Question

Write an equation of the line passing through the point (5, 1) that is perpendicular to the line 5x+3y=15. An equation of the line is y= x+ .

Answer 1

Answer:

$\displaystyle y = \frac{3}{5}\, x + (-2)$.

Step-by-step explanation:

The slope of a line in a plane would be $m$ if the equation of that line could be written in the slope-intercept form $y = m\, x + b$ for some constant $b$.

Find the slope of the given line by rearranging its equation into the slope-intercept form.

$3\, y = (-5)\, x + 15$.

$\displaystyle y = \frac{(-5)}{3} \, x + 5$.

Thus, the slope of the given line would be $(-5) / 3$.

Two lines in a plane are perpendicular to one another if and only if the product of their slopes is $(-1)$.

Let $m_{1}$ and $m_{2}$ 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, $m_{1}\, m_{2} = (-1)$. Apply the fact that the slope of the given line is $m_{1} = (-5) / 3$ and solve for $m_{2}$, the slope of the line in question.

$\begin{aligned}m_{2} &= \frac{(-1)}{m_{1}} \\ &= \frac{(-1)}{(-5) / 3} = \frac{3}{5}\end{aligned}$.

In other words, the slope of the line perpendicular to $5\, x + 3\, y = 15$ would be $(3 / 5)$.

If the slope of a line in a plane is $m$, and that line goes through the point $(x_{0},\, y_{0})$, the equation of that line in point-slope form would be:

$(y - y_{0}) = m\, (x - x_{0})$.

Since the slope of the line in question is $(3 / 5)$ and that line goes through the point $(5,\, 1)$, the equation of that line in point-slope form would be:

$\begin{aligned} (y - 1) = \frac{3}{5}\, (x - 5) \end{aligned}$.

Rearrange this equation as the question requested:

$\begin{aligned} y - 1 = \frac{3}{5}\, x - 3 \end{aligned}$.

$\begin{aligned} y = \frac{3}{5}\, x - 2 \end{aligned}$.

$\begin{aligned} y = \frac{3}{5}\, x + (-2) \end{aligned}$.