What is the equation of the line that is perpendicular to the line passing through #(5,12)# and #(6,14)# at midpoint of the two points?

1 Answer
Nov 26, 2017

Answer:

In point-slope form:

#y-13=-\frac{1}{2}(x-\frac{11}{2})#

Explanation:

First, we need to find the slope of the original line from the two points.

#\frac{y_2-y_1}{x_2-x_1}#

Plugging in corresponding values yields:

#\frac{14-12}{6-5}#

#=\frac{2}{1}#

#=2#

Since the slopes of perpendicular lines are negative reciprocals of each other, the slope of the lines we’re looking for is going to be the reciprocal of #2#, which is #-\frac{1}{2}#.


Now we need to find the midpoint of those two points, which will give us the remaining information to write the equation of the line.

The midpoint formula is:

#(\frac{x_1+x_2}{2}\quad,\quad\frac{y_1+y_2}{2})#

Plugging in yields:

#(\frac{5+6}{2}\quad,\quad\frac{12+14}{2})#

#=(\frac{11}{2},13)#

Therefore, the line we’re trying to find teh equation of passes through that point.

Knowing the slope of the line, as well as a point where it passes through, we can write its equation in point-slope form, denoted by:

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

Plugging in yields:

#y-13=-\frac{1}{2}(x-\frac{11}{2})#