Question

Find a relation between x and y such that the point (x, y) is equidistant from
the points (6, 1) and (3,5).​

Answer 1

Answer:

Step-by-step explanation:

Let the point P(x,y) is equidistant from the points A(3,6)and B(−3,4)

∴PA=PB

By using the distance formula

=  

(x  

2

​  

−x  

1

​  

)  

2

+(y  

2

​  

−y  

1

​  

)  

2

 

​  

    we have

⇒  

(x−3)  

2

+(y−6)  

2

 

​  

=  

(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

⇒3x+y−5=0

Hence this is a relation between x and y.

Answered By

Answer 2

Solution:-

Given points :-

$\rm \to A = (6,1) \: and \: B = (3,5) \: \: are \: equidistant \:$

$\rm \: \to \: p \: = (x,y)$

To find :-  

$\to \rm \: relation \: between \: x \: and \: y$

Formula

$\boxed{\rm{d=\sqrt{{(x_2-x_1)}^{2}+{( y_2 - y_1)}^{2}}}}$

Now we can write as

$\to \rm \: AP = BP$

$\rm \to {AP}^{2} = {BP}^{2}$

For equidistant

$\rm\to{(x_2-x_1)}^{2}+{( y_2-y_1)}^{2} ={(x_2 - x_1)}^{2}+{( y_2 - y_1)}^{2}$

$\rm \to \: (x - 6) {}^{2} + (y - 1) {}^{2} = (x - 3) {}^{2} + (y - 5) {}^{2}$

Using this identity

$\rm \: {a}^{2} + {b}^{2} - 2ab =( {a} - b)^{2}$

Now we can write

$\rm \to \: {x}^{2} + 36 - 12x + {y}^{2} + 1 - 2y = {x}^{2} + 9 - 6y + {y}^{2} + 25 - 10y$

$\rm \to \: \cancel{x}^{2} + 36 - 12x + \cancel{y}^{2} + 1 - 2y = \cancel{x}^{2} + 9 - 6y + \cancel{y}^{2} + 25 - 10y$

$\rm \: \to 36 - 12x + 1 - 2y = 9 - 6x + 25 - 10y$

$\rm \: \to \: 37 - 12x - 2y = 34 - 6x - 10y$

$\rm \to \: 37 - 34 - 2y + 10y = - 6x + 12x$

$\rm \to \: 3 + 8y = 6x$

$\rm \to \: 6x - 8y = 3$

So relation between x and y is

$\rm \to \: 6x - 8y = 3$