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The straight lines 4a x+3b y+c=0 , where...

The straight lines `4a x+3b y+c=0` , where `a+b+c` `(4,3)` (b) `(1/4,1/3)` `(1/2,1/3)` (d) none of these

A

(4,3)

B

(1/4,1/3)

C

(1/2,1/3)

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
B
The set of lines is 4ax+3by+c=0, where a+b+c=0.
Eliminating c, we get
4ax+3by-(a+b)=0
a(4x-1)+b(3y-1)=0
This passes through the intersection of the lines 4x-1=0 and 3y-1=0, i.e., x 1/4, y=1/3, i.e.,(1/4, 1/3).
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