Question

if the point p(x,y) are equidistant from the point a(5,1) and b(-1,5), prove that 3x=2y

Answer 1

A ___________P__________B
(5,1) (x,y) (-1,5)

If P is equidistant from point A and B
then,AP:PB=1:1

Distance AP=√(x2-x1)²+(y2-y1)²
=√(x-5)²+(y-1)²
AP²=x²+25-10x+y²+1-2y
AP²=x²+y²+26-10x-2y

Distance BP=√(-1-x)²+(5-y)²
BP²=1+x²+2x+25+y²-10y
BP²=26+x²+y²+2x-10y

since P is mid point.
AP²=BP²

x²+y²+26-10x-2y=x²+y²+2x-10y+26

x²-x²+y²-y²+26-26-10x+2x=-10y+2y
-12x=-8y
-3x=-2y
3x=2y

hence proved

@Altaf

Answer 2

heya friends ...

let the point p(x,y)are equidistant from the point a(5,1) and b(-1,5)

like it..a___________p_____________b

now,,,We will find the distance between ap and ab by using distance formulae.
.so,

ap≡√(5-x)^2+(1-y)^2

=>√25+x^2-10x+1+y^2-2y..

=>√26+x^2+y^2-10x+2y...---------1)

now,. we find bp≡√(-1-x)^2+(5-y)^2

=>√x^2+1+2x+25+y^2-10y

=>√26+x^2+y^2+2x-10y------2)

now,squaring the 1) and 2) equation we will get .

=>26+x^2+y^2-10x-2y=26+x^2+y^2+2x-10y

=>-12x=-8y

=>3x=2y prooved. here...

hope it help you.

@rajukumar☺☺