Maths-
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Question

Statement- 1 open parentheses S subscript 1 end subscript close parentheses:If A open parentheses x subscript 1 end subscript comma y subscript 1 end subscript close parentheses comma B open parentheses x subscript 2 end subscript comma y subscript 2 end subscript close parentheses comma C open parentheses x subscript 3 end subscript comma y subscript 3 end subscript close parentheses are non-collinear points. Then every point left parenthesis x comma y right parenthesis in the plane of capital delta to the power of text le  end text end exponent A B C, can be expressed in the form open parentheses fraction numerator k x subscript 1 end subscript plus l x subscript 2 end subscript plus m x subscript 3 end subscript over denominator k plus l plus m end fraction comma fraction numerator k y subscript 1 end subscript plus l y subscript 2 end subscript plus m y subscript 3 end subscript over denominator k plus l plus m end fraction close parentheses
Statement- 2 open parentheses S subscript 2 end subscript close parentheses:The condition for coplanarity of four points A left parenthesis stack a with ‾ on top right parenthesis comma B left parenthesis stack b with ‾ on top right parenthesis comma C left parenthesis stack c with ‾ on top right parenthesis comma D left parenthesis stack d with ‾ on top right parenthesis is that there exists scalars 1 comma m comma n comma p not all zeros such that  l a with ‾ on top plus m b with ‾ on top plus n c with ‾ on top plus p d with ‾ on top equals 0 with minus on top where l plus m plus n plus p equals 0.

  1. Both S subscript 1 end subscript comma S subscript 2 end subscript are true & S subscript 2 end subscript is reason for S subscript 1 end subscript.    
  2. Both S subscript 1 end subscript comma S subscript 2 end subscript are true & but S subscript 2 end subscript is not reason for S subscript 1 end subscript.    
  3. Both S subscript 1 end subscript comma text end text S subscript 2 end subscript are false    
  4. S subscript 1 end subscript is true, S subscript 2 end subscript is false    

The correct answer is: Both S subscript 1 end subscript comma S subscript 2 end subscript are true & S subscript 2 end subscript is reason for S subscript 1 end subscript.

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