# If the three distinct lines \[x + 2ay + a = 0\], \[x + 3by + b = 0\] and \[x + 4ay + a = 0\] are concurrent, then the point $(a,b)$ lies on a:

$A$. Circle

$B$. Hyperbola

$C$. Straight line

$D$. Parabola

Answer

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327.9k+ views

Hint: - Determinant of a concurrent line is always zero. By solving the determinant we get the relation between point (a, b). That relation gives us the result where the point lies on.

We know that when the lines are concurrent it means that the determinant of the coefficient of the line must be zero.

Here the given equation of lines are

\[x + 2ay + a = 0\]

\[x + 3by + b = 0\]

\[x + 4ay + a = 0\]

By the property of concurrence of line,

$\left| \begin{gathered}

1{\text{ 2a a}} \\

{\text{1 3b b}} \\

{\text{1 4a a}} \\

\end{gathered} \right| = 0$

By applying row transformation \[{R_{1}} \to {R_{1}} - {R_3}\]

$\left| \begin{gathered}

{\text{0 }} - {\text{2a 0}} \\

{\text{1 3b b}} \\

{\text{1 4a a}} \\

\end{gathered} \right| = 0$

And now we open the determinant and the formula for opening the determinant is

\[0 \cdot \left| \begin{gathered}

3{\text{b b}} \\

{\text{4a a}} \\

\end{gathered} \right| - ( - 2{\text{a)}} \cdot \left| \begin{gathered}

1{\text{ b}} \\

{\text{1 a}} \\

\end{gathered} \right| + 0 \cdot \left| \begin{gathered}

1{\text{ 3b}} \\

{\text{1 4a}} \\

\end{gathered} \right| = 0\]

$0\left( {(a \times 3b) - (4a \times b)} \right) - ( - 2a)\left( {(a \times 1) - (b \times 1)} \right) + 0\left( {(4a \times 1) - (1 \times 3b)} \right)$

$ \Rightarrow 2a(a - b) = 0$

But, $a$ cannot be $0$ as lines 1 and 3 becomes identical.

$ \Rightarrow a - b = 0$

Hence,$(a,b)$ lies on the line $x - y = 0$

So option C is the correct answer.

Note: - In such a type of question we have to always apply the property that is given as a hint in question. In this question the hint is concurrence of lines. So we apply the property of concurrence of line to get the relation between the points where they lie.

We know that when the lines are concurrent it means that the determinant of the coefficient of the line must be zero.

Here the given equation of lines are

\[x + 2ay + a = 0\]

\[x + 3by + b = 0\]

\[x + 4ay + a = 0\]

By the property of concurrence of line,

$\left| \begin{gathered}

1{\text{ 2a a}} \\

{\text{1 3b b}} \\

{\text{1 4a a}} \\

\end{gathered} \right| = 0$

By applying row transformation \[{R_{1}} \to {R_{1}} - {R_3}\]

$\left| \begin{gathered}

{\text{0 }} - {\text{2a 0}} \\

{\text{1 3b b}} \\

{\text{1 4a a}} \\

\end{gathered} \right| = 0$

And now we open the determinant and the formula for opening the determinant is

\[0 \cdot \left| \begin{gathered}

3{\text{b b}} \\

{\text{4a a}} \\

\end{gathered} \right| - ( - 2{\text{a)}} \cdot \left| \begin{gathered}

1{\text{ b}} \\

{\text{1 a}} \\

\end{gathered} \right| + 0 \cdot \left| \begin{gathered}

1{\text{ 3b}} \\

{\text{1 4a}} \\

\end{gathered} \right| = 0\]

$0\left( {(a \times 3b) - (4a \times b)} \right) - ( - 2a)\left( {(a \times 1) - (b \times 1)} \right) + 0\left( {(4a \times 1) - (1 \times 3b)} \right)$

$ \Rightarrow 2a(a - b) = 0$

But, $a$ cannot be $0$ as lines 1 and 3 becomes identical.

$ \Rightarrow a - b = 0$

Hence,$(a,b)$ lies on the line $x - y = 0$

So option C is the correct answer.

Note: - In such a type of question we have to always apply the property that is given as a hint in question. In this question the hint is concurrence of lines. So we apply the property of concurrence of line to get the relation between the points where they lie.

Last updated date: 01st Jun 2023

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