MCQ
The straight lines $4ax + 3by + c = 0$ where $a + b + c = 0$, will be concurrent, if point is
  • A
    $(4,\, 3)$
  • $(1/4, \,1/3)$
  • C
    $(1/2, \,1/3)$
  • D
    None of these

Answer

Correct option: B.
$(1/4, \,1/3)$
b
(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.$x = \frac{1}{4},\,y = \frac{1}{3}$i.e., $\left( {\frac{1}{4},\,\frac{1}{3}} \right)$.

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