If the distance between the parallel lines x+2y+3=0 and x+2y+k=0 is √5, then the value of k is?

Q.If the distance between the parallel lines x+2y+3=0 and x+2y+k=0 is √5, then the value of k is?
Solution:

 As shown in the graph the given two lines, their slopes, the perpewndicular distance between two lines and the coordinates of various points which would immideatly help solving this question given,
              The perpendicular distance between two lines(AB)=√5

               The value of tanθ= 1/2 = slope of two lines
In the triangle ABC,      tanθ=1/2=[√5]/BC, Therefore, BC=2√5
                                       AC^2    =AB^2+BC^2
                                       (k-3)^2 =(√5)^2+(2√5)^2
                Therefore,                k  = 8(ans)


                                                                                                                      -Mimansa