Consider the family of lines #(4a+3)x - (a+1)y - (2a+1) = 0# where #ainR#. Minimum area of the triangle which a member of this family with negative gradient can make with the positive semi axes?

A) 8
B) 6
C) 4
D) 2

1 Answer
Apr 1, 2018

Option (C)

Explanation:

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Given family of lines #(4a+3)x - (a+1)y - (2a+1) = 0#,where #ainR#

Rearranging we get

#4ax+3x - ay-y - 2a-1 = 0#

#or,(4x- y - 2)a+(3x-y-1) = 0#

Obviously the all lines of the family mus pass through the point of intersection of two lines represented by the following two equations

#4x- y - 2=0.....[1]#

and

#3x-y-1=0........[2]

Subtracting [2] from [1] we get #x=1#

Inserting #x=1# in [1] we get #y=2#

So the coordinates of common point of intersection of all lines of the family will be #(1,2)#

Any line passing through this common point #(1,2)# may be represented as #y-2=m(x-1)....[3]#,where m is the gradient of the line , a variable one.
Now rearranging [3] in intercept form we get

#y-2=m(x-1)#

#=>mx-y=m-2#

#=>x/((m-2)/m)+y/(2-m)=1#

So area of the triangle made by this line with the positive semi axes will be given by

#A=1/2xx(m-2)/mxx(2-m)=(-m^2+4m-4)/(2m)#

#=>A=-m/2+2-2/m#

Differentiating w r, to m we get

#(dA)/(dm)=-1/2+0+2/m^2#

Imposing the condition of minimization of #A#i.e . #(dA)/(dm)=0# we get

#-1/2+2/m^2=0#

#=>m^2=4#

#=>m=-2#, as per given condition the sraight line which forms minimum area with the positive semi axes must have negative gradient.

Hence minimum area of the the triangle should be for m=-2##

#A_"min"=(-(-2)^2+4(-2)-4)/(2(-2))=4#

[It matches with Option (C)]