For what value of k, the following system of equations will represent the coincident lines?
$
x + 2y + 7 = 0 \\
2x + ky + 14 = 0 \\
$
Answer
644.4k+ views
Hint- Here, we will proceed by comparing the given pair of linear equations with any general pair of linear equations i.e., ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$. Then using the condition for having infinitely many solutions i.e., $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$.
Complete Step-by-Step solution:
Given system of linear equations is $x + 2y + 7 = 0{\text{ }} \to {\text{(1)}}$ and $2x + ky + 14 = 0{\text{ }} \to {\text{(2)}}$
As we know that for any pair of linear equations ${a_1}x + {b_1}y + {c_1} = 0{\text{ }} \to {\text{(3)}}$ and ${a_2}x + {b_2}y + {c_2} = 0{\text{ }} \to {\text{(4)}}$ to be coincident lines, they should have infinitely many solutions. Here the condition which must be satisfied is that the ratio of the coefficients of x should be equal to the ratio of the coefficients of y which further should be equal to the ratio of the constant terms in the pair of linear equations.
The condition is $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}{\text{ }} \to (5{\text{)}}$
By comparing equations (1) and (3), we get
${a_1} = 1,{b_1} = 2,{c_1} = 7$
By comparing equations (2) and (4), we get
${a_2} = 2,{b_2} = k,{c_2} = 14$
For the given pair of linear equations to have inconsistent solution, equation (5) must be satisfied
By equation (5), we can write
\[
\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}} \\
\Rightarrow \dfrac{1}{2} = \dfrac{2}{k} = \dfrac{7}{{14}} \\
\Rightarrow \dfrac{1}{2} = \dfrac{2}{k} = \dfrac{1}{2}{\text{ }} \to {\text{(6)}} \\
\]
By equation (6), we can write
\[
\Rightarrow \dfrac{1}{2} = \dfrac{2}{k} \\
\Rightarrow k = 2 \times 2 = 4 \\
\]
Therefore, the required value of k for which the given system of linear equations will represent the coincident lines (or has infinitely many solutions) is 4.
Note- Any general pair of linear equations which are given by ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$ can also have unique solution (consistent solution) if the condition $\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}$ is satisfied. Also, for these pair of linear equations to have no solution, the condition $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$ should always be satisfied.
Complete Step-by-Step solution:
Given system of linear equations is $x + 2y + 7 = 0{\text{ }} \to {\text{(1)}}$ and $2x + ky + 14 = 0{\text{ }} \to {\text{(2)}}$
As we know that for any pair of linear equations ${a_1}x + {b_1}y + {c_1} = 0{\text{ }} \to {\text{(3)}}$ and ${a_2}x + {b_2}y + {c_2} = 0{\text{ }} \to {\text{(4)}}$ to be coincident lines, they should have infinitely many solutions. Here the condition which must be satisfied is that the ratio of the coefficients of x should be equal to the ratio of the coefficients of y which further should be equal to the ratio of the constant terms in the pair of linear equations.
The condition is $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}{\text{ }} \to (5{\text{)}}$
By comparing equations (1) and (3), we get
${a_1} = 1,{b_1} = 2,{c_1} = 7$
By comparing equations (2) and (4), we get
${a_2} = 2,{b_2} = k,{c_2} = 14$
For the given pair of linear equations to have inconsistent solution, equation (5) must be satisfied
By equation (5), we can write
\[
\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}} \\
\Rightarrow \dfrac{1}{2} = \dfrac{2}{k} = \dfrac{7}{{14}} \\
\Rightarrow \dfrac{1}{2} = \dfrac{2}{k} = \dfrac{1}{2}{\text{ }} \to {\text{(6)}} \\
\]
By equation (6), we can write
\[
\Rightarrow \dfrac{1}{2} = \dfrac{2}{k} \\
\Rightarrow k = 2 \times 2 = 4 \\
\]
Therefore, the required value of k for which the given system of linear equations will represent the coincident lines (or has infinitely many solutions) is 4.
Note- Any general pair of linear equations which are given by ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$ can also have unique solution (consistent solution) if the condition $\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}$ is satisfied. Also, for these pair of linear equations to have no solution, the condition $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$ should always be satisfied.
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