Question

Two lines are given by \[{\left( {x - 2y} \right)^2} + k\left( {x - 2y} \right) = 0\]. The value of \[k\] so that the distance between them is \[3,\]is
A. \[\dfrac{1}{{\sqrt 3 }}\]
B. \[\dfrac{2}{5}\]
C. $\pm 3\sqrt{5}$
D. None of these

Answer 1

Hint:
In this question, we have to find the value of k for this, first we will find the equations of the two parallel lines that are represented by the given equation and then find the value of using distance between the two using distance formula for parallel lines and the given distance.

Formula used:
The distance formula for two parallel lines are given as:
1. \[\left| {\dfrac{{\left( {{c_1} - {c_2}} \right)}}{{\sqrt {\left( {{a^2} + {b^2}} \right)} }}} \right| = d\]
2. \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\]

Complete step-by-step solution:
Given that \[{\left( {x - 2y} \right)^2} + k\left( {x - 2y} \right) = 0\]…(1)
Firstly, we will write the given equation as for finding the equations of respective pair lines by using the formula \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\].
\[\begin{array}{l}{x^2} + 2x\left( { - 2y} \right) + {\left( { - 2y} \right)^2} + k\left( {x - 2y} \right) = 0\\{x^2} - 4xy + 4{y^2} + k\left( {x - 2y} \right) = 0\\{x^2} - 2xy - 2xy + 4{y^2} + k\left( {x - 2y} \right) = 0\\x\left( {x - 2y} \right) - 2y\left( {x - 2y} \right) + k\left( {x - 2y} \right) = 0\end{array}\]
Now, we will take out \[\left( {x - 2y} \right)\] as the common factor, we get
\[\left( {x - 2y} \right)\left( {x - 2y + k} \right) = 0\]
Now, we will compare each equation of left hand side with right hand side, we get
\[\begin{array}{c}x - 2y = 0\\x - 2y + k = 0\end{array}\]
As we know, we are given that the distance between two lines is \[3\].
Further, we will use the formula of distance between two parallel lines \[ax + by + {c_1} = 0\] and \[ax + by + {c_2} = 0\]is given by\[\left| {\dfrac{{\left( {{c_1} - {c_2}} \right)}}{{\sqrt {\left( {{a^2} + {b^2}} \right)} }}} \right|\], we get
\[3 = \left| {\dfrac{{\left( {k - 0} \right)}}{{\sqrt {\left( {{1^2} + {2^2}} \right)} }}} \right|\]
Furthermore, we will simplify the above equation, we get
\[\begin{array}{l} \pm 3 = \dfrac{k}{{\sqrt 5 }}\\k = \pm 3\sqrt 5 \end{array}\]
Therefore, the value of \[k\] for the given two lines \[{\left( {x - 2y} \right)^2} + k\left( {x - 2y} \right) = 0\] whose distance between them is \[3\] is \[ \pm 3\sqrt 5 \].
Hence, the option (C) is correct.

Note
To check whether our answer is correct or not by checking that the two lines obtained are parallel or not is from the fact that Coefficients of \[x\] and \[y\] are the same for both the lines but the constant is different.