Question

The lines $15x-18y+1=0,12x+10y-3=0$and $6x+66y-11=0$ are

• A. Parallel
• B. Perpendicular
• C. Concurrent
• D. None of these

Answer 1

The correct option is C

Concurrent

Find the nature of the given lines

The equation of the three lines is,

$$\begin{gathered} 15x-18y+1=0, \\ 12x+10y-3=0 \end{gathered}$$

$6x+66y-11=0$

For, the lines to be the concurrent, determinant of all the coefficient $=0$

So, the determinant is

$$\begin{gathered} \left|\begin{array}{ccc}15& -18& 0\\ 12& 10& 0\\ 6& 66& 0\end{array}\right| \\ =15(0-0)+18(0-0)+0\left(\right(12\times 66)-\left(6\times 10\right)) \\ =0 \end{gathered}$$

$\therefore$ The lines are concurrent

Hence, the correct option is $C$.