How to find out if a point is on a line with an equation - Algebra

To find out if a point is on a line, you can plug the points back into an equation. If the values equal one another, then the point must be on a line. In this case, the only equation where (6,5) would correctly fit as an value is .

To find out if a point (x, y) is on the graph of a line, we plug in the values and see if we get a true statement, such as 10 = 10. If we get something different, like 6 = 4, we know that the point is not on the line because it does not satisfy the equation. In the given choices, when we plug in (1, 10) we get 10 = 7 + 2, which is false, making this is the desired answer.

y = 7x + 2

(2, 16) gives 16 = 7(2) + 2 = 14 + 2 = 16

(–1, –5) gives –5 = 7(–1) + 2 = –7 + 2 = –5

(0, 2) gives 2 = 7(0) + 2 = 0 + 2 = 2

(–2, –12) gives –12 = 7(–2) + 2 = –14 + 2 = –12

All of these are true.

(1, 10) gives 10 = 7(1) + 2 = 7 + 2 = 9

10 = 9 is a false statement.

Lines that have the same slope are parallel (unless the two lines are identical) and lines with slopes that are opposite-reciprocals are perpendicular. So, the only statements left to evaluate are the two that contain a set of points.

Consider and .

So the slope, or , is 2.

Plugging the point into the half-finished equation gives us a value of . So that statement is true and the only one that could be the answer is the statement containing and .

Let's check it just in case.

gives us a slope value of 6, so we can already tell the equation for the line will not be . We have found our answer.

To determine whether a point is on a line, simply plug the points back into the equation. When we plug in (2,7) into the equation of as and respectively, the equation works out, which indicates that the point is located on the line.

The easiest way to find out if a point falls on a specific line is to plug the first value of the point in for and the second value for .

If we do this for , we find that

which is true.

The equation also holds true for , but is false for the other values. So, two of the answer choices are correct.

To determine which ordered pair satisfies the equation, it would help to rearrange the equation to slope-intercept form.

Then, plug in each ordered pair and see if it satisfies the equation. We are looking for an value that produces the desired answer.

satisfies the equation. All of the other points do not.

(Note: you could also use the original equation in standard form).

To determine whether a point is on a line, you can plug it into the equation to see if the equation remains valid/equal with the point.

Plugging the point (3,2) into the equation gives you

which works out. None of the other equations would remain equal after pluggin in (3,2).

To determine whether a point is located in a given line, simply plug in the coordinates of the point into the line. In this case, plugging in the coordinates into the only line where you can plug in the coordinates and have a valid equation is . Plugging in (2,7) would give you an equation of , which works out to .

To find out if a point is on a line with an equation, we just need to substitute in the point's and values and see if the equation holds true. For example, let's look at the point . Substitution into the equation gives us

or

, which is true.

So, this point does fall on the line. Doing the same with the other two points shows us that yes, all three of them fall on the line expressed by this equation.

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 .

To determine if a point is on a line you can simply subsitute the x and y coordinates into the equation. Another way to solve the problem would be to graph the line and see if it falls on the line. Plugging in will give which is a true statement, so it is on the line.

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 .

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 .