How do you write the equation that represents the line perpendicular to y=-3x+ 4 and passing through the point (-1, 1)?

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Answer 1 · 2016-12-13T21:59:04

#y - 1 = 1/3(x + 1)# or #y = 1/3x + 4/3#

First, we need to determine the slope of the perpendicular line.

If the slope of a line is #a/b# then the slope of a line perpendicular to this is #-b/a#.

Because the equation for the line we are given is already in slope-intercept form we can obtain the slope.

Slope-intercept form is #y = mx + c# where #m# is slope.

So, the slope of the line given is:

#-3# or #-3/1# converting this to the slope of a perpendicular line using the rule above gives:

#- -1/3 -> 1/3#

Now that we have the slope and a point we can use the point-slope formula to determine the equation for the perpendicular line.

The point-slope formula is:

#y - y_1 = m(x - x_1)#

Where #m# is the slope and

#(x_1, y1)# is the point which is given. Substituting the information gives:

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

#y - 1 = 1/3(x + 1)#

To convert to slope-intercept form we can solve for #y# while keeping the equation balanced:

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

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

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

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

#y = 1/3x + 4/3#