Question

How do you find equation of a line, L which passes through the point (3, -1) and perpendicular to the line with equation 9x - 3y = 2?

Answer 1

`y + = -1/3(x - 3)` or `y = -1/3x`

Explanation:

To solve this we must first determine the slope of the given line by converting the equation to the slope-intercept form which is: `color(red)(y = mx - b)` where `color(red)(m)` is the slope.

`9x - 9x - 3y = -9x + 2`

`0 - 3y = -9x + 2`

`-3y = -9x + 2`

`(-3y)/-3 = (-9x + 2)/-3`

`y = (-9x)/-3 + 2/-3`

`y = 3x - 2/3`

So the slope of the given line is `3`.

The slope of a perpendicular line is the negative inverse of the line give. So if `m` is the slope of the given line the `-1/m` is the slope of a perpendicular line.

So, in this case the slope of the perpendicular line is `-1/3`

We now have a slope and a point so we can use the point-slope to determine the equation for the perpendicular line.

The point-slope formula states: `color(red)((y - y_1) = m(x - x_1))`
Where `color(red)(m)` is the slope and `color(red)((x_1, y_1))` is a point the line passes through.

Substituting the information we were given and calculated gives:

`y - -1 = -1/3(x - 3)`

#y + 1 = -1/3x + 1/3 * 3

`y + 1 = -1/3x + 1`

or converting to the more familiar point-intercept form gives:

`y + 1 - 1 = -1/3x + 1 - 1`

`y + 0 = -1/3x + 0`

`y = -1/3x`