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

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

 

thanks....got it