Question

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

Answer 1

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Given lines are,
$$\begin{gathered} ax+2y+1=0...\left(1\right) \\ bx+y+1=0...\left(2\right) \\ cx+4y+1=0...\left(3\right)\mathrm{are}\mathrm{concurrent} \end{gathered}$$

Solving (1) and (2),

$$\begin{gathered} \frac{x}{2-1}=\frac{-y}{b-a}=\frac{1}{a-2b} \\ \mathrm{i}.\mathrm{e}.x=\frac{1}{a-2b},y=\frac{a-b}{a-2b} \end{gathered}$$

Since (x, y) lie on (3) (∵ lines are concurrent)

$$\begin{gathered} \therefore \frac{c}{a-2b}+4\left(\frac{a-b}{a-2b}\right)+1=0 \\ c+4a-4b+a-2b=0 \\ \mathrm{i}.\mathrm{e}.3a-6b+c=0 \end{gathered}$$