How do you write an equation of a line passing through (2, 3), perpendicular to #y = (3/2)x − 4#?

1 Answer
Apr 24, 2018

Answer:

We find the line with negative reciprocal slope through the point, which is #y = -2/3 x + 13/3 #. Another way to go is to swap the coefficients on #x# and #y# and negate one, and compute the constant from the point, giving #2x + 3y = 13# which is the same line.

Explanation:

When we multiply perpendicular slopes together we get #-1,# meaning perpendicular slopes are the negative reciprocals of each other.

We can read off the slope of #y = 3/2 x - 4 # as the coefficient on #x#, which is #3/2#. The slope of the perpendiculars will be the negative reciprocal, #-2/3.#

We fit the constant to the line.

#y = -2/3 x + b#

#3 = -2/3 (2) + b#

#b = 13/3#

So the equation we seek is

#y = -2/3 x + 13/3 #


I prefer not to deal with the slope if I can avoid it. I like first writing the line in standard form:

# - 3x + 2y = -8#

Then the perpendiculars are gotten by swapping the coefficients on #x# and #y#, and negating one. The constant is quickly determined by the point:

#2 x + 3 y = 2(2) + 3(3) = 13#

#2x + 3y = 13#