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

`\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}`.