Question

Write the equation of a line PERPENDICULAR to y = -1/3x + 5 and passes through
the point (6,4)

Answer 1

`y = 3\, x - 14`.

Explanation:

The slope-intercept form equation of a slanting line is in the form
`y = m\, x + b`, where
`m` would be the slope of that line.

The equation of the original line is given in the slope-intercept form:
`y = (-1/3)\, x + 5`. The slope of that line would thus be
`(-1/3)`.

Two slanted lines in a plane are perpendicular to one another if and only if their slopes are inverse reciprocals.

In other words, if the slope of two slanted lines are
`m_(1)` and
`m_(2)`, those two lines would be perpendicular to one another if and only if
`m_(1) \cdot m_(2) = (-1)`.

In this question, the slope of the given line is
`m_(1) = (-1/3)`. Rearrange the equation
`m_(1) \cdot m_(2) = (-1)` to find
`m_(2)`, the slope of the line perpendicular to the given line:

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

In other words, the slope of the line perpendicular to the given line would be
`3`.

If a line of slope
`m` goes through the point
`(x_(0),\, y_(0))`, the point-slope equation of that line would be
`y - y_(0) = m\, (x - x_(0))`.

In this question, the requested line goes through the point
`(6,\, 4)`. It was also deduced that the slope of this requested line is be
`3`. The equation of this line in point-intercept form would be:

`y - 4 = 3\, (x - 6)`.

Rearrange to find the equation of this line in slope-intercept form:

`y = 3\, x - 14`.