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) → \(k\ne6\)
Each pipeline is a straight line. Two lines \(a_1x+b_1y+c_1=0\) and \(a_2x+b_2y+c_2=0\) cross at one point only when they are not parallel, that is, when \(\displaystyle\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\).
Here \(a_1=k,\ b_1=3\) and \(a_2=2,\ b_2=1\)
Condition for crossing: \(\displaystyle\frac{k}{2}\ne\frac{3}{1}\)
So \(k\ne6\)
For \(k=6\) the first line becomes \(6x+3y+1=0\), which has the same slope as \(2x+y+3=0\) but a different constant, so the two pipelines run parallel and never meet.