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text If  end text P equals open parentheses fraction numerator 1 over denominator x subscript p end subscript end fraction comma p close parentheses semicolon Q equals open parentheses fraction numerator 1 over denominator x subscript q end subscript end fraction comma q close parentheses semicolon R equals open parentheses fraction numerator 1 over denominator x subscript r end subscript end fraction comma r close parentheses blank text where  end text x subscript k end subscript not equal to 0 text  , dentoes the  end text k to the power of text sh  end text end exponent terms of H.P. for k N element of ,then

  1. area of capital delta P Q R equals open parentheses fraction numerator p to the power of 2 end exponent q to the power of 2 end exponent r to the power of 2 end exponent over denominator 2 end fraction square root of left parenthesis p minus q right parenthesis to the power of 2 end exponent plus left parenthesis q minus r right parenthesis to the power of 2 end exponent plus left parenthesis r minus p right parenthesis to the power of 2 end exponent end root close parentheses    
  2. DPQR is a right angled triangle.    
  3. The points P,Q,R are collinear    
  4. None of these.    

The correct answer is: The points P,Q,R are collinear


    To find the correct option.
    We know that 1 over x subscript 1 comma 1 over x subscript 2 space a r e space i n space H P space t h e n space a subscript 1 comma a subscript 2 space a r e space i n space A P space w h e r e space 1 over x subscript 1 equals a subscript 1 comma space 1 over x subscript 2 equals a subscript 2

1 over x subscript p equals a subscript p equals a subscript 1 plus left parenthesis p minus 1 right parenthesis d

1 over x subscript q equals a subscript q equals a subscript 1 plus left parenthesis q minus 1 right parenthesis d

1 over x subscript r equals a subscript r equals a subscript 1 plus left parenthesis r minus 1 right parenthesis d

S l o p e space o f space P Q space equals space fraction numerator left parenthesis q minus p right parenthesis over denominator left parenthesis q minus p right parenthesis end fraction d space equals space 1 over d

S l o p e space o f space Q R space equals space fraction numerator left parenthesis r minus q right parenthesis over denominator left parenthesis r minus q right parenthesis end fraction d space equals space 1 over d

S l o p e space o f space P R space equals space fraction numerator left parenthesis r minus p right parenthesis over denominator left parenthesis r minus p right parenthesis end fraction d space equals space 1 over d

    All have same slope therefore, given points re collinear.

    Hence, The points P,Q,R are collinear.

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