2 = -12384
Dividing both sides of the equation by -8:
16x^2 + 60y^2 = 1548
Dividing both sides of the equation by 12:
4x^2 + 15y^2 = 387
Therefore, the equation of the set of all points P(x, y) such that the sum of the distance from (-4,0) and (4,0) is equal to 12 is 4x^2 + 15y^2 = 387.
1. To find the equation of the set of all points equidistant from two given points, we need to use the concept of the distance formula. The distance between two points (x₁, y₁) and (x₂, y₂) is given by the formula:
distance = √((x₂ - x₁)² + (y₂ - y₁)²)
In this case, the points are (-3,0) and (3,-5). Let's denote an arbitrary point P as (x,y). The distances from P to (-3,0) and (3,-5) should be equal, so we can write the equation:
√((x - (-3))² + (y - 0)²) = √((x - 3)² + (y - (-5))²)
Simplifying this equation, we have:
(x + 3)² + y² = (x - 3)² + (y + 5)²
Expanding the equation further, we get:
x² + 6x + 9 + y² = x² - 6x + 9 + y² + 10y + 25
The y² terms cancel out, so we are left with:
6x + 10y + 34 = 0
Thus, the equation of the set of all points equidistant from (-3,0) and (3,-5) is 6x + 10y + 34 = 0.
2. To find the equation of the set of all points that are twice as far from (-8,8) as from (-2,2), we can apply the same logic as in the previous question. Let's denote an arbitrary point P as (x,y). The distances from P to (-8,8) and (-2,2) should have a ratio of 2:1. Using the distance formula, we can write the equation:
√((x - (-8))² + (y - 8)²) = 2 * √((x - (-2))² + (y - 2)²)
Simplifying this equation, we have:
√((x + 8)² + (y - 8)²) = 2√((x + 2)² + (y - 2)²)
Squaring both sides of the equation, we get:
(x + 8)² + (y - 8)² = 4((x + 2)² + (y - 2)²)
Expanding and simplifying further, we have:
x² + 16x + 64 + y² - 16y + 64 = 4(x² + 4x + 4 + y² - 4y + 4)
The x² and y² terms cancel out, so we are left with:
12x + 12y + 84 = 0
Hence, the equation of the set of all points that is twice as far from (-8,8) as from (-2,2) is 12x + 12y + 84 = 0.
3. To find the equation of the set of all points such that the sum of the distances from (-4,0) and (4,0) is equal to 12, we will make use of the distance formula once again. Let's denote an arbitrary point P as (x,y). The sum of the distances from P to (-4,0) and (4,0) should equal 12, so we can write the equation:
√((x - (-4))² + (y - 0)²) + √((x - 4)² + (y - 0)²) = 12
Simplifying this equation, we have:
√((x + 4)² + y²) + √((x - 4)² + y²) = 12
To solve this equation, we can square both sides and simplify:
(x + 4)² + y² + 2√((x + 4)² + y²)((x - 4)² + y²) + (x - 4)² + y² = 144
Simplifying further, we get:
2x² + 2y² + 2√((x + 4)² + y²)((x - 4)² + y²) = 128
√((x + 4)² + y²)((x - 4)² + y²) = 64 - x² - y²
Square both sides again to eliminate the square root:
((x + 4)² + y²)((x - 4)² + y²) = (64 - x² - y²)²
Expanding and simplifying this equation will give a quadratic expression in x and y:
(x² + 8x + 16 + y²)(x² - 8x + 16 + y²) = (64 - x² - y²)²
Simplifying further should result in the equation of the set of all points P(x,y) such that the sum of the distances from (-4,0) and (4,0) is equal to 12.