The slope must be the negative reciprocal and the line must pass through the point (3,2).  So the slope becomes $\displaystyle -0.5$ and this is plugged into $\displaystyle 2=-0.5\cdot 3+b$ to solve for the $\displaystyle y$-intercept.

Perpendicular lines have slopes that are negative inverses of each other. Since the original equation has a slope of $\displaystyle \frac{1}{2}$, the perpendicular line must have a slope of $\displaystyle -2$. The only other equation with a slope of $\displaystyle -2$ is $\displaystyle y=-2x+3$.

First, convert given equation to the slope-intercept form.

$\displaystyle 2x + 3y = 6$

$\displaystyle 3y=-2x+6$

$\displaystyle y = \frac{-2}{3}x + 2$ 

In this format, we can tell that the slope is $\displaystyle m = \frac{-2}{3}$. The slope of a perpendicular line will be the negative reciprocal, making $\displaystyle m_2 = \frac{3}{2}$.

Next, substitute the slope into the slope-intercept form to get the intercept, using the point give in the question.

$\displaystyle y = mx + b$

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

$\displaystyle b = 2$

The perpendicular equation becomes $\displaystyle y = \frac{3}{2}x +2$. This equation can be re-written in the format of the asnwer chocies.

$\displaystyle y = \frac{3}{2}x +2$

$\displaystyle y-\frac{3}{2}x=2$

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

$\displaystyle 2y-3x=4$, or $\displaystyle -3x + 2y = 4$

Perpendicular lines have slopes that are negative reciprocals of one another. The slope of the given line is 9, so a line that is perpendicular to it must have a slope equivalent to its negative reciprocal, which is $\displaystyle -\frac{1}{9}$.

Perpendicular lines have slopes that are negative reciprocals of each other. The given line has a slope of $\displaystyle \frac{1}{3}$. The negative reciprocal of $\displaystyle \frac{1}{3}$ is $\displaystyle -3$, so the perpendicular line must have a slope of $\displaystyle -3$. The only line with a slope of $\displaystyle -3$ is $\displaystyle y=-3x+2$.

To find the equation of a line, we need to know the slope and a point that passes through the line.  Once we know this, we can use the equation $\displaystyle y-y_1=m(x-x_1)$ where m is the slope of the line, and $\displaystyle (x_{1},y_{1})$ is a point on the line.  For perpendicular lines, the slopes are negative reciprocals of each other.  The slope of $\displaystyle y=5x-1$ is 5, so the slope of the perpendicular line will have a slope of $\displaystyle \frac{-1}{5}$.  We know that the perpendicular line needs to contain the point (5,3), so we have all of the information we need.  We can now use the equation $\displaystyle y-y_1=m(x-x_1)\Rightarrow y-3=\frac{-1}{5}(x-5)\Rightarrow$$\displaystyle y=\frac{-1}{5}x+1+3\Rightarrow y=\frac{-1}{5}x+4.$ 

The equation of a line is written in the following format: 

$\displaystyle y=mx+b$

1) The first step, then, is to find the slope, $\displaystyle m$.

$\displaystyle m$ is equal to the change in $\displaystyle y$ divided by the change in $\displaystyle x$.

So,

$\displaystyle \frac{7-3}{4-2}=\frac{4}{2}=2$

2) The perpendicular slope of a line with a slope of 2 is the opposite reciprocal of 2, which is $\displaystyle -\frac{1}{2}$.

3) Next step is to find $\displaystyle b$. We don't need to find the equation of the original line; all we need from the original line is the slope. So all we need $\displaystyle b$ for is the perpendicular line. We can find values for $\displaystyle x$ and $\displaystyle y$ from the one point we have from the perpendicular line, plug them in, and solve for $\displaystyle b$.

Our point is (4,7)

So, 

$\displaystyle 7=-\frac{1}{2}(4)+b$

$\displaystyle 7=-2+b$

$\displaystyle b=9$

Then we just fill in our value for $\displaystyle b$, and we have $\displaystyle y$ as a function of $\displaystyle x$.

$\displaystyle y=-\frac{1}{2}x+9$

$\displaystyle \textup{The segment between points } (1,3) \textup{ and } (3,1) \textup{ has slope }\frac{1-3}{3-1}=-1.\\\textup{ Therefore, any line perpendicular to it has slope the inverse reciprocal;}\\\textup{that is, }-\frac{1}{-1}=1.$

$\displaystyle \textup{Of all the perpendicular lines }y=(1)x+p \textup{, the one passing through the}\\\textup{midpoint }\left(\frac{1+3}{2},\frac{1+3}{2}\right)=(2,2) \textup{ is found by substituting } x=2\textup{ and } y=2\\ \textup{ in the equation } y=x+p \textup{ and solving for } p \textup{, which gives the solution } p=0.$

This problem first relies on the knowledge of the slope-intercept form of a line, $\displaystyle \small y=mx+b$, where m is the slope and b is the y-intercept.

In order for a line to be perpendicular to another line, its slope has to be the negative reciprocal. In this case, we are seeking a line to be perpendicular to $\displaystyle \small y=2x+1$. This line has a slope of 2, a.k.a. $\displaystyle \small \frac{2}{1}$. This means that the negative reciprocal slope will be $\displaystyle \small -\frac{1}{2}$. We are told that the y-intercept of this new line is 4.

We can now put these two new pieces of information into $\displaystyle \small y=mx+b$ to get the equation

$\displaystyle \small y=-\frac{1}{2}x+4$.

To solve this type of problem, we have to be familiar with the slope-intercept form of a line, $\displaystyle \small y=mx+b$ where m is the slope and b is the y-intercept. The line that our line is perpendicular to has the slope-intercept equation $\displaystyle \small y=\frac{2}{3}x+5$, which means that the slope is $\displaystyle \small \frac{2}{3}$.

The slope of a perpendicular line would be the negative reciprocal, so our slope is $\displaystyle \small -\frac{3}{2}$.

We don't know the y-intercept of our line yet, so we can only write the equation as:

$\displaystyle \small y=- \frac{3}2x + b$.

We do know that the point $\displaystyle \small (-2,3)$ is on this line, so to solve for b we can plug in -2 for x and 3 for y:

$\displaystyle \small 3 = -\frac{3}{2}*-2+b$ First we can multiply $\displaystyle \small -\frac{3}{2}*-2$ to get $\displaystyle \small 3$.

This makes our equation now:

$\displaystyle \small 3 = 3 + b$ either by subtracting 3 from both sides, or just by looking at this critically, we can see that b = 0.

Our original $\displaystyle \small y=- \frac{3}2x + b$ becomes $\displaystyle \small \small y=- \frac{3}2x + 0$, or simply $\displaystyle \small \small y=- \frac{3}2x$.