the equation of the line parallel to the lines L1 : x + 2y - 5 =0 and L2 : x +2y + 9 =0 and dividing the distance between the lines L1 and L2 in the ratio  1 : 6 ( internally ) is

choose a pt on any line draw a line perpendicular to it using the slope pt form

ull get pt of intersections thus use the section formula

A line parallel to the given two lines would be of the form x+2y+a=0

And a line perpendicular to all three would be of the form y=2x+c..now solving this line with all three lines we get the points of intersection as..

With L1 : ((5-2c)/5,(10+c)/5)...let's say point A

With L2 : ((-9-2c)/5,(-36+2c)/5).....ptB

 

With x+2y+a=0 : ((-a-2c)/5, (-4a+2c)/10).......ptC

now ptB divides the line AC in the ration 1:6 internally ..

.: using section formula

(-a-2c)/5 = ((-9-2c)/5 + 6((5-2c)/5)) / 7

=>-7a-14c=-9+30-2c-12c

=>a=-3..

so the rqd line is x+2y-3=0...(ans)

yea..