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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+ .

User Jim Wooley
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7 votes

Answer:


\displaystyle y = (3)/(5)\, x + (-2).

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 = ((-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) &= ((-1))/(m_(1)) \\ &= ((-1))/((-5) / 3) = (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) = (3)/(5)\, (x - 5) \end{aligned}.

Rearrange this equation as the question requested:


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


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


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

User Wthamira
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7.7k points

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