Cartesian Coordinate System - Pre-Calculus

First, solve for : .

Then, graph the at .

Since the slope of the line is , you can graph the point as well.

There is only one graph that fits these requirements.

To determine if a point lies on a line, plug in the x-value and y-value to see if the equation is satisfied. We can do this for each choice to check.

For example:

:

Since both sides are equivalent, this point does lie on the line.

We can continue to do this for each of the points until one point does not work out.

Thus, this point does not lie on the line. Thus, this must be the solution.

In order to determine in which quadrant the point lies, we must remember the order of the quadrants. The first quadrant is that where x and y are both positive, to the upper right of the origin. To move sequentially to the final quadrant, we go counterclockwise from the first quadrant, which means the second is where x is negative and y is positive, the third is where x and y are both negative, and the fourth is where x is positive and y is negative. We can see from our point (-3,-8) that x and y are both negative, which means the point lies in the third quadrant.

When looking at the graph, it is clear that when , has a value less than . If we were to plug in the value of , our equation would come out as such:

Therefore, at , we get a , providing the coordinate .

When looking at the graph, it is clear that when , has a value greater than . When we plug in both and values into the function, it is clear that these values do not work for the function:

If we are to plug into our function, the values would not work and both sides of the equation would not be equal:

Therefore, we know that these coordinates do not lie on the graph of the function.

If we were to plug in the coordinate into the function, we will find that it does not equate properly:

Since these values do not equate properly when plugged into the function, we now know that does not fit on the provided graph.

Use the distance formula to find the length of .

.

Since the length of is that of the diameter, the radius of the circle is .

Thus, the area of the circle is

.

First, solve for : .

Then, graph the at .

Since the slope of the line is , you can graph the point as well.

There is only one graph that fits these requirements.

To determine if a point lies on a line, plug in the x-value and y-value to see if the equation is satisfied. We can do this for each choice to check.

For example:

:

Since both sides are equivalent, this point does lie on the line.

We can continue to do this for each of the points until one point does not work out.

Thus, this point does not lie on the line. Thus, this must be the solution.

In order to determine in which quadrant the point lies, we must remember the order of the quadrants. The first quadrant is that where x and y are both positive, to the upper right of the origin. To move sequentially to the final quadrant, we go counterclockwise from the first quadrant, which means the second is where x is negative and y is positive, the third is where x and y are both negative, and the fourth is where x is positive and y is negative. We can see from our point (-3,-8) that x and y are both negative, which means the point lies in the third quadrant.

When looking at the graph, it is clear that when , has a value less than . If we were to plug in the value of , our equation would come out as such:

Therefore, at , we get a , providing the coordinate .

When looking at the graph, it is clear that when , has a value greater than . When we plug in both and values into the function, it is clear that these values do not work for the function:

If we are to plug into our function, the values would not work and both sides of the equation would not be equal:

Therefore, we know that these coordinates do not lie on the graph of the function.

If we were to plug in the coordinate into the function, we will find that it does not equate properly:

Since these values do not equate properly when plugged into the function, we now know that does not fit on the provided graph.

Use the distance formula to find the length of .

.

Since the length of is that of the diameter, the radius of the circle is .

Thus, the area of the circle is

.

First, solve for : .

Then, graph the at .

Since the slope of the line is , you can graph the point as well.

There is only one graph that fits these requirements.

To determine if a point lies on a line, plug in the x-value and y-value to see if the equation is satisfied. We can do this for each choice to check.

For example:

:

Since both sides are equivalent, this point does lie on the line.

We can continue to do this for each of the points until one point does not work out.

Thus, this point does not lie on the line. Thus, this must be the solution.

In order to determine in which quadrant the point lies, we must remember the order of the quadrants. The first quadrant is that where x and y are both positive, to the upper right of the origin. To move sequentially to the final quadrant, we go counterclockwise from the first quadrant, which means the second is where x is negative and y is positive, the third is where x and y are both negative, and the fourth is where x is positive and y is negative. We can see from our point (-3,-8) that x and y are both negative, which means the point lies in the third quadrant.

When looking at the graph, it is clear that when , has a value less than . If we were to plug in the value of , our equation would come out as such:

Therefore, at , we get a , providing the coordinate .