Question

The straight lines $4ax+3by+c=0$, where $a+b+c=0$, are concurrent at the point

• A. $(4,3)$
• B. $(\frac{1}{4},\frac{1}{3})$
• C. $(\frac{1}{2},\frac{1}{3})$
• D. None of these

Answer 1

The correct option is B $(\frac{1}{4},\frac{1}{3})$

$4ax+3by+c=\mathrm{0.........}\left(i\right)$ and $a+b+c=\mathrm{0..........}\left(ii\right)$
Putting (ii) in (i) we get
$4ax+3by-a-b=0$
$⟹a(4x-1)+b(3y-1)=0$ this always passes through $x=\frac{1}{4},y=\frac{1}{3}$