Question

If two straight lines $3x+4y+2=0\And 9x+12y+k=0$ represent coincident lines, then find k.

Answer 1

Hint: When two lines are coincident then the ratio of their coefficients of x is equal to the ratio of coefficients of y and both these ratios are equal to the ratio of the constants of the equation. Then applying this property in the given equations, we have $\dfrac{3}{9}=\dfrac{4}{12}=\dfrac{2}{k}$.From this relation; we can find the value of k.

Complete step-by-step solution -
Let us suppose that two equations of the straight lines are given as:
 $\begin{align}
  & {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0 \\
 & {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0 \\
\end{align}$
When the two straight lines are given then the following three conditions could be possible:
First is that the two straight lines intersect each other. When two straight lines intersect each other then we will get a unique solution. The condition for the unique solution is:
$\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}$
Second is that the two straight lines are parallel to each other then we have no solution for which the condition is:
$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}$
Third is that when the two straight lines are coincident meaning the two straight lines are overlapping with each other then we have infinitely many solutions and condition for the infinitely many solutions is:
$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$………..Eq. (1)
It is given that the two straight lines represent the coincident lines so we are going to use the third condition that we have discussed above.
 Now, we are going to compare these straight lines with the straight lines given in the question i.e.
$\begin{align}
  & 3x+4y+2=0 \\
 & 9x+12y+k=0 \\
\end{align}$
 then the values of ${{a}_{1}},{{b}_{1,}}{{a}_{2}},{{b}_{2}},{{c}_{1}},{{c}_{2}}$ are as follows:
$\begin{align}
  & {{a}_{1}}=3,{{a}_{2}}=9 \\
 & {{b}_{1}}=4,{{b}_{2}}=12 \\
 & {{c}_{1}}=2,{{c}_{2}}=k \\
\end{align}$
Substituting these values in the eq. (1) we get,
$\dfrac{3}{9}=\dfrac{4}{12}=\dfrac{2}{k}$
From the above relation the simplification of first two ratios is giving the value of:
$\dfrac{3}{9}=\dfrac{4}{12}=\dfrac{1}{3}$
Now, equating the above value with $\dfrac{2}{k}$ we get,
$\dfrac{1}{3}=\dfrac{2}{k}$
On cross-multiplication of the above relation we get,
$k=6$
Hence, the value of k that we have got from above is equal to 6.

Note: If the question is saying that the given straight line equations have infinitely many solutions instead of representing coincident lines then also the two straight line equations are following the same relation as that we have shown above.
$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$.