Question

If 4a+5b+6c=0 then the set of lines ax+by+c=0 are concurrent at the point

• A. $(\frac{2}{3},\frac{5}{6})$
• B. $(\frac{1}{3},\frac{1}{2})$
• C. $(\frac{1}{2},\frac{4}{3})$
• D. $(\frac{1}{3},\frac{7}{3})$

Answer 1

The correct option is A

$(\frac{2}{3},\frac{5}{6})$

4a+5b+6c=0
(6x)a+(6y)b+6c=0 $\to$ [Multiply given set of lines ax+by+c=0 with '6']
Now on comparing 6x=4 and 6y =5
⇒ (x,y) = $(\frac{2}{3},\frac{5}{6})$
$\therefore$ ax+by+c=0 must passes through $(\frac{2}{3},\frac{5}{6})$