Question

The set of lines $ax+by+c=0$ where $a+2b+3c=0$ are concurrent at the point:

• A. $(1,1)$
• B. $(2,3)$
• C. $(3,\frac{2}{3})$
• D. $(\frac{1}{3},\frac{2}{3})$

Answer 1

Answer: (A)

$(1,1)$

$\Rightarrow c=-ax-by$

$\Rightarrow a+2b-3(ax+by)=0$

$\Rightarrow a(1-3x)+b(2-3y)=0$

As the equation is identical

$\Rightarrow 1-3x=0,2-3y=0$

$\Rightarrow x=\frac{1}{3},y=\frac{2}{3}$

$\Rightarrow$ Concurrency point $=(\frac{1}{3},\frac{2}{3})$