Distance possible to draw a line which belongs to all the given family of lines `y-2x+1+lambda_1(2y-x-1)=0,3y-x-6+lambda_2(y-3x+6)=0,a x+y-2+lambda_3(6x+a y-a)=0` , then `a=4` (b) `a=3` `a=-2` (d) `a=2`

Options

A. a=4
B. a=3
C. a=-2
D. a=2

Answer

A

Explanation

The first two families of lines pass through (1,1) and (3,3), respectively. The point of intersection of the lines belonging to the third family of lines will lie on line y=x. Hence,
ax+x-2=0 and 6x+ax-a=0
`"or " (2)/(a+1) = (a)/(6+a)`
`"or " a^(2)-a-12=0 or (a-4)(a+3)=0`