Question

The value of $\left| l \right|+\left| k \right|$ if the equation $4{{x}^{2}}+2lxy+{{y}^{2}}+6x+ky-10=0$ represents a pair of parallel lines.
A. 3
B. 4
C. 2
D. 5

Answer 1

Hint: To solve this question, we should use the properties of pair of parallel lines. We know that the parallel lines have the corresponding coefficients equal and the only difference is the constant term. Let us assume the parallel lines in our question are $ax+by+{{c}_{1}}=0,ax+by+{{c}_{2}}=0$. By multiplying them and equating the coefficient terms with the given equation $4{{x}^{2}}+2lxy+{{y}^{2}}+6x+ky-10=0$, we can get the required values.

Complete step-by-step solution:
We are given the equation $4{{x}^{2}}+2lxy+{{y}^{2}}+6x+ky-10=0$ which represents a pair of parallel lines. We know that the parallel lines have the corresponding coefficients equal and the only difference is the constant term. Let us assume the parallel lines in our question are $ax+by+{{c}_{1}}=0,ax+by+{{c}_{2}}=0$. Let us consider the product of them.
\[\begin{align}
  & \left( ax+by+{{c}_{1}} \right)\times \left( ax+by+{{c}_{2}} \right)=0 \\
 & {{a}^{2}}{{x}^{2}}+2abxy+{{b}^{2}}{{y}^{2}}+a\left( {{c}_{1}}+{{c}_{2}} \right)x+b\left( {{c}_{1}}+{{c}_{2}} \right)y+{{c}_{1}}{{c}_{2}}=0 \\
\end{align}\]
We know that the above equation should be equal to the given equation. Comparing the coefficients of different terms, we get
$\begin{align}
  & {{a}^{2}}=4\Rightarrow a=\pm 2 \\
 & {{b}^{2}}=1\Rightarrow b=\pm 1 \\
 & 2ab=2l \\
 & ab=l \\
 & \left( \pm 2 \right)\left( \pm 1 \right)=l \\
 & l=\pm 2 \\
 & \left| l \right|=2 \\
\end{align}$
Comparing the coefficients of x and y terms, we get
$\begin{align}
  & a\left( {{c}_{1}}+{{c}_{2}} \right)=6\to \left( 1 \right) \\
 & b\left( {{c}_{1}}+{{c}_{2}} \right)=k\to \left( 2 \right) \\
\end{align}$
Dividing both the equations, we get
$\dfrac{a}{b}=\dfrac{6}{k}$
By cross multiplying, we get
$k=6\dfrac{b}{a}$
Using the values of b and a in the above equation, we get
$\begin{align}
  & k=6\dfrac{\pm 1}{\pm 2} \\
 & k=\pm 3 \\
 & \left| k \right|=3 \\
\end{align}$
We can write the required value $\left| l \right|+\left| k \right|$ as
$\left| l \right|+\left| k \right|=2+3=5$
$\therefore $The required value $\left| l \right|+\left| k \right|=5$, the answer is option-D.

Note: We have to be careful while multiplying the equations of lines. The basic rule is that R.H.S should be equal to 0. If we multiply $ax+by={{c}_{1}},ax+by={{c}_{2}}$, we get a total different expression which is wrong. This question can be solved using the formulae related to a pair of parallel lines. That is if the equation $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$ represents a pair of parallel lines, then the condition is ${{h}^{2}}=ab$ and $a{{f}^{2}}=b{{g}^{2}}$. Using these two conditions, we can get the values of l and k and the required result.