Question

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?

Answer 1

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}`.

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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})`