Step \(1\)
Find the slope of the given line
Let
\(A(-3,2)\ B(2,-1)\)
slope mAB is equal to
\(mAB=\frac{(y2-y1)}{(x2-x1)} \\ \\ mAB=\frac{(-1-2)}{(2+3)} \\ \\ mAB=-\frac{3}{5}\)
Step \(2\)
Find the slope of the line that is perpendicular to the given line
Let
CD ------> the line that is perpendicular to the given line
we know that
If two lines are perpendicular, then the product of their slopes is equal to \(-1\)
so
\(mAB*mCD=-1\\ mAB=-\frac{3}{5} \\ mCD=-\frac{1}{mAB} \\ mCD=\frac{5}{3}\)
Step \(3\)
Find the equation of the line with mCD and the point (3,0)
we know that
the equation of the line in the form point-slope is equal to
\(y-y1=m(x-x1)\\\\ y-0=\frac{5}{3} *(x-3)\\\\ y=\frac{5}{3} x-5\)
Multiply by \(3\) both sides
\(3y=5x-15\)
\(5x-3y=15\)
therefore
the answer is
the equation of the line that is perpendicular to the given line is the equation \(5x-3y=15\)
