Question

If the lines `a x+2y+1=0,b x+3y+1=0a n dc x+4y+1=0` are concurrent, then `a ,b ,c` are in a. A.P. b. G.P. c. H.P. d. none of these

Answer 1

Correct Answer - False
Given lines are
`ax+2y+1=0`
`bx+3y+1=0`
From Eq. (i), on putting `y=(-ax-1)/(2)` in Eq. (ii), we get
`bx-(3)/(2)(ax+1)+1=0`
`rArr 2bx-3ax-3+2=0`
`rArr x(2b-3a)=1rArrx=(1)/(2b-3a)`
Now, using `x=(1)/(2b-3a)` in Eq. (i), we get
`(a)/(2b-3a)+2y+1=0`
`rArr 2y=-[(a+2b-3a)/(2b-3a)]`
`rArr 2y=(-(-2b-2a))/(2b-3a)`
`rArr y=((a-b))/(2b-3a)`
So, the point of intersection is `((1)/(2b-3a),(a-b)/(2b-3a))`
Since, this point lies on `cx+4y+1=0`, then
`(c)/(2b-3a)+(4(a-b))/(2b-3a)+1=0`
`rArr c+4a-4b+2b-3a=0`
`rArr -2b+a+c=0rArr2b=a+c`
Hence, the given statement is false.