Question

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

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

Answer 1

The correct option is D

$\left(\frac{3}{4},\frac{1}{2}\right)$

Find the point of concurrency

Given, $ax+by+c=0....\left(1\right)$

Now given, $3a+2b+4c=0$

Divide it by $4$

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

comparing $\left(1\right)$ and $\left(2\right)$,

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

Hence the correct answer is $D$, $\left(\frac{3}{4},\frac{1}{2}\right)$.