Question

find a relation between xand y such that the points (x, y) is equidistant from the point (3, 6) and (-3, 4)​

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.

Answer 2

Answer:

Hii,

Step-by-step explanation:

HEY SIS...YOUR DP IS SO CUTE....

$Solution$

solution : using distance formula,

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

(x

2

−x

1

)

2

+(y

2

−y

1

)

2

let P = (x, y) , Q = (3, 6) and R = (-3, 4).

distance between P and R = \sqrt{(x+3)^2+(y-4)^2}

(x+3)

2

+(y−4)

2

similarly, distance between P and Q = \sqrt{(x-3)^2+(y-6)^2}

(x−3)

2

+(y−6)

2

a/c to question, (x, y) is the equidistant from the points (3,6) and (-3,4)

i.e.,PQ = PR

⇒\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

squaring both sides we get,

⇒(x - 3)² + (y - 6)² = (x + 3)² + (y - 4)²

⇒x² + 9 - 6x + y² + 36 - 12y = x² + 9 + 6x + y² + 16 - 8y

⇒-6x + 45 - 12y = 6x - 8y + 25

⇒-6x - 6x - 12y + 8y = 25- 45

⇒-12x - 4y = -20

⇒3x + y = 5

Therefore the relation between x and y is 3x + y = 5.

Hope it will help you...

# TrueLover ❤️