For what value of \(k\) are the lines \(3x-ky+4=0\) and \(2x+6y-1=0\) perpendicular to each other?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(1\)
Two lines are perpendicular when the product of their slopes is \(-1\). The line \(ax+by+c=0\) has slope \(-\frac{a}{b}\).
Slope of \(3x-ky+4=0\): \(\displaystyle -\frac{3}{-k}=\frac{3}{k}\)
Slope of \(2x+6y-1=0\): \(\displaystyle -\frac{2}{6}=-\frac{1}{3}\)
Product of the slopes: \(\displaystyle \frac{3}{k}\times\left(-\frac{1}{3}\right)=-\frac{1}{k}\)
\(\displaystyle -\frac{1}{k}=-1\), so \(k=1\)
Why the other options are incorrect
Option (2) \(-9\) — Incorrect. With \(k=-9\) the first slope is \(-\frac{1}{3}\), equal to the second, so the lines are parallel.
Option (3) \(9\) — Incorrect. The slopes are \(\frac{1}{3}\) and \(-\frac{1}{3}\); their product is \(-\frac{1}{9}\), not \(-1\).
Option (4) \(-1\) — Incorrect. The slopes are \(-3\) and \(-\frac{1}{3}\); their product is \(+1\), not \(-1\).