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
Morgan S.
Nov 26, 2017

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})

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