Question

find the relation between X and Y such that the point x and y is equidistant from the .3 to 6 and -3 to 4

Answer 1

The way the problem is worded, it appears you have complete 

flexibility in choosing coordinates to satisfy the problem requirements.  

For illustrative purposes, let's choose coordinates on a coordinate grid

that make your diagram simple, recognizable, and your calculations

easy.  Graph the following explanation as we proceed.  Let's assign

the following coordinates to the points...

 

A (1,0)

B (7,0)

C (4,0).....C is midpoint between A and C

D (4,6).....is "r=6 units" above C and equidistant from A and B

E (4,-6)...is "r=6 units" below C and als equidistant from A and B

 

Recognize that there can be two points (D and E) which satisfy the 

problem statement requirements.  Segment DE is perpendicular to

segment AB and passes through point C.

 

In your proof, write the equations (both in slope-intercept form)

for line AB (y=0), write the equation for line DE (x=4) and show

that one slope is the inverse reciprocal of the other (definition of

perpendicularity).  Calculate the distance CD and CE using the

distance between two points formula (d=√((x2-x1)+(y2-y1)).  

Show that the distances are equal.

 

Then, in your proof, calculate the distance between AD, BD, AE,

and BE using the distance between two points formula.  Show that 

all distances are equal (you will be calculating the hypotenuses of 

each of four right triangles shown on your graph).

 

As an added challenge, assign points that form lines which are 

neither horizontal nor vertical.  The above explanation should

hold true as well

I think this answer is help yu

Answer 2

Answer:

The answer is 3x + y - 5 = 0 .

Step-by-step explanation:

Let the point P(x,y) be the equidistant from the points A(3,6) and B(-3,4).

therefore, PA = PB.

By using distance formula,

$\sqrt{(x2-x1)^{2}\:+\: (y2-y1)^{2} } \:we\:have,\\\\\sqrt{(x-3)^{2}\:+\: (y-6)^{2} }\:=\:\sqrt{(x+3)^{2}\:+\: (y-4)^{2} }\\\\{(x-3)^{2}\:+\: (y-6)^{2} }\:=\:{(x+3)^{2}\:+\: (y-4)^{2} }\\\\x^{2} \:-\:6x\:+\:9\:+\:y^{2} \:-\:12y\:+\:36\:=\:x^{2} \:+\:6x\:+\:9\:+\:y^{2} \:-\:8y\:+\:16\\\\-\:6x\:-\:6x\:-\:12y\:+\:8y\:+\:36\:-\:16\:=\:0\\\\-\:12x\:-\:4y\:+\:20\:=\:0\\\\therefore,\:3x\:+\:y\:-\:5\:=\:0$

Hope it helps you :)