
The lines $2x + y - 1 = 0,ax + 3y - 3 = 0$ and $3x + 2y - 2 = 0$ are concurrent for
1. All $a$
2. $a = 4$ only
3. $ - 1 \leqslant a \leqslant 3$
4. $a > 0$ only
Answer
162k+ views
Hint: Concurrent lines are lines that intersect at the same precise place. The definition of concurrent is that anything is happening at the same moment or point. Here, in this question, make the matrix of given lines. As the lines are concurrent, the determinant of the required matrix will be equal to zero. Solve and find the value of $a$.
Formula used:
Determinant –
$\left| {\begin{array}{*{20}{c}}
a&b&c \\
d&e&f \\
g&h&i
\end{array}} \right| = a\left( {ei - fh} \right) - b\left( {di - fg} \right) + c\left( {dh - eg} \right)$
Complete step by step solution:
Given that,
The lines $2x + y - 1 = 0,ax + 3y - 3 = 0$ and $3x + 2y - 2 = 0$ are concurrent.
$ \Rightarrow \left| A \right| = 0$ where,$A$is the required matrix of the given lines
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
2&1&{ - 1} \\
a&3&{ - 3} \\
3&2&{ - 2}
\end{array}} \right| = 0$
\[2\left( {\left( {3 \times \left( { - 2} \right) - \left( { - 3} \right) \times 2} \right) - 1\left( {a \times \left( { - 2} \right) - \left( { - 3} \right) \times 3} \right)} \right) + \left( { - 1} \right)\left( {a \times \left( 2 \right) - 3 \times 3} \right) = 0\]
\[2\left( { - 6 + 6} \right) - 1\left( { - 2a + 9} \right) - 1\left( {2a - 9} \right) = 0\]
\[0 = 0\]
Here both sides are zero.
It implies that the given lines are concurrent for all values of $a$
Hence, Option (1) is the correct answer.
Note: The key concept involved in solving this problem is the good knowledge of determinant. Students must know that Determinant of $3 \times 3$formula is derived from the $2 \times 2$ matrix formula only.
Let, $P = \left[ {\begin{array}{*{20}{c}}
a&b&c \\
d&e&f \\
g&h&i
\end{array}} \right]$be the $3 \times 3$matrix
Then, $\left| P \right| = \left| {\begin{array}{*{20}{c}}
a&b&c \\
d&e&f \\
g&h&i
\end{array}} \right|$
$ = a\left| {\begin{array}{*{20}{c}}
e&f \\
h&i
\end{array}} \right| - b\left| {\begin{array}{*{20}{c}}
d&f \\
g&i
\end{array}} \right| + c\left| {\begin{array}{*{20}{c}}
d&e \\
g&h
\end{array}} \right|$
$ = a\left( {ei - fh} \right) - b\left( {di - fg} \right) + c\left( {dh - eg} \right)$
Formula used:
Determinant –
$\left| {\begin{array}{*{20}{c}}
a&b&c \\
d&e&f \\
g&h&i
\end{array}} \right| = a\left( {ei - fh} \right) - b\left( {di - fg} \right) + c\left( {dh - eg} \right)$
Complete step by step solution:
Given that,
The lines $2x + y - 1 = 0,ax + 3y - 3 = 0$ and $3x + 2y - 2 = 0$ are concurrent.
$ \Rightarrow \left| A \right| = 0$ where,$A$is the required matrix of the given lines
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
2&1&{ - 1} \\
a&3&{ - 3} \\
3&2&{ - 2}
\end{array}} \right| = 0$
\[2\left( {\left( {3 \times \left( { - 2} \right) - \left( { - 3} \right) \times 2} \right) - 1\left( {a \times \left( { - 2} \right) - \left( { - 3} \right) \times 3} \right)} \right) + \left( { - 1} \right)\left( {a \times \left( 2 \right) - 3 \times 3} \right) = 0\]
\[2\left( { - 6 + 6} \right) - 1\left( { - 2a + 9} \right) - 1\left( {2a - 9} \right) = 0\]
\[0 = 0\]
Here both sides are zero.
It implies that the given lines are concurrent for all values of $a$
Hence, Option (1) is the correct answer.
Note: The key concept involved in solving this problem is the good knowledge of determinant. Students must know that Determinant of $3 \times 3$formula is derived from the $2 \times 2$ matrix formula only.
Let, $P = \left[ {\begin{array}{*{20}{c}}
a&b&c \\
d&e&f \\
g&h&i
\end{array}} \right]$be the $3 \times 3$matrix
Then, $\left| P \right| = \left| {\begin{array}{*{20}{c}}
a&b&c \\
d&e&f \\
g&h&i
\end{array}} \right|$
$ = a\left| {\begin{array}{*{20}{c}}
e&f \\
h&i
\end{array}} \right| - b\left| {\begin{array}{*{20}{c}}
d&f \\
g&i
\end{array}} \right| + c\left| {\begin{array}{*{20}{c}}
d&e \\
g&h
\end{array}} \right|$
$ = a\left( {ei - fh} \right) - b\left( {di - fg} \right) + c\left( {dh - eg} \right)$
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