Two water pipelines are represented by the equations $kx + 3y +1 = 0$ and $2x + y + 3 = 0$. For what value of $k$, the pipelines cross each other?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $k≠6$
For two straight lines to cross each other, they must not be parallel, i.e., their slopes must be different.
Given equations:
1. $kx + 3y + 1 = 0$
$3y = -kx - 1 \Rightarrow y = -\frac{k}{3}x - \frac{1}{3}$
Slope $m_1 = -\frac{k}{3}$
2. $2x + y + 3 = 0$
$y = -2x – 3$
Slope $m_2 = -2$
For the lines to cross:
$m_1 \ne m_2$
$-\frac{k}{3} \ne -2$
$k \ne 6$