The equation of the line that passes through \((-1,3)\) and is perpendicular to the straight line \(5 x+3 y=-1\) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(3 x-5 y+18=0\)
Two lines are perpendicular when the product of their slopes is −1. Find the slope of the given line, take the negative reciprocal, and use the point-slope form.
\(5x+3y=-1\) gives \(y=-\frac{5}{3}x-\frac{1}{3}\), so its slope is \(-\frac{5}{3}\)
Slope of the perpendicular line \(=\frac{3}{5}\)
Through \((-1,3)\): \(y-3=\frac{3}{5}(x+1)\)
\(5y-15=3x+3\), that is, \(3x-5y+18=0\)
Shortcut:
A line perpendicular to \(ax+by=c\) has the form \(bx-ay+k=0\): here \(3x-5y+k=0\)
Putting \((-1,3)\): \(-3-15+k=0\), so \(k=18\)