Maths-
General
Easy

Question

In what direction a line be drawn through the point (1, 2) so that its points of intersection with the line x plus y equals 4 blankis at a distance square root of 6 divided by 3 from the given point

  1. 30 to the power of ring operator end exponent    
  2. 45 to the power of ring operator end exponent    
  3. 60 to the power of ring operator end exponent    
  4. 75 to the power of ring operator end exponent    

The correct answer is: 75 to the power of ring operator end exponent


    Let the required line through the point (1, 2) be inclined at antext  angle  end text theta to the axis of x. Then its equation is
    fraction numerator x minus 1 over denominator cos invisible function application theta end fraction equals fraction numerator y minus 2 over denominator sin invisible function application theta end fraction equals r
    where r is the distance of any point (x, y) on the line from the point (1, 2). The coordinates of any point on the line given in Eq. (1) are left parenthesis 1 plus r c o s invisible function application theta comma 2 plus r s i n invisible function application theta right parenthesis If this point is at a distance
    square root of 6 divided by 3 text  from the point  end text left parenthesis 12 right parenthesis text  , then  end text r equals square root of 6 divided by 3 text  . Therefore, the point is end text
    open parentheses 1 plus fraction numerator square root of 6 over denominator 3 end fraction c o s invisible function application theta comma 2 plus fraction numerator square root of 6 over denominator 3 end fraction s i n invisible function application theta close parentheses
    However, this point lies on the line x y + = 4. So,
    fraction numerator square root of 6 over denominator 3 end fraction left parenthesis c o s invisible function application theta plus s i n invisible function application theta right parenthesis equals 1 text  or  end text s i n invisible function application theta plus c o s invisible function application theta equals fraction numerator 3 over denominator square root of 6 end fraction
    rightwards double arrow fraction numerator 1 over denominator square root of 2 end fraction s i n invisible function application theta plus fraction numerator 1 over denominator square root of 2 end fraction c o s invisible function application theta equals fraction numerator square root of 3 over denominator 2 end fraction
    text (Dividing both sides by  end text square root of 2 text  ) end text
    table row cell rightwards double arrow s i n invisible function application open parentheses theta plus 45 to the power of ring operator end exponent close parentheses equals s i n invisible function application 60 to the power of ring operator end exponent text end text text o end text text r end text text end text s i n invisible function application 120 to the power of ring operator end exponent end cell row cell rightwards double arrow theta equals 15 to the power of ring operator end exponent text end text text o end text text r end text text end text 75 to the power of ring operator end exponent end cell end table

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