Because the two lines are parallel, we know that the slope of the line we need to find must also be \(\displaystyle -1\).

Next, plug in the point given by the question to find the y-intercept of the line.

\(\displaystyle 5=-1(-2)+b\)

\(\displaystyle b+2=5\)

\(\displaystyle b=3\)

Now, we can write the equation for the line: \(\displaystyle y=-x+3\)

Because the two lines are parallel, we know that the slope of the line we need to find must also be \(\displaystyle \frac{1}{2}\).

Next, plug in the point given by the question to find the y-intercept of the line.

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

\(\displaystyle b-1=3\)

\(\displaystyle b=4\)

Now, we knwo the equation of the line must be \(\displaystyle y=\frac{1}{2}x+4\).

Because the two lines are parallel, we know that the slope of the line we need to find must also be \(\displaystyle 4\).

Next, plug in the point given by the question to find the y-intercept of the line.

\(\displaystyle 8=4(2)+b\)

\(\displaystyle b+8=8\)

\(\displaystyle b=0\)

Thus, the equation of the line must be \(\displaystyle y=8x\).

Because the two lines are parallel, we know that the slope of the line we need to find must also be \(\displaystyle \frac{1}{5}\).

Next, plug in the point given by the question to find the y-intercept of the line.

\(\displaystyle 2=\frac{1}{5}(10)+b\)

\(\displaystyle b+2=2\)

\(\displaystyle b=0\)

The equation of the line is \(\displaystyle y=\frac{1}{5}x\).

Because the two lines are parallel, we know that the slope of the line we need to find must also be \(\displaystyle \frac{1}{3}\).

Next, plug in the point given by the question to find the y-intercept of the line.

\(\displaystyle 1=\frac{1}{3}(3)+b\)

\(\displaystyle b+1=1\)

\(\displaystyle b=0\)

The equation of the line is \(\displaystyle y=\frac{1}{3}x\).

Because the two lines are parallel, we know that the slope of the line we need to find must also be \(\displaystyle 2\).

Next, plug in the point given by the question to find the y-intercept of the line.

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

\(\displaystyle b+1=4\)

\(\displaystyle b=3\)

Now, we know the equation of the line must be \(\displaystyle y=2x+3\).

Because the two lines are parallel, we know that the slope of the line we need to find must also be  \(\displaystyle \frac{1}{8}\).

Next, plug in the point given by the question to find the y-intercept of the line.

\(\displaystyle \frac{1}{4}=\frac{1}{8}(2)+b\)

\(\displaystyle b+\frac{1}{4}=\frac{1}{4}\)

\(\displaystyle b=0\).

We can then write the equation of the line: \(\displaystyle y=\frac{1}{8}x\)

This is a two-step problem. First, the slope of the original line needs to be found. The slope will be represented by "\(\displaystyle m\)" when the line is in \(\displaystyle y\)-intercept form \(\displaystyle (y=mx+b)\).

\(\displaystyle 2x +3y = 16\)

\(\displaystyle 3y = -2x +16\)

\(\displaystyle y = \left(\frac{-2}{3}\right)x +\left(\frac{16}{3}\right)\)

So the slope of the original line is \(\displaystyle -\frac{2}{3}\). A line with perpendicular slope will have a slope that is the inverse reciprocal of the original. So in this case, the slope would be \(\displaystyle \frac{3}{2}\). The second step is finding which line will give you that slope. For the correct answer, we find the following:

\(\displaystyle 6y - 9x = 16.5\)

\(\displaystyle 6y = 9x +16.5\)

\(\displaystyle y= \left(\frac{3}{2}x\right) +\left(\frac{16.5}{6}\right)\)

So, the slope is \(\displaystyle \frac{3}{2}\), and this line is perpendicular to the original.

For lines to be parallel, they must have the same slope. The slope of the line we are looking for then must be \(\displaystyle 4\).

The point that's given in the equation is also the y-intercept.

Using these two pieces of information, we know that the equation for the line must be \(\displaystyle y=4x+4\)

In this kind of problem, it's important to keep track of information given about your line of interest. In this case, the coordinates given set up the stage for us to be able to get to our line of focus - the line perpendicular to the tangent line. In order to determine the perpendicular line's slope, the tangent line's slope must be calculated. Keeping in mind that:

\(\displaystyle slope =\frac{rise}{run}=\frac{\Delta y}{\Delta x} =\frac{y_2-y_1}{x_2-x_1}\)

where y2,x2 and y1,x1 are assigned arbitrarily as long as the order of assignment is maintained. 

\(\displaystyle \frac{y_2-y_1}{x_2-x_1}=\frac{7-0}{5-1}=\frac{7}{4}\)  which is the slope of the tangent line.

To calculate the perpendicular line, we have to remember that the product of the tangent slope and the perpendicular slope will equal -1.

\(\displaystyle \frac{7}{4} x \frac{?}{?} = -1\), the perpendicular slope can then be calculated as \(\displaystyle \frac{-4}{7}\)