Answer:
[tex]\displaystyle y = \frac{3}{5}\, x + (-2)[/tex].
Step-by-step explanation:
The slope of a line in a plane would be [tex]m[/tex] if the equation of that line could be written in the slope-intercept form [tex]y = m\, x + b[/tex] for some constant [tex]b[/tex].
Find the slope of the given line by rearranging its equation into the slope-intercept form.
[tex]3\, y = (-5)\, x + 15[/tex].
[tex]\displaystyle y = \frac{(-5)}{3} \, x + 5[/tex].
Thus, the slope of the given line would be [tex](-5) / 3[/tex].
Two lines in a plane are perpendicular to one another if and only if the product of their slopes is [tex](-1)[/tex].
Let [tex]m_{1}[/tex] and [tex]m_{2}[/tex] denote the slope of the given line and the slope of the line in question, respectively.
Since the two lines are perpendicular to each other, [tex]m_{1}\, m_{2} = (-1)[/tex]. Apply the fact that the slope of the given line is [tex]m_{1} = (-5) / 3[/tex] and solve for [tex]m_{2}[/tex], the slope of the line in question.
[tex]\begin{aligned}m_{2} &= \frac{(-1)}{m_{1}} \\ &= \frac{(-1)}{(-5) / 3} = \frac{3}{5}\end{aligned}[/tex].
In other words, the slope of the line perpendicular to [tex]5\, x + 3\, y = 15[/tex] would be [tex](3 / 5)[/tex].
If the slope of a line in a plane is [tex]m[/tex], and that line goes through the point [tex](x_{0},\, y_{0})[/tex], the equation of that line in point-slope form would be:
[tex](y - y_{0}) = m\, (x - x_{0})[/tex].
Since the slope of the line in question is [tex](3 / 5)[/tex] and that line goes through the point [tex](5,\, 1)[/tex], the equation of that line in point-slope form would be:
[tex]\begin{aligned} (y - 1) = \frac{3}{5}\, (x - 5) \end{aligned}[/tex].
Rearrange this equation as the question requested:
[tex]\begin{aligned} y - 1 = \frac{3}{5}\, x - 3 \end{aligned}[/tex].
[tex]\begin{aligned} y = \frac{3}{5}\, x - 2 \end{aligned}[/tex].
[tex]\begin{aligned} y = \frac{3}{5}\, x + (-2) \end{aligned}[/tex].
