Find the equation of the locus of points equidistant from (-5,0) and (0,4)?

Answer 1

perpendicular bisector:
mdpoint(-2.5,2)
slope of perpendicular: - 5/4

y -2 = (-5/4)(x+2.5)

d:

Answer 2

In general, an equation of a straight line (or a straight locus in this case) has the formula:
y - y1 = m(x - x1), where,
x1 is the x-coordinate of a point;
y1 is the y-coordinate of the same point; and
m is the gradient of the straight line.

So, gradient, m = (0 - 4)/(-5 - 0) = -4/-5 = 4/5
Choose any point; either (-5,0) or (0,4). Let's select (0,4) as it will be easier to work with numbers. Therefore, x1 = 0 and y1 = 4.

Substitute m = 4/5, x1 = 0 and y1 = 4 into the above equation, you'll get:
y - 4 = 4/5(x - 0)
y = 4/5x + 4
5y = 4x + 20

Hence, the equation of the locus of points equidistant from (-5,0) and (0,4) is 5y = 4x + 20.

Answer 3

The locus of points equidistant from two given points is the line which is the perpendicular bisector to the segment they define.
So, first, find the midpoint:
((-5+0)/2),((0+4)/2)=
(-2.5, 2)
Then the slope of the given line is:
4/(0--5)=4/5
so the slope of our line will be the negative reciprocal, or:
-5/4
since y=mx+b
2=-5/4*-2.5+b, hence...
b=-1.125 (or -9/8)
So the equation for the locus of points is:
y= -5/4x - 9/8

Answer 4

Locus of points is perpendicular bisector of line thro` given points.
m = 4 / 5 so gradient of perpendicular is - 5 / 4
Mid point is (- 5/2 , 2)
Equation of perpendicular is:-
y - 2 = - 5/4 (x + 5/2)
4y - 8 = - 5x - 25/2
4y = - 5x - 9/2
y = (- 5/4) x - 9/8

Answer 5

squared distance from (-5,0):
R1^2= (x+5)^2 + y^2

squared distance from (0,4):
R2^2 = x^2 + (y-4)^2

So:
x^2 + (y-4)^2 = R2^2 = R1^2 = (x+5)^2 + y^2

x^2 - (x+5)^2 + (y-4)^2 - y^2 = 0


x^2 - x^2 - 10x - 25 + y^2 - 8y + 16 - y^2 = 0

-10x -25 -8y + 16 = 0

-10x - 8y - 9 = 0
or
10x + 8y = -9

This is a straight line, and passes through the mid-point, (-2.5,2), as would be expected.

Answer 6

(((-5-0)^2+(0-4)^2))^1\2
=(25+16)^1/2
=(41)^1/2