Final answer:
To find the equation of the line that is equidistant from (-2,4) and (-4,-4) using the locus definition method, we find the midpoint of the line segment and then use the distance formula. The equation of the line in general form is x^2 + y^2 + 4x - 8y + 3 = 0.
Step-by-step explanation:
To find the equation of the line that is equidistant from (-2,4) and (-4,-4) using the locus definition method, we first need to find the midpoint of the line segment joining the two points. The midpoint formula is given by:
(x,y) = ((x1+x2)/2, (y1+y2)/2)
Substituting the given coordinates into the formula, we get:
(x,y) = ((-2+(-4))/2, (4+(-4))/2) = (-3, 0)
Therefore, the midpoint of the line segment is (-3, 0).
Since the line we are looking for is equidistant from the two given points, the distance between the line and each point will be the same. We can use the distance formula to calculate this distance. The distance formula is given by:
d = sqrt((x-x1)^2 + (y-y1)^2)
Substituting (-3, 0) for (x, y) and (-2, 4) for (x1, y1), we get:
d = sqrt(((-3)-(-2))^2 + ((0)-4)^2) = sqrt(((-1)^2 + (-4)^2) = sqrt(1+16) = sqrt(17)
So, the distance between the line and the point (-2,4) is sqrt(17).
Now, we can write the equation of the line equidistant from the two points using the distance formula:
sqrt((x-x1)^2 + (y-y1)^2) = sqrt(17)
Squaring both sides of the equation to eliminate the square root, we get:
(x-x1)^2 + (y-y1)^2 = 17
Substituting (-2,4) for (x1, y1), we get:
(x-(-2))^2 + (y-4)^2 = 17
Simplifying the equation, we have:
(x+2)^2 + (y-4)^2 = 17
Converting the equation to general form, we expand the squares:
x^2 + 4x + 4 + y^2 - 8y + 16 = 17
Combining like terms, we get:
x^2 + y^2 + 4x - 8y + 3 = 0
Therefore, the equation of the line equidistant from the points (-2,4) and (-4,-4) in general form is x^2 + y^2 + 4x - 8y + 3 = 0.