Question

find the relation between X and Y such that point X, Y is equal distance from 3, 6 and - 3, 4​

Answer 1

Answer:

5 = 3X + Y

Step-by-step explanation:

Using distance formula :

Distance between ( 3, 6 ) and ( X, Y ) = Distance between ( - 3, 4 ) and ( X, Y )

= > √[ ( 3 - X )^2 + ( 6 - Y )^2 ] = √[ ( - 3 - X )^2 + ( 4 - Y )^2 ]

= > ( 3 - X )^2 + ( 6 - Y )^2 = ( - 3 - X )^2 + ( 4 - Y )^2

= > [ 9 + X^2 - 6X ] + [ 36 + Y^2 - 12Y ] = [ 9 + X^2 + 6X ] + [ 16 + Y^2 - 8Y ]

= > 9 - 6X + 36 - 12Y = 9 + 6X + 16 - 8Y

= > 45 - 6X - 12Y = 25 + 6X - 8Y

= > 45 - 25 = 6X + 6X - 8Y + 12Y

= > 20 = 12X + 4Y

= > 5 = 3X + Y

Answer 2

To Find -

A relation between X and Y such that point X, Y is equal distance from 3, 6 and - 3, 4 .

Solution -

First let me define the distance Formula -

Distance between two points -

$\sf{ \sqrt { { ( x_{2} - x_{1} ) }^2 + { ( y_{2} - y_{1} ) }^2 } }$ $\sf{ \bold{ Where \ - }} \\ \\ \sf{ x_{1} \ and \ x_{2} \ are \ the \ corresponding \ x \ coordinates } \\ \\ \sf{ y_{1} \ and \ y_{2} \ are \ the \ corresponding \ y \ coordinates }$ $\sf{ \bold { Here \ - }} \\ \\ \sf{ x_{1} = 3 } \\ \\ \sf{ x_{2} = -6 } \\ \\ \sf{ y_{1} = 6 } \\ \\ \sf{ y_{2} = 4 }$

Now, suppose there exists a point P , whose coordinates are X and Y .

This point is equidistant from two points A ( 3, 6 ) and B ( -3, 4 )

So ,

$\sf{ PA = PB } \\ \\ \sf{ Squaring \ Both \ Sides \ - } \\ \\ \sf{ {PA}^2 + {PB}^2 } \\ \\ \sf{ { ( x - 3 ) }^2 + { ( y - 6 ) }^2 = { ( x + 3 ) }^2 + { ( y - 4 ) }^2 } \\ \\ \sf{ => {x}^2 + {y}^2 - 6x - 12y + 45 = {x}^2 + {y}^2 + 6x - 8y + 25 } \\ \\ \sf{ Cancelling \ the \ like \ terms \ - } \\ \\ \sf{ 12x = -4y + 20 } \\ \\ \sf{ 3x + y = 5 }$

Hence , the required relation between x and y is 3x + y = 5 .

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