To solve the problem of finding the equation of the line that is perpendicular to the given line [tex]\( 3x + 5y = -9 \)[/tex] and passes through the point [tex]\( (3, 0) \)[/tex], follow these steps:
1. Determine the slope of the given line:
The given line is [tex]\( 3x + 5y = -9 \)[/tex]. First, we rewrite this equation in slope-intercept form ( [tex]\( y = mx + b \)[/tex] ).
[tex]\[
3x + 5y = -9 \implies 5y = -3x - 9 \implies y = -\frac{3}{5}x - \frac{9}{5}
\][/tex]
The slope ([tex]\(m\)[/tex]) of the given line is [tex]\(-\frac{3}{5}\)[/tex].
2. Find the slope of the perpendicular line:
Perpendicular lines have slopes that are negative reciprocals of each other. Therefore, the slope of the line perpendicular to [tex]\( y = -\frac{3}{5}x - \frac{9}{5} \)[/tex] is:
[tex]\[
m_{\perpendicular} = -\left( -\frac{5}{3} \right) = \frac{5}{3}
\][/tex]
3. Use the point-slope form to find the equation of the perpendicular line:
The perpendicular line must pass through the point [tex]\( (3, 0) \)[/tex]. Using the point-slope form ([tex]\( y - y_1 = m(x - x_1) \)[/tex]):
[tex]\[
y - 0 = \frac{5}{3}(x - 3) \implies y = \frac{5}{3}x - 5
\][/tex]
4. Convert the equation to standard form:
To convert [tex]\( y = \frac{5}{3}x - 5 \)[/tex] into standard form ([tex]\( Ax + By = C \)[/tex]):
[tex]\[
y = \frac{5}{3}x - 5 \implies 3y = 5x - 15
\][/tex]
Simplifying, we get:
[tex]\[
5x - 3y = 15
\][/tex]
Thus, the equation of the line that is perpendicular to [tex]\( 3x + 5y = -9 \)[/tex] and passes through the point [tex]\( (3, 0) \)[/tex] is [tex]\( \boxed{5x - 3y = 15} \)[/tex].
