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?

1 Answer
Dec 17, 2016

Answer:

#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#