Question

find a relation between x and y such that the points P(x,y) is equidistant from the points A.(2,5) and B.(-3,7)
And. 2x+y+3=0 how to get this answer only ​

Answer 1

Given that

 ap=bp

$\sqrt{ (3-y)^{2}+ (-5-x)^{2}} = \sqrt{ (7-x)^{2} + (2-y)^{2} }$

By Squaring both the sides,

(-5-x)^{2} + (3-y)^{2}= (7-x)^{2}+ (2-y)^{2}

25+$x^{2}$+10x+9+$y^{2}$-6y=49+$x^{2}$-14x+4+$y^{2}$-4y

24x=19+2y

24x-2y=19

This the relation between x and y for the given condition .

Answer 2

Answer:

Given that

ap=bp

\sqrt{ (3-y)^{2}+ (-5-x)^{2}} = \sqrt{ (7-x)^{2} + (2-y)^{2} }

(3−y)

2

+(−5−x)

2

=

(7−x)

2

+(2−y)

2

By Squaring both the sides,

(-5-x)^{2} + (3-y)^{2}= (7-x)^{2}+ (2-y)^{2}

25+x^{2}x

2

+10x+9+y^{2}y

2

-6y=49+x^{2}x

2

-14x+4+y^{2}y

2

-4y

24x=19+2y

24x-2y=19

This the relation between x and y for the given condition .