answer give me, u give me proof their solution
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Hi,
Let P(x,y)
Given it is equidistant so,
PA=PB
take square both side
PA^2=PB^2
now use distance
formula ,
{x-(a+b)}^2+{y-(b-a)}^2={x-(a-b)}^2+{y-(a+b)}^2
=>x^2+(a+b)^2-2x(a+b)+y^2+(b-a)^2-2y(b-a)y=x^2+(a-b)^2-2x(a-b)+y^2+(a+b)^2-2y(a+b)
=>2x(a-b)-2x(a+b)=2y(b-a)-2y(a+b)
=>2x{a-b-a-b}=2y{b-a-a-b}
=>2x(-2b)=2y(-2a)
=>bx=ay
hence proved
Hope this helps u!! brainliest??
Let P(x,y)
Given it is equidistant so,
PA=PB
take square both side
PA^2=PB^2
now use distance
formula ,
{x-(a+b)}^2+{y-(b-a)}^2={x-(a-b)}^2+{y-(a+b)}^2
=>x^2+(a+b)^2-2x(a+b)+y^2+(b-a)^2-2y(b-a)y=x^2+(a-b)^2-2x(a-b)+y^2+(a+b)^2-2y(a+b)
=>2x(a-b)-2x(a+b)=2y(b-a)-2y(a+b)
=>2x{a-b-a-b}=2y{b-a-a-b}
=>2x(-2b)=2y(-2a)
=>bx=ay
hence proved
Hope this helps u!! brainliest??
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