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Q . EQUATION OF THE STRAIGHT LINE BELONGING TO THE FAMILY OF LINES :- ( x + y ) + k( 2x – y +1) =0 , THAT IS FARTHEST FROM ( 1 , – 3) IS :- 13y – 6x = 7 13y + 6x = 0 15y + 6x =7 15y – 6x = 7

 
Q . EQUATION OF THE STRAIGHT LINE BELONGING TO THE FAMILY OF LINES :-
( x + y ) + k( 2x – y +1) =0 , THAT IS FARTHEST FROM ( 1 , – 3) IS :-
  1. 13y – 6x = 7
  2. 13y + 6x = 0
  3. 15y + 6x =7
  4. 15y – 6x = 7

Grade:Select Grade

2 Answers

Himaja
40 Points
6 years ago
The point of intersection of given family of lines is the point of intersection of x+y=0 and 2x-y+1=0 and it is  (-1/3 1/3).
The required line belongs to the given family so it must pass through (-1/3,1/3) and it must be farthest from (1,-3). Tht means the perpendicular distance from the point to the line is maximum.
This occurs when the req. line is perpendicular to the line joining (1,-3) and (-1/3,1/3)
Therfore the eqn of req line is 15y-6x=7
 
Bharat Makkar
34 Points
6 years ago
thanks....got it

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