Questions
The straight lines 4ax + 3by + c = 0, where a + b + c = 0, are concurrent at the point
a
(4, 3)
b
(1/4, 1/3)
c
(1/2,1/3)
d
None of these
detailed solution
Correct option is C
The set of lines is 4ax + 3by + c = 0, where a + b + c = 0. Eliminating c, we get 4ax+3by−(a+b)=0 or a(4x−1)+b(3y−1)=0This 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) .Talk to our academic expert!
Similar Questions
If the lines ax+by+c = 0, bx+cy+a = 0 and cx+ay+b=0 are concurrent then the point of concurrency is
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