The equation of a line is  where  is the slope.

Rearrange the equation to match this:

For the perpendicular line, the slope is the negative reciprocal;

therefore 

A slope perpendicular to another line can be found by taking the reciprocal of the orignal slope and changing the sign.

If you solve for  in the given equation,

divide by three on each side

.

There is a slope of  because the equation is in slope-intercept form where m represents the slope,

.

The reciprocal of that with a changed sign is

.

The line  is a vertical line.  The line perpendicular to a vertical line must always have a slope of zero.  

The only valid answer with a slope of zero is:   

Let's write the equation of the given line in slope-intercept form.

The slope of the given line is expressed by the coefficient of x. The slope here is .

To find the slope of the perpendicular line, we take the negative reciprocal of the given line's slope. Therefore, the slope of the perpendicular line is . 

Two lines are perpendicular to each other if their slopes are negative reciprocals. For example, if one line has a slope of 2, the line perpendicular to it has a slope of –1/2. One easy way to eliminate answer choices is to check if the slopes have the same sign, i.e. both positive or both negative. If so, they cannot be perpendicular. Several of the lines in the answer choices are of the form y = mx + b, where m is the slope and b is the y-intercept. We are only worried about the slope for the purposes of this question. 

y = 3x + 5 and y = 5x + 3 both have positive slopes (m = 3 and m = 5, respectively), so they aren't perpendicular. 

y = 3x/5 – 3 and y = 5x/3 + 3 both have positive slopes, so again they aren't perpendicular.

y = x – 1/2 and y = –x + 1/2 have slopes of m = 1 and m = –1, respectively. One is positive and one is negative, so that is a good sign. Let's take the negative reciprocal of 1. 1 /–1 = –1. So these two slopes are in fact negative reciprocals, and these two lines are perpendicular to each other. Even though we have found the correct answer, let's go through the other two choices to be sure.

The line between the points (1,3) and (3,5), and y = 4x + 7: We need to find the slope of the first line. slope = rise / run = (y₂ – y₁) / (x₂ – x₁) = (5 – 3) / (3 – 1) = 1. The slope of y = 4x + 7 is also positive (m = 4), so the lines are not perpendicular.

The line between the points (7,4) and (4,7), and the line between the points (3,9) and (4,8): the first slope = (7 – 4) / (4 – 7) = –1 and the second slope = (8 – 9) / (4 – 3) = –1. They have the same slope, making them parallel, not perpendicular.

Perpendicular lines have slopes that are the negative reciprocals of each other. 

Perpendicular lines will have slopes that are negative reciprocals of one another. Our first step will be to find the slope of the given line by putting the equation into slope-intercept form.

The slope of this line is . The negative reciprocal will be , which will be the slope of the perpendicular line.

Now we need to find the answer choice with this slope by converting to slope-intercept form.

This equation has a slope of , and must be our answer.

When determining if a two lines are perpindicular, we are only concerned about their slopes. Consider the basic equation of a line, , where m is the slope of the line. Two lines are perpindicular to each other if one slope is the negative and reciprocal of the other.

The first step of this problem is to get it into the form, , which is . Now we know that the slope, m, is . The reciprocal of that is , and the negative of that is . Therefore, any line that has a slope of  will be perpindicular to the original line.

If lines are perpendicular, then their slopes will be negative reciprocals.

First, we need to find the slope of the given line.

Because we know that our given line's slope is , the slope of the line perpendicular to it must be .

Perpendicular lines have slopes that are negative reciprocals. The given equation has a slope of ; therefore, any lime that is perpendicular to it will have a slope of .