Question

if a point p (x , y) is equidistant from the points A (6, -1 ) and B(2,3) , find the relation between x and y.

Answer 1

If point p (x , y) is equidistant from the points A (6, -1 ) and B(2,3), Find the distance AP and BP and equate them.


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

(36 + x^2 - 12x) + (1 + y^2 + 2y) = (4 + x^2 -4x) +(9+y^2 - 6y)

x^2 + y^2 -12x + 2y +37 = x^2 + y^2 -4x - 6y + 13

-12x + 2y + 37 = -4x - 6y + 13

-12x + 2y + 37 + 4x + 6y - 13 = 0

-8x +8y + 24 = 0

-x + y + 3 = 0

x = y+3

Relation between x and y is x = y+3

Answer 2

Step-by-step explanation:

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MATHS

Find a relation between x and y such that the point (x,y) is equidistant from the point (3,6) and (−3,4).

January 17, 2020avatar

Pratibha Sarkarmondal

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ANSWER

Distance between two points (x

1

,y

1

) and (x

2

,y

2

) can be calculated

using the formula

(x

2

−x

1

)

2

+(y

2

−y

1

)

2

Distance between P(x,y) and A(3,6)=PA=

(3−x)

2

+(6−y)

2

=

3

2

+x

2

−6x+6

2

+y

2

−12y

=

x

2

−6x+y

2

−12y+45

Distance between P(x,y) and B(−3,4)=PB=

(−3−x)

2

+(4−y)

2

=

9+x

2

+6x+16+y

2

−8y

=

x

2

+6x+y

2

−8y+25

Given, PA = PB

x

2

−6x+y

2

−12y+45

=

x

2

+6x+y

2

−8y+25

=>x

2

−6x+y

2

−12y+45=x

2

+6x+y

2

−8y+25

12x+4y−20=0

3x+y−5=0