# 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?

Nov 26, 2017

In point-slope form:

$y - 13 = - \setminus \frac{1}{2} \left(x - \setminus \frac{11}{2}\right)$

#### Explanation:

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

$\setminus \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$

Plugging in corresponding values yields:

$\setminus \frac{14 - 12}{6 - 5}$

$= \setminus \frac{2}{1}$

$= 2$

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 $- \setminus \frac{1}{2}$.

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:

$\left(\setminus \frac{{x}_{1} + {x}_{2}}{2} \setminus \quad , \setminus \quad \setminus \frac{{y}_{1} + {y}_{2}}{2}\right)$

Plugging in yields:

$\left(\setminus \frac{5 + 6}{2} \setminus \quad , \setminus \quad \setminus \frac{12 + 14}{2}\right)$

$= \left(\setminus \frac{11}{2} , 13\right)$

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:

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

Plugging in yields:

$y - 13 = - \setminus \frac{1}{2} \left(x - \setminus \frac{11}{2}\right)$