Question

Find the types of the lines \[15x - 18y + 1 = 0\], \[12x + 10y - 3 = 0\] and \[6x + 66y - 11 = 0\]
A. Parallel
B. Perpendicular
C. Concurrent
D. None of these

Answer 1

Hint First, form a square matrix by using the coefficients of the variables of the given equations of lines. Then calculate the determinant of the square matrix. Check the sign of the determinant to determine the type of the lines.

Formula used:
The lines \[{a_1}x + {b_1}y + {c_1} = 0\], \[{a_2}x + {b_2}y + {c_2} = 0\] and \[{a_3}x + {b_3}y + {c_3} = 0\] are concurrent, if
\[\left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}} \right| = 0\]

Complete step by step solution:
The given equations of lines are \[15x - 18y + 1 = 0\], \[12x + 10y - 3 = 0\] and \[6x + 66y - 11 = 0\].
Let’s formed a matrix using the coefficients of the variables of the given equation.
\[A = \left[ {\begin{array}{*{20}{c}}{15}&{ - 18}&1\\{12}&{10}&{ - 3}\\6&{66}&{ - 11}\end{array}} \right]\]
We know that if the lines are parallel, then the lines have the same coefficients of \[x\] and \[y\] but the constant term is different.
In the above matrix, the coefficients of \[x\] and \[y\] are not equal.
So, the given lines are not parallel.
Hence, the option A is incorrect.
Also, if the lines are perpendicular, then the coefficient of \[y\] term in the first line is equal to the coefficient of \[x\] term in the second line and vice versa with one negative sign.
In the above matrix, the coefficients of \[y\] term in the first line is not equal to the coefficient of \[x\] term in the second line.
So, the given lines are not perpendicular.
Hence, the option B is incorrect.
Now calculate the determinant of the above matrix.
\[\left| A \right| = \left| {\begin{array}{*{20}{c}}{15}&{ - 18}&1\\{12}&{10}&{ - 3}\\6&{66}&{ - 11}\end{array}} \right|\]
\[ \Rightarrow \]\[\left| A \right| = 15\left[ {\left( {10} \right)\left( { - 11} \right) - \left( {66} \right)\left( { - 3} \right)} \right] - \left( { - 18} \right)\left[ {\left( {12} \right)\left( { - 11} \right) - \left( 6 \right)\left( { - 3} \right)} \right] + 1\left[ {\left( {12} \right)\left( {66} \right) - \left( 6 \right)\left( {10} \right)} \right]\]
\[ \Rightarrow \]\[\left| A \right| = 15\left[ { - 110 + 198} \right] + 18\left[ { - 132 + 18} \right] + 1\left[ {792 - 60} \right]\]
\[ \Rightarrow \]\[\left| A \right| = 15\left[ {88} \right] + 18\left[ { - 114} \right] + 1\left[ {732} \right]\]
\[ \Rightarrow \]\[\left| A \right| = 1320 - 2052 + 732\]
\[ \Rightarrow \]\[\left| A \right| = - 732 + 732\]
\[ \Rightarrow \]\[\left| A \right| = 0\]
Therefore, the given lines are concurrent.
Hence the correct option is C.

Note: Students often confused with the formula \[\left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}} \right| = 0\] and \[\left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&1\\{{a_2}}&{{b_2}}&1\\{{a_3}}&{{b_3}}&1\end{array}} \right| = 0\].
To check whether three lines \[{a_1}x + {b_1}y + {c_1} = 0,{a_2}x + {b_2}y + {c_2} = 0,{a_3}x + {b_3}y + {c_3} = 0\] are concurrent we use the formula \[\left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}} \right| = 0\].
To check whether the points\[\left( {{a_1},{b_1}} \right)\], \[\left( {{a_2},{b_2}} \right)\] and \[\left( {{a_3},{b_3}} \right)\] are collinear, we use \[\left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&1\\{{a_2}}&{{b_2}}&1\\{{a_3}}&{{b_3}}&1\end{array}} \right| = 0\].