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Question

Find the relationship between x and y so that the point (x,y) is equidistant from the two points(-2,1) and (2,4). also explain! : )

Answer 1

If you need to find a single point, just use the point halfway between the two. (e.g average the x-coords and the y-coords):

Average of -2, 2 = (2 + -2) / 2 = 0
Average of 1, 4 = (1 + 4) / 2 = 2.5

Midpoint: (0, 2.5)

However, if you need the equation for all points that are equidistant, that would be a line through that point perpendicular to the original line.

For a line perpendicular, it is simply the opposite slope.

The slope of the line through the first two points is given by the equation:
m = (y2-y1) / (x2-x1).
m = ( 4 - 1 ) / (2 - (-2) )
m = 3 / 4

The opposite slope (of a line perpendicular) is the negative reciprocal.
m' = -1 / m
m' = -1 / (3/4)
m' = -4 / 3

So what's a line through (0, 2.5) with a slope of -4/3?

Standard form of a line:
y = mx + b

In this case, b is 2.5, since (0, 2.5) is the y-intercept.

y = (-4/3)x + 2.5

So any points that satisfy this equation will be equidistant from the two points (-2,1) and (2,4)...

Another way to solve this is take a point (x, y). The distance from (-2, 1) to x,y can be figured using the pythagorean theorem:
d² = (x- (-2))² + (y - 1)²
d² = (x + 2)² + (y - 1)²

Similarly from (x, y) to (2, 4):
d² = (x - 2)² + (y - 4)²

Now just equate these and solve for y:
(x + 2)² + (y - 1)² = (x - 2)² + (y - 4)²
(x² + 4x + 4) + (y² -2y + 1) = (x² - 4x +4) + (y² - 8y +16)
x² + 4x + 4 + y² -2y + 1 = x² - 4x +4 + y² - 8y +16

Cancel x² and y² from both sides:
4x - 2y + 5 = -4x - 8y + 20

Group x and y and the numbers together:
6y = -8x + 15

Divide by 6:
y = (-4/3)x + 5/2

This is the same answer as before, so either method will work.

Answer 2

You need two things to determine the equation from the straight line equidistant from (-2,1) and (2,4), namely the slope and a point on that line.

The slope is calculated from the slope of the line through (-2,1) and (2,4). This slope is (y2-y1)/(x2-x1) = 3/4
The line we are looking for makes a straight angle with this line.
Its slope is therefore -4/3 or -1.33

A point on the line is in the middle between (-2,1) and (2,4).
This point is (0,2.5)

The equidistant line obeys the equation y = -4x/3 + c
with x=0 and y=2.5 we obtain c=2.5

=> The answer is y = -4x/3 + 2.5

Answer 3

for (x,y) to be equidistant, the line in which (x,y) lies must be midpoint of the points (-2,1) and (2,4) and perpendicular to the gradient between them.

finding gradient:
m = (4-1)/(2+2) = 3/4

gradient of (x,y):
M= -4/3

midpoint:
{(2-2)/2 , (4+1)/2} = (0, 5/2)

EQN:
-4/3=(y-5/2)/(x-0)
-4/3x=y-5/2

therefore the relationship between x and y is y= -4/3x + 5/2

Answer 4

Draw a line between the two points and find it's center.
Draw another line through the center point that you have just found at right angles to the line that you just drew.
Every point on the second line is equidistant from the two points.

Answer 5

what does it mean to be equidistant ? well we want the set of points (x,y) for which distance from (x,y) to (-2,1) is the same as the distance from (x,y) to (2,4). since this involves disatnce we use the distance formula d=sqrt((x2-x1)^2+(y2-y1)^2). the key here is to use a generic point (X,Y) and each of the given points in an equation which we will solve for Y to get the relation you seek

d=sqrt((X-(-2))^2+(Y-1)^2) and d=sqrt((X-2)^2+(Y-4)^2) moreover problem says these two distances must be equal so we set these two expreesions equal to each other and solve for Y

sqrt((X-(-2))^2+(Y-1)^2) = sqrt((X-2))^2+(Y-4)^2)
square both sides to get
(X-(-2))^2+(Y-1)^2 = (X-2)^2+(Y-4)^2
(X+2)^2+(Y-1)^2=(X-2)^2+(Y-4)^2

(X+2)^2=X^2+4X+4
(Y-1)^2=Y^2-2Y+1
so left side is
X^2+4x-2Y+5 after combining like terms
and right side is
(X-2)^2=X^2-4X+4
(Y-4)^2=Y^2-8Y+16
X^2-4X-8Y+20
so we have
X^2+4x-2Y+5=X^2-4X-8Y+20
simplify to get
4X-2Y+5=-4X-8Y+20
solve for Y to get
Y=-(4/3)*X+(3/2)
check my algebra but the locution to get the answer is correct

ALL PROBLEMS WHICH ASK YO TO FIND REALTIONSHIP SUBJECT TO GIVEN CONDITIONS ARE SOLVED THIS WAY USE A GENERIC (X,Y) AND THE GIVEN POINTS IN SOME FORMULA WHICH YOU THEN SOLVE FOR Y TO GET THE RELATION YOU SEEK

good luck