Suppose two straight lines $3 x - 2 y = 5 a n d 2 x + k y + 7 = 0$ .Find the value of $k$ for which the given lines are parallel to each other.

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Solution

Step 1: Finding the slope of two lines:
The general equation of a straight line is given by $y = m x + c$
where, $m$ = slope of the line
$c$ = constant term
Here we will use the concept that any two straight lines are said to parallel if their slopes are equal.
The given straight lines are $3 x - 2 y = 5 a n d 2 x + k y + 7 = 0$
From the equation, $3 x - 2 y = 5$ we can write
$- 2 y = - 3 x + 5 \Rightarrow 2 y = 3 x - 5 \Rightarrow y = \frac{3}{2} x - \frac{5}{2} . . . . . . . . . . . . . . (1)$
From the equation, $2 x + k y + 7 = 0$ we can write
$2 x + k y = - 7 \Rightarrow k y = - 2 x - 7 \Rightarrow y = - \frac{2}{k} x - \frac{7}{k} . . . . . . . . . . . . . . . . . . . . . . . . (2)$
On comparing equation (1) and (2) with $y = m_{1} x + c_{1} a n d y = m_{2} x + c_{2}$ respectively,
we get,
$m_{1} = \frac{3}{2} a n d m_{2} = - \frac{2}{k}$
Step 2: Finding the value of $k$ :
As the lines are given parallel. So, their slopes must be equal.
$\Rightarrow m_{1} = m_{2} \Rightarrow \frac{3}{2} = - \frac{2}{k} \Rightarrow 3 k = - 4 \Rightarrow k = - \frac{4}{3}$
Therefore, the value of $k i s - \frac{4}{3} .$

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