Question

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

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

Answer 1

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

3a+2b+4c=03a+2b+4c=0

3434a+2b4+4cca+2b4+4cc=0=0

=> 3434a+12a+12b+c=0b+c=0

=> The set of lines ax+by+c=0ax+by+c=0 passes through (34,12)