Question

If the point [x,y]is equidistant from points [a+b,a-b] and [a-b,a+b], then prove that bx=ay​

Answer 1

Step-by-step explanation:

See the attachment

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Answer 2

Given :-

P ( x,y ) is equidistant from points A [ (a+b),(b-a) ] and B [ (a-b),(a+b) ] .

T. P. :- bx = ay

Proof :-

Let ( x1,y1 ) and ( x2,y2 ) be A [ (a+b),(a-b) ] and B [ (a-b),(a+b) ] respectively.

By distance formula,

Distance = √ ( x - x1 )² + ( y -y1 )

PA = √ [ x - ( a+b ) ]² + [ y - ( b-a ) ]²

squaring both sides....

=> PA² = [ x - ( a+b ) ]² + [ y - ( b-a ) ]²

=> PA² = x² + ( a+b )² - 2x( a+b ) + y² + ( a-b )² - 2y( b-a ) ..... (i)

Distance = √ ( x - x2 )² + ( y -y2 )

PB = √ [ x - ( a-b ) ]² + [ y - ( a+b ) ]²

squaring both sides....

squaring both sides.... => PB² = [ x - ( a-b ) ]² + [ y - ( a+b ) ]²

b ) ]²=> PB² = x² + ( a-b )² - 2x( a-b ) + y² + ( a+b )² - 2y( a+b ) ...... (ii)

Now,

PA = PB

=> PA² = PB²

=> x² + ( a+b )² - 2x( a+b ) + y² + ( a-b )² - 2y( b-a ) = x² + ( a-b )² - 2x( a-b ) + y² + ( a+b )² - 2y( a+b )

=> - 2x( a+b ) - 2y( b-a ) = - 2x( a-b ) - 2y( a+b )

( cancelling x² , y² , ( a-b )² , ( a+b )² from both sides )

=> - 2x( a+b ) + 2x( a-b ) = - 2y( a+b ) + 2y( b-a )

=> - x ( a+b ) + x ( a-b ) = - y ( a+b ) + y ( b-a )

( cancelling 2 from both sides )

=> x ( a-b ) - x ( a+b ) = y ( b-a ) - y ( a+b )

=> x ( a - b - a - b ) = y ( b - a - a - b )

=> x (-2b) = y (-2a)

=> bx = ay

( cancelling -2 from both sides)

hence proved.....

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