Question

If the lines $ax+2y+1=0,bx+3y+1=0$ and $cx+4y+1=0$ are concurrent, then $a,b,c$ are in:

• A. A.P.
• B. G.P.
• C. H.P.
• D. None of these

Answer 1

The correct option is A

A.P.

Explanation for the correct option:

Given lines $ax+2y+1=0,bx+3y+1=0$ and $cx+4y+1=0$ are concurrent.

For the lines to be concurrent there determinant should be zero.

So, the determinant will be,

$\left|\begin{array}{ccc}a& 2& 1\\ b& 3& 1\\ c& 4& 1\end{array}\right|=0$

Upon solving we get,

$$\begin{gathered} \Rightarrow a(3-4)-2(b-c)+1(4b-3c)=0 \\ \Rightarrow -a-2b+2c+4b-3c=0 \\ \Rightarrow -a+2b-c=0 \\ \Rightarrow 2b=a+c \end{gathered}$$

Thus, the result is the the condition for arithmetic progression (A.P.). So, $a,b,c$ are in an A.P.

Therefore, option (A) is the correct option.