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.
1. (-3,0), (x,y), (3,-5).
(3-(-3) = 2(x-(-3), 6 = 2x+6, X = 0.
(-5-0) = 2(y-0), -5 = 2y, Y = -2.5.
(-3,0), (0,-2.5), (3,-5).
m = (-2.5-0)/(0-(-3)) = -5/6.
Y = mx + b, -5x/6 + b = 0, -5*(-3)/6 + b = 0, b = -5/2 = -2.5.
Eq: Y = (-5/6)x - 5/2.
2. (-8,8), (x,y), (2,2).
x-(-8) = 2(-2-x), x+8 = -4-2x, X = -4.
y-8 = 2(2-y), Y = 4.
(-8,8), (-4,4), (-2,2).
m = (4-8)/(4-(-8)) = -1.
Y = mx+b = 8, -1*(-8)+b = 8, b = 0. Eq: Y = -x.
To find the equation of the set of all points equidistant from two given points, you can use the concept of distance formula. The distance formula between two points (x1, y1) and (x2, y2) is given by:
d = √((x2 - x1)^2 + (y2 - y1)^2)
Now, let's solve the given questions step by step:
1. Find the equation of the set of all points equidistant from (-3, 0) and (3, -5):
Let P(x, y) be a point equidistant from (-3, 0) and (3, -5). The distance from P(x, y) to (-3, 0) must be equal to the distance from P(x, y) to (3, -5).
Using the distance formula, we have:
√((x - (-3))^2 + (y - 0)^2) = √((x - 3)^2 + (y - (-5))^2)
Simplifying the equation, we get:
(x + 3)^2 + y^2 = (x - 3)^2 + (y + 5)^2
Expanding and rearranging terms further, we have:
x^2 + 6x + 9 + y^2 = x^2 - 6x + 9 + y^2 + 10y + 25
Simplifying the equation, we get:
12x + 10y = -16
Thus, the equation of the set of all points equidistant from (-3, 0) and (3, -5) is 12x + 10y = -16.
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):
Let P(x, y) be a point that is twice as far from (-8, 8) as from (-2, 2). The distance from P(x, y) to (-8, 8) must be equal to twice the distance from P(x, y) to (-2, 2).
Using the distance formula, we have:
√((x - (-8))^2 + (y - 8)^2) = 2√((x - (-2))^2 + (y - 2)^2)
Simplifying the equation, we get:
(x + 8)^2 + (y - 8)^2 = 4((x + 2)^2 + (y - 2)^2)
Expanding and rearranging terms further, we have:
x^2 + 16x + 64 + y^2 - 16y + 64 = 4(x^2 + 4x + 4 + y^2 - 4y + 4)
Simplifying the equation, we get:
3x^2 - 4x + 3y^2 + 8y - 20 = 0
Thus, the equation of the set of all points P(x, y) that is twice as far from (-8, 8) as from (-2, 2) is 3x^2 - 4x + 3y^2 + 8y - 20 = 0.
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:
Let P(x, y) be a point such that the sum of the distance from (-4, 0) and (4, 0) is equal to 12. The sum of the distances from P(x, y) to (-4, 0) and (4, 0) must be equal to 12.
Using the distance formula, we have:
√((x - (-4))^2 + (y - 0)^2) + √((x - 4)^2 + (y - 0)^2) = 12
Simplifying the equation, we get:
√((x + 4)^2 + y^2) + √((x - 4)^2 + y^2) = 12
Squaring both sides of the equation to eliminate the square roots, we have:
(x + 4)^2 + y^2 + 2√((x + 4)^2 + y^2)√((x - 4)^2 + y^2) + (x - 4)^2 + y^2 = 144
Expanding and simplifying the equation further, we get:
2x^2 + 2y^2 - 32x + 32 = 0
Thus, 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 2x^2 + 2y^2 - 32x + 32 = 0.