Question

Find the value of k, if 2x+3y-1=0 and 4x+ky+2=0 are parallel lines​

Answer 1

$\textbf{Concept used:}$

$\textbf{If the lines are parallel, their slopes are equal}$

$\text{Given lines are}$

$\textbf{2x+3y-1=0 and 4x+ky+2=0}$

$\text{Slope of 2x+3y-1=0 is \bf\,m_1=\frac{-2}{3} }$

$\text{Slope of 4x+ky+2=0 is \bf\,m_2=\frac{-4}{k} }$

$\text{Since the lines are parallel, we have}\;m_1=m_2$

$\implies\displaystyle\frac{-2}{3}=\frac{-4}{k}$

$\implies\displaystyle\frac{1}{3}=\frac{2}{k}$

$\implies\boxed{\bf\,k=6}$

$\therefore\textbf{The value of k is 6}$

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Answer 2

The value of k for the parallel lines condition is  6  

Step-by-step explanation:

Given as :

The two lines equation are

2 x + 3 y - 1 = 0                   ...........1

4 x + k y + 2 = 0                .............2

The two lines are parallel to each other

According to question

The two lines are parallel to each other

So, The product of their slope are equal

The standard equation of line is  y = m x + c

where m is the slope of line

For line eq 1

3 y = 1 - 2 x

Or, y = $\dfrac{-2}{3}$ x + $\dfrac{1}{3}$

So, The slope of line 1  = $m_1$ = $\dfrac{-2}{3}$

For line eq 2

k y = - 2 - 4 x

Or, y = $\dfrac{-4}{k}$ x - $\dfrac{2}{k}$

So, The slope of line 2 = $m_2$ = $\dfrac{-4}{k}$

Now, from the condition of parallel slope

        $m_1$ = $m_2$  

i.e     ( $\dfrac{-2}{3}$ ) = ( $\dfrac{-4}{k}$ )

By cross multiplication

i.e     - 2 k = - 12

∴            k = $\dfrac{-12}{-2}$

Or,          k = 6

So, The value of k for the parallel lines condition  = k = 6

Hence, The value of k for the parallel lines condition is  6  Answer