arora007 wrote:
the shaded region can be taken as all the values of y which are lesser than (5/4)x-10 and greater than -(1/2)x-3.
I was unruffled by this step... somehow found it tough!
arora i hope u got the solution until the 2 equations.
Further to that:
Consider line l, there are two regions, one above it (where (0,0) lies) and one below it. Similarly for line k there are two regions, above and below.
Now the requirement in our question is to find out the equation for the shaded region, so we can very well establish one fact that the required region is the common part of area below line l and above line k.
now consider any particular point which you know will lie below line l. for instance (8,-1) does not lie on line l and is below it.
put this value in the equation for line l (y=(5/4)x-10).
we will get
L.H.S. =-1
R.H.S. = (5/4)*8-10 = 0
hence L.H.S < R.H.S.
this means that the equation
y<(5/4)x-10 represents the area below line l.
similarly for line k we can find out the equation for the region above line k. Try with point (0,0).
so we get the required solution.
But i still have the confusion if the equality should be there or not, because there are points in line l and in line k which can never come in the shaded region hence no equality.
Ishaan
Practice makes does a man perfect, but a planned practice takes the man beyond perfection!!!
