Question

The lines "ax+by+c=0" ,where 3a+2b+4c=0 are concurrent at which point?

Answer 1

The lines $ax+by+c=0$ , where $3a+2b+4c=0$ re concurrent.

We have to identify the point at which the given lines are concurrent.

Consider the line $3a+2b+4c=0$

Dividing the above equation by '4', we get

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

$\frac{3a}{4} + \frac{b}{2}+ \frac{c}{1}=0$

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

Comparing this equation to the line ax+by+c=0, we get a= $\frac{3}{4}$ and b= $\frac{1}{2}$

Therefore, the lines ax+by+c=0 and 3a+2b+4c=0 are concurrent at the point $(\frac{3}{4}, \frac{1}{2})$.

Answer 2

Step-by-step explanation:

so this is the answer please mark as well