Find the slope of the line through the two points. It is \(\displaystyle -\frac{4}{5}\).

Since the slope of a perpendicular line is the negative reciprocal of the original line, the new line's slope is \(\displaystyle \frac{5}{4}\). Plug the slope and one of the points into the point-slope formula  \(\displaystyle y-y_{1}=m(x-x_{1})\). Isolate for \(\displaystyle y\). 

Since a perpendicular line has a slope that is the negative reciprocal of the original line, the new slope is \(\displaystyle \frac{1}{3}\). There is only one answer with the correct slope.

First, find the slope of the original line, which is \(\displaystyle -\frac{3}{2}\). You can do this by isolating for \(\displaystyle y\) so that the equation is in slope-intercept form. Once you find the slope, just replace the \(\displaystyle m\) in the original equation withe the negative reciprocal (perpendicular lines have a negative reciprocal slope for each other). Thus, your answer is 

\(\displaystyle y=\frac{2}{3}X+\frac{11}{2}\)

In order for two lines to be perpendicular, their slopes must be opposites and recipricals of each other. The first step is to find the slope of the given equation:

\(\displaystyle 2x+4y=-12\)

\(\displaystyle 4y=-2x-12\)

\(\displaystyle y=-\frac{1}{2}x-3\)

Therefore, the slope of the perpendicular line must be \(\displaystyle 2\). Using the point-slope formula, we can find the equation of the new line:

\(\displaystyle y-2=2(x-1)\)

\(\displaystyle y-2=2x-2\)

\(\displaystyle y=2x\)

Perdendicular lines have slopes that are opposite reciprocals.  The slope of the old line is \(\displaystyle \frac{1}{2}\), so the new slope is \(\displaystyle -2\).

Plug the new slope and the given point into the slope intercept equation to calculate the intercept:

\(\displaystyle y=mx+b\) or \(\displaystyle 5=-2(6)+b\), so \(\displaystyle b=17\).

Thus \(\displaystyle y=-2x+17\), or \(\displaystyle y+2x=17\).

First, calculate the slope of the line segment between the given points.

\(\displaystyle m=\frac{y_2-y_1}{x_2-x_1}=\frac{6-2}{8-2}=\frac{4}{6}=\frac{2}{3}\)

We want a line that is perpendicular to this segment and passes through its midpoint. The slope of a perpendicular line is the negative inverse. The slope of the perpendicular bisector will be \(\displaystyle m=-\frac{3}{2}\).

Next, we need to find the midpoint of the segment, using the midpoint formula.

\(\displaystyle mid=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{x})=(\frac{2+8}{2},\frac{2+6}{2})=(5,4)\)

Using the midpoint and the slope, we can solve for the value of the y-intercept.

\(\displaystyle y=mx+b\)

\(\displaystyle 4=(-\frac{3}{2})(5)+b\)

\(\displaystyle 4=\frac{8}{2}=-\frac{15}{2}+b\)

\(\displaystyle \frac{23}{2}=b\)

Using this value, we can write the equation for the perpendicular bisector in slope-intercept form.

\(\displaystyle y=-\frac{3}{2}x+\frac{23}{2}\)

The equation is given in the slope-intercept form, so we know the slope is \(\displaystyle 2\).  To have perpendicular lines, the new slope must be the opposite reciprocal of the old slope, or\(\displaystyle \frac{-1}{2}\)

Then plug the new slope and the point into the slope-intercept form of the equation:

\(\displaystyle -2=\frac{-1}{2}(6)+b\) so \(\displaystyle -2=-3+b\) so \(\displaystyle b=1\) 

So the new equation becomes:  \(\displaystyle y=\frac{-1}{2}x+1\) and in standard form \(\displaystyle x+2y=2\)