What is the value of x and y if the point (x, y) is equidistant from (3, 6) and (4,8)?

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We are given two points ( 3,6) and (4,8) and the point (x, y) is equidistant from the two.

Now we can use the formula to find the distance between two points (x1, y1) and (x2, y2) which gives the distance as sqrt[(x2 – x1)^2 + ( y2 – y1)^2]

Now the distance between (3,6) and (x,y) is equal to distance between ( 4, 8) and (x, y)

=> sqrt[(3 – x)^2 + ( 6 – y)^2] = sqrt[(x – 4)^2 + ( y – 8)^2]

take the square of both the sides

=> (3 – x)^2 + ( 6 – y)^2 = (x – 4)^2 + ( y – 8)^2

=> (3 – x)^2 - (x – 4)^2 = = ( y – 8)^2 - ( 6 – y)^2

=> (3 – x + x - 4)(3 –x + -x + 4 ) = ( y – 8 – 6 + y)(y – 8 + 6 – y)

=> -1 (7 – 2x) = (2y – 14)(-2)

=> 7 – 2x = 4y – 28

=> 2x + 4y – 35 =0

Therefore we get the equation of a line. This line passes between the given points and is perpendicular to the line joining the two points.

**The required values of (x , y) can be expressed as the equation 2x + 4y – 35 =0.**

The distance d between the points (x1,y1) and (x2,y2) is given by:

d^2 = (x2-x1)^2+(y2-y1)^2.

Given that (x,y) is equidistant from the points (3,6) and (4,8), we have to find (x,y). But we know by geometry that a lot of solutions will be there. Actually every point on the perpendicular bisector of the line segment joining (3,6) and (4,8) is equidistant from (3,6) and (4,8)

The distance between (x,y) and (3,6) is given by:

d^2 = (3-x)^2+(6-y)^2....(1)

The distance between (x,y) and the point (4,8) is given by:

d^2= (4-x)^2+(8-y)^2....(2)

The distances at (1) and (2) are equal by data. Therefore ,

(3-x)^2+(6-y)^2 = (4-x)^2+(8-y)^2.

9 -6x+x^2 +36-12y +y^2 = 16 -8x+x^2+64-16y+y^2.

45-6x-12y = 80 -8x -16y . Other terms cancel.

8x-6x + 16y-12y = 80-45

2x+ 4y = 35.

Therefore all the points on 2x+4y - 10 are equidistant from (3,6) and (4,8).

Since the point (x,y) is equidistant from the endpoints A(3, 6) and B(4,8), that means that the point is located on the midperpendicular of the segment AB.

The slope of the midperpendicular multiplied by the slope of the segment AB is -1.

mAB*m = -1

We'll write the equation of the segment line AB.

(yB - yA)/(y-yA) = (xB-xA)/(x-xA)

We'll substitute the coordinates of A and B:

(8-6)/(y-6) = (4-3)/(x-3)

2/(y-6) = 1/(x-3)

We'll cross multiply and we'll get:

y - 6 = 2(x-3)

The slope of the segment AB is mAB = 2.

mAB*m = -1

2*m = -1

m = -1/2

The equation of the midperpendicular is:

y - yM = (-1/2)(x-xM)

We'll calculate the coordinates of th midpoint of the segment AB:

xM = (xA+xB)/2

xM = (3+4)/2

xM = 7/2

yM = (yA+yB)/2

yM = (6+8)/2

yM = 7

y - 7 = (-1/2)(x - 7/2)

y - 7 = -x/2 + 7/4

4y - 28 = -2x + 7

2x - 4y - 35 = 0

The equation of the midperpendicular is: 2x - 4y - 35 = 0, so that the coordinates that verify the equation of the midperpendicular are equidistant from the endpoints A and B.

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