This problem tests two key principles related to lines in the coordinate plane. For one, the slopes of perpendicular lines are negative reciprocals of each other (e.g. $\displaystyle a$ and $\displaystyle -\frac{1}{a}$. So here since you need to find a line perpendicular to a line with slope of $\displaystyle 3$, you're looking for the negative reciprocal $\displaystyle -\frac{1}{3}$. You can therefore eliminate choices $\displaystyle y=-3x+4$ and $\displaystyle y=-3x+\frac{3}{4}$, which do not have the proper slope.

The second key concept relates to what a point in the coordinate plane is. Because points are all given in the form $\displaystyle (x,y)$ you know that for a point to pass through the point $\displaystyle (1,2)$ it will need to allow for $\displaystyle x$ to equal $\displaystyle 1$ when $\displaystyle y$ equals $\displaystyle 2$. To find the $\displaystyle b$ term (in $\displaystyle y=mx+b$), take what you do know about the line:

$\displaystyle y=-\frac{1}{3}x+b$

From here, add $\displaystyle \frac{1}{3}$ to each side to isolate $\displaystyle b$ and you'll have your answer:

$\displaystyle \frac{7}{3}=b$, so the correct equation is $\displaystyle y=-\frac{1}{3}x+\frac{7}{3}$.

In order for two lines to be perpendicular, the slopes need to be negative reciprocals of each other. Since the given slope is $\displaystyle \frac{1}{2}$, you're looking for a slope of $\displaystyle -2$.

To find the slope, you need to put the equations in point-slope form, $\displaystyle y=mx+b$, where $\displaystyle m$ is the slope. Checking the answer choices quickly:

Choices $\displaystyle y-2x=8$ and $\displaystyle 4y-8x=16$ will each give $\displaystyle x$ a positive coefficient, as your first step is to get $\displaystyle y$ on the opposite side of the equation from $\displaystyle x$.

Only choice $\displaystyle 3y+6x=12$ simplifies to a coefficient of $\displaystyle -2$:

$\displaystyle 3y+6x=12$

$\displaystyle 3y=-6x+12$

$\displaystyle y=-2x+4$

You know that the line goes through point $\displaystyle (a,b)$ and the y-intercept of 6 tells you that the line also goes through point $\displaystyle (0,6)$. So you can use the "rise over run" slope formula to calculate the slope. Recognizing that the five answer choices focus on $\displaystyle a,b$, and 6, you should see that you should subtract 0 and 6 (subtracting 0 is the same as doing nothing at all) to get your math looking like the answer choices. So you'll set up the equation:

$\displaystyle \frac{b-6}{a-0}$

That Line J passes through the origin gives you a very helpful bit of information regarding point-slope form $\displaystyle (y=mx+b)$. That tells you that when $\displaystyle x=0,\, y=0$, meaning that $\displaystyle 0=0+b\cdots$ meaning that $\displaystyle b$ also equals $\displaystyle 0$. That takes a variable away, allowing you to get to work on the two points knowing that the line has a constant slope, $\displaystyle m$:

For point $\displaystyle (a,2)$, that means that $\displaystyle 2=ma$.

For point $\displaystyle (8,a)$, that means that $\displaystyle a=8m$.

This means that you can plug in $\displaystyle a=8m$ into the first equation, allowing you to solve for $\displaystyle m$ (the slope):

$\displaystyle 2=8m^{2}$

$\displaystyle \frac{1}{4}=m^{2}$

$\displaystyle m=\frac{1}{2}$ (or $\displaystyle -\frac{1}{2}$).

When attempting to find where two lines intersect, it is typically best to get the lines in the form $\displaystyle y=mx+b$. For Line A, you know that the slope is -1, so you have a head start in that $\displaystyle m=-1$. So you're starting with $\displaystyle y=-x+b$. And then remember - the x-intercept is the point at which $\displaystyle y=0$, so that point is $\displaystyle (1,0)$. You can then plug that point into the equation to find $\displaystyle b$:

$\displaystyle 0=-(1)+b$, so $\displaystyle b=1$. You know now that Line A has the equation $\displaystyle y=-x+1$.

For Line B, you know that the slope is 2 (so $\displaystyle y=2x+b$) and that when $\displaystyle x$ is $\displaystyle 0$, $\displaystyle y=-2$. Plug that into the line equation to solve for b and you have -2 = 2(0) + b, so b = -2. Now you know the equation for Line B: $\displaystyle y=2x-2$.

Since $\displaystyle y=-x+1$ and $\displaystyle y=-2x-2$, the two lines will intersect where $\displaystyle -x+1=2x-2$. Algebraically that leads you to $\displaystyle 3=3x$, so $\displaystyle x=1$. Plug that back into either line to find y, and you'll find that $\displaystyle y=0$. Since the point of intersection is $\displaystyle (1,0)$, the sum $\displaystyle a+b=1$ makes $\displaystyle 1$ the correct answer.

We’ll need to use the graph of the linear equation to better understand the existing line, so that we can draw conclusions about the new line. Since the existing line moves vertically by one for every three it moves to the right, our slope, or $\displaystyle \frac{rise}{run}$ is $\displaystyle \frac{1}{3}$. Since our line crosses the y axis at the point (0, 4), our full equation is $\displaystyle y=\frac{1}{3}x+4$.

Thus, our new line will have a slope of $\displaystyle \frac{2}{3}$ since the question told us that the slope of our new line, $\displaystyle y=ax+b$, will be twice the slope of the pictured line. Now, to find the full equation of the new line, and thus its y-intercept, we’ll want to plug the known point of the line $\displaystyle (3,4)$ into the equation.

If $\displaystyle 4=\frac{2}{3}(3)+b$, we can solve for $\displaystyle b$!

$\displaystyle 4=2+b$, thus $\displaystyle 2=b$.

So, our new line is $\displaystyle y=\frac{2}{3}x+2$ (shown below in blue) and its y-intercept is $\displaystyle (0,2)$.

To identify the slope of the line, we’ll want to identify two points on the line to find the $\displaystyle \frac{change\, in\, y}{change \, in\, x}$ otherwise known as the $\displaystyle \frac{rise}{run}$. One point, $\displaystyle (a,b)$ is given to us. Our other point can be determined by looking to the y-intercept, in this case, $\displaystyle (0,6)$. So, if we identify our $\displaystyle \frac{change\, in\, y}{change \, in\, x}$ by taking $\displaystyle \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$, we arrive at $\displaystyle \frac{b-6}{a-0}$ or $\displaystyle \frac{b-6}{a}$, our correct answer.

In order to attack this question, we can plug in the value of x for each coordinate to see if the proper value of y matches. for 

(2, 7) if we plug in 2 for x, we can see that

$\displaystyle y = \frac{3}{2}(2)+4$

which does in fact = 7. 

For (-5, -3.5), if we take the same step we can see that 

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

does give us a y value of -3.5.

For (14, 25), plugging in x = 14 gets us to

$\displaystyle y = \frac{3}{2}(14)+4$

which simplifies to 25.

So, by default we can see that our correct answer must be 

(6, 12), since an input of 6 gives us 

$\displaystyle y = \frac{3}{2}(6)+4$

which equals 13, not 12. So, (6, 12) is the only point *not* on the line. 

In order to find the area of the triangle, we can use relevant points such as the point of intersection and the x intercepts. In this case, the point of intersection is the point at which the equations of the lines are equal. So, if 

$\displaystyle 2x-2 = -\frac{1}{2}x+8$

then x = 4, and if we plug 4 back into our equation we can see that y = 6. So, the point of intersection is (4, 6). 

If we set y = 0 for each equation, we can also find the x intercepts, in this case, (1, 0) and (16, 0).

The triangle will appear as follows

At this point, we can use the x intercepts to find the base, 16-1 or 15, and the point of intersection to find the height, in this case 6.

If the area of a triangle is 

$\displaystyle \frac{1}{2}(b)(h)$

we can fill in the unknowns with

$\displaystyle \frac{1}{2}(15)(6)$

to arrive at an area of 45.

We can quickly visualize this line to draw conclusions.

In order to do so, plug in 0 for x to find a point on the line:

3(0)+2y=6

y=3

Thus, (0,3) is a point on the line.

Plug in 0  for y to find a second point on the line:

3x+2(0)=6

x=2

(2,0) is another point on the line.

Now we know that the line passes through the points (2,0) and (0,3).  

A quick sketch of the two points reveals that the line passes through all but the third quadrant.

*note - a line with a positive slope will always pass through quadrants I and III, while a line with a negative slope will always pass through quadrants II and IV.*