6. For what value of \(k\), do the equations \(3x - y + 8 = 0\) and \(6x - ky = -16\) represent coincident lines?
\(\dfrac{1}{2}\)
\(-\dfrac{1}{2}\)
2
-2
Step 1: Write the equations in standard form.
First equation: \(3x - y + 8 = 0\)
Here \(a_1 = 3\), \(b_1 = -1\), \(c_1 = 8\).
Second equation: \(6x - ky = -16\)
Bring RHS to LHS: \(6x - ky + 16 = 0\)
Here \(a_2 = 6\), \(b_2 = -k\), \(c_2 = 16\).
Step 2: Condition for coincident lines.
For two lines to be coincident:
\( \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} \)
Step 3: Compute the ratios.
\(\dfrac{a_1}{a_2} = \dfrac{3}{6} = \dfrac{1}{2}\)
\(\dfrac{c_1}{c_2} = \dfrac{8}{16} = \dfrac{1}{2}\)
So, \(\dfrac{b_1}{b_2}\) must also equal \(\dfrac{1}{2}\).
Step 4: Solve for \(k\).
\(\dfrac{b_1}{b_2} = \dfrac{-1}{-k} = \dfrac{1}{k}\)
Equating: \(\dfrac{1}{k} = \dfrac{1}{2}\)
\(\Rightarrow k = 2\)
Final Answer: The lines are coincident when \(k = 2\).