To see if a point is directly on a line, plug that point into the equation replacing the x in the slope intercept equation by the x coordinate and the y with the y coordinate respectively and then simplify. If the equation is a true statement like 1=1 or 5=5 then that coordinate pair is on the line.

Since we ended up with a true statement, the point  indeed is on the line  .

To ascertain if a point lies on a line, substitute the coordinate pair into the equation of the line and simplify. If the equation yields a true statement then that point lies on the line.

Unfortunately none of the points lie on our line, but this is how one would find out:

Based on the graph of this line it appears (1,1) is on this line. 

Because this statement is not true, (1,1) is not on the line. Try this same method with the rest of the coordinate pairs to see that they do not lie on our line.

This question asks which of the given points belongs on the line . This is another way of asking which of the points does the line pass through.

This question can be quickly solved for by substituting in the given points. This may also be known as the "plug and chug" method. 

Each point (coordinate) represents an  and a  value through this format: 

Simply by arbitrarily substituting in the  or  into the equation  and solving for  or , you can determine if the point belongs on the line if you are left with the given point. 

For example, using the point that does belong on the line: 

substituting in the  value from , where , into the equation , we can solve for 

 and given that  in the coordinate , we know that this coordinate would belong. 

If we did not receive the anticipated  value from the coordinate, we can automatically deduce that the point does not belong on the line. For example, using  and substituting in the  value, 

Because , we can deduce that  does not belong on the line .

This question asks which of the given points does not belong on the line . This is another way of asking which of the points does the line not pass through.

This question can be quickly solved for by substituting in the given points. This may also be known as the "plug and chug" method. But first, let's rewrite the equation in a more comfortable format with a positive . This can be achieved by multiplying  to both sides of the equation so the result will be: 

Each point (coordinate) represents an  and a  value through this format: 

Simply by arbitrarily substituting in the  or  into the equation  and solving for  or , you can determine if the point belongs on the line if you are left with the given point. 

For example, using a point that does belong on the line: 

substituting in the  value from , where , into the equation , we can solve for 

 and given that  in the coordinate , we know that this coordinate would belong. 

If we did not receive the anticipated  value from the coordinate, we can automatically deduce that the point does not belong on the line. For example, using  and substituting in the  value, 

Because , we can deduce that  does not belong on the line .

This question asks which of the given points belongs on the line . This is another way of asking which of the points does the line pass through.

This question can be quickly solved for by substituting in the given points. This may also be known as the "plug and chug" method. 

Each point (coordinate) represents an  and a  value through this format: 

Simply by arbitrarily substituting in the  or  into the equation  and solving for  or , you can determine if the point belongs on the line if you are left with the given point. 

For example, using the point that does belong on the line: 

substituting in the  value from , where , into the equation , we can solve for 

 and given that  in the coordinate , we know that this coordinate would belong. 

If we did not receive the anticipated  value from the coordinate, we can automatically deduce that the point does not belong on the line. For example, using  and substituting in the  value, 

Because , we can deduce that  does not belong on the line .

At first glance, the given function looks very intimidating due to its length and the inclusion of many fractions. One should realize, however, that if like terms are combined, the equation quickly condenses to the standard form of a line. Additionally, the concept of an "x-intercept" might not be immediately familiar, but the student should intuit that the x-intercept is the value of x, where the line crosses the x-axis. Another way of saying this is, "the x-intercept is the x value when y=0". Therefore, plug zero in for "y" and eliminate those "y" terms immediately. That leaves: 

.

The quickest way to finish this problem is to convert all fractions on the left-hand side to decimal form. (A student should quickly recognize that all fractions on the left-hand side are easily converted to decimals even without the use of a calculator).

The conversion to decimal form yields: 

.

Now, combine the like x terms to obtain: 

.

Finally, divide both sides of this equation by -2, to get:

.

Recall that this is the value of x when y=0 (because we have already plugged zero in for y) and therefore, this answer represents the x-intercept of the original line.

Which of the following points is on the line of f(x)?

We can solve this problem by plugging in all of the options and seeing which one works. However, we can probably work more quickly by trying the easier options first.

Let's begin with the options including 0, 0 usually makes an equation easier to look at by simplifying things.

So  is out.

Next up, try 

We are good, the point  is on our line, so it is our answer.

To determine if a point works, plug it in and see if it makes a true statement.

The correct answer does:

Answers that don't work include :

NOT TRUE.

In order to determine which point will satisfy the equation, we will have to plug in each value of , solve the right side of the equation, and see if the  will match for the order pairs given.

The only order pair that will satisfy this criteria is  since:

The answer is:  

The -coordinate of the -intercept of a line can be found by substituting 0 for  in its equation and solving for :

The -intercept is the point .