In order for a line \(\displaystyle c\) to be perpendicular to another line \(\displaystyle a\) defined by the equation \(\displaystyle y=mx+b\) , the slope of line \(\displaystyle c\) must be a negative reciprocal of the slope of line \(\displaystyle a\). Since line \(\displaystyle a\)'s slope is \(\displaystyle m\) in the slope-intercept equation above, line \(\displaystyle c\)'s slope would therefore be \(\displaystyle -\frac{1}{m}\).

Given that we have a line \(\displaystyle a\) with a slope \(\displaystyle m=6\), we can therefore conclude that any perpendicular line would have a slope \(\displaystyle -\frac{1}{m}=-\frac{1}{6}\).

In order for a line \(\displaystyle c\) to be perpendicular to another line \(\displaystyle a\) defined by the equation \(\displaystyle y=mx+b\) , the slope of line \(\displaystyle c\) must be a negative reciprocal of the slope of line \(\displaystyle a\). Since line \(\displaystyle a\)'s slope is \(\displaystyle m\) in the slope-intercept equation above, line \(\displaystyle c\)'s slope would therefore be \(\displaystyle -\frac{1}{m}\).

Since in this instance the slope \(\displaystyle m=\frac{5}{6}\), \(\displaystyle -\frac{1}{m}=-\frac{1}{\frac{5}{6}}=-\frac{6}{5}\). Two of the above answers have this as their slope, so therefore that is the answer to our question.

If two lines intersect at a ninety-degree angle, they are said to be perpendicular. Two lines are perpendicular if their slopes are opposite reciprocals. In this case:

\(\displaystyle f(x)\:slope=\frac{3}{1}\)

\(\displaystyle g(x)\:slope=-\frac{1}{3}\)

The two lines' slopes are reciprocals with opposing signs, so the answer is yes. Of our two yes answers, only one has the right explanation. Eliminate the option dealing with \(\displaystyle y\)-intercepts.

To find the slope \(\displaystyle m\) of the line running through \(\displaystyle (7,4)\) and \(\displaystyle (5,8)\), we use the following equation:

\(\displaystyle m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{8-4}{5-7}=\frac{4}{-2}=-2\)

The slope of any line perpendicular to the given line would have a slope that is the negative reciprocal of \(\displaystyle m\), or \(\displaystyle -\frac{1}{m}\). Therefore, \(\displaystyle -\frac{1}{-2}=\frac{1}{2}\)

Given a line \(\displaystyle a\) defined by the equation \(\displaystyle y=mx+b\) with a slope of \(\displaystyle m\), any line perpendicular to \(\displaystyle a\) would have a slope that is the negative reciprocal of \(\displaystyle m\), \(\displaystyle -\frac{1}{m}\). Given our equation  \(\displaystyle y=\frac{3}{4}x+16\), we know that \(\displaystyle m=\frac{3}{4}\) and that \(\displaystyle -\frac{1}{m}=-\frac{1}{\frac{3}{4}}=-\frac{4}{3}\). 

The only answer choice with this slope is \(\displaystyle y=-\frac{4}{3}x+12\). 

Given a line \(\displaystyle a\) defined by the equation \(\displaystyle y=mx+b\) with a slope of \(\displaystyle m\), any line perpendicular to \(\displaystyle a\) would have a slope that is the negative reciprocal of \(\displaystyle m\), \(\displaystyle -\frac{1}{m}\). Given our equation  \(\displaystyle y=5x-12\), we know that \(\displaystyle m=5\) and that \(\displaystyle -\frac{1}{m}=-\frac{1}{5}\). 

There are two answer choices with this slope, \(\displaystyle y=-\frac{1}{5}x-4\) and \(\displaystyle y=-\frac{1}{5}x+7\) . 

For any line \(\displaystyle a\) with an equation \(\displaystyle y=mx+b\) and slope \(\displaystyle m\), a line that is perpendicular to \(\displaystyle a\) must have a slope of \(\displaystyle -\frac{1}{m}\), or the negative reciprocal of \(\displaystyle m\). Given \(\displaystyle y=2x+5\), we know that \(\displaystyle m=2\) and therefore know that \(\displaystyle -\frac{1}{m}=-\frac{1}{2}\). 

Only one equation above has a slope of \(\displaystyle -\frac{1}{2}\): \(\displaystyle y=-\frac{1}{2}x+7\). 

For any line \(\displaystyle a\) with an equation \(\displaystyle y=mx+b\) and slope \(\displaystyle m\), a line that is perpendicular to \(\displaystyle a\) must have a slope of \(\displaystyle -\frac{1}{m}\), or the negative reciprocal of \(\displaystyle m\). Given the equation \(\displaystyle y=-\frac{1}{7}x+9\), we know that \(\displaystyle m=-\frac{1}{7}\) and therefore know that \(\displaystyle -\frac{1}{m}=-\frac{1}{-\frac{1}{7}}=7\).

For any line \(\displaystyle a\) with an equation \(\displaystyle y=mx+b\) and slope \(\displaystyle m\), a line that is perpendicular to \(\displaystyle a\) must have a slope of \(\displaystyle -\frac{1}{m}\), or the negative reciprocal of \(\displaystyle m\). Given the equation \(\displaystyle y=4x+8\), we know that \(\displaystyle m=4\) and therefore know that \(\displaystyle -\frac{1}{m}=-\frac{1}{4}\). 

Given a slope of \(\displaystyle -\frac{1}{4}\), we know that there are two solutions provided: \(\displaystyle y=-\frac{1}{4}x+5\) and \(\displaystyle y=-\frac{1}{4}x-4\). 

First, we need to rearrange the equation into slope-intercept form.  \(\displaystyle y=mx+b\).

\(\displaystyle 6y-3x=2->y=\frac{3x+2}{6}.\)  Therefore, the slope of this line equals \(\displaystyle \frac{1}{2}.\) Perpendicular lines have slope that are the opposite reciprocal, or \(\displaystyle -2.\)