Question

The set of lines $ax+by+c=0$, where $3a+2b+4c=0$, is concurrent at the point :

• A. $(\frac{3}{4},\frac{3}{4})$
• B. $(\frac{1}{2},\frac{1}{2})$
• C. $(\frac{3}{4},\frac{1}{2})$
• D. $(1,1)$

Answer 1

The correct option is C $(\frac{3}{4},\frac{1}{2})$

$3a+2b+4c=0$

$\therefore \frac{3}{4}a+\frac{2b}{4}+c=0$

$\therefore \frac{3a}{4}+\frac{b}{2}+c=0$ .......... $\left(1\right)$

Now $ax+by+c=0$ .......... $\left(2\right)$

From $\left(1\right)$ & $\left(2\right)$

$x=\frac{3}{4};y=\frac{1}{2}$

$\therefore$ the lines Concurrent at $(\frac{3}{4},\frac{1}{2})$