Maths-
General
Easy

Question

Consider A(0,1) and B(2,0) and P be a point on the line 4x+3y+9=0, co-ordinates of P such vertical line PA minus PB vertical line is maximum is

  1. open parentheses fraction numerator negative 12 over denominator 5 end fraction comma 17 over 5 close parentheses
  2. open parentheses fraction numerator negative 84 over denominator 5 end fraction comma 13 over 5 close parentheses
  3. open parentheses fraction numerator negative 6 over denominator 5 end fraction comma 17 over 5 close parentheses
  4. open parentheses fraction numerator negative 6 over denominator 5 end fraction comma fraction numerator negative 12 over denominator 5 end fraction close parentheses

hintHint:

The point p should satisfy the line AP and BP then find the lines passing through AP and BP gives point of intersection which is nothing but P.

The correct answer is: open parentheses fraction numerator negative 84 over denominator 5 end fraction comma 13 over 5 close parentheses


    Given That:
                         Consider A(0,1) and B(2,0) and P be a point on the line 4x+3y+9=0, co-ordinates of P such vertical line PA minus PB vertical line is maximum is
     
    >>> Given:PAPB is maximum.
    >>> A line passes through P having the equation 4x+3y+9=0
    >>> Maximum value of PAPB=AB
    >>> Equation of the line AB=y0=2001(x0)
                                               ⇒2y=x or x+2y=0
    >>> The intersection of the lines x=2y and 4x+3y+9=0 is
                                        4×2y+3y+9=0
                                             ⇒8y+3y=9
                                             ⇒5y=9
                                             ⇒y=9 over 5
                         ∴x=2y=2×9 over 5=negative 18 over 5
    >>> Hence the point is (negative 18 over 5, 9 over 5).

     Hence the point is (negative 18 over 59 over 5).

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