How to find out if a point is on a line with an equation - GRE Quantitative Reasoning

Since the vertical coordinates of each point are given by y, solve each equation for y and plug in 6 for x, as follows:

Taking the difference of the resulting y -values give the vertical distance between the points (6,27) and (6,48), which is 21.

Since we have –2x and +2x in the equations, it makes sense to add the equations together to give 8y = 16 yielding y = 2. Then we substitute y = 2 into one of the original equations to get x = –2. So the solution to the system of equations is (–2, 2)

When the line crosses the x-axis, the y-coordinate is 0. Substitute 0 into the equation for y and solve for x.

.25(x – 20) + 12 = 0

.25_x_ – 5 = –12

.25_x_ = –7

x = –28

The answer is the point (–28,0).

BD+AC is the length of the line, except that BC is double counted. By subtracting BC, we get the length of the line, or AD.

(2,2) when plugged in for y and x make the linear equation true, therefore those coordinates fall on that line.

y=3x-4

Because this equation is true, the point must lie on the line. The other given answer choices do not result in true equalities.

We are looking for an ordered pair that makes the given equation true. To solve, plug in the various answer choices to find the true equality.

Because this equality is true, we can conclude that the point lies on this line. None of the other given answer options will result in a true equality.

You can solve for the within these two equations by eliminating the . By doing this, you get .

Solve for to get and plug back into either equation to get the value of as 1.

Solve for by setting the two equations equal to one another:

Plugging back into either equation gives .

These are the coordinates for the intersection of the two lines.

The easy way to solve this question is to take each set of points and put it into the equation. When we do this, we find the only time the equation balances is when we use the points .

For practice, try graphing the line to see which of the points fall on it.

To test the coordinates, plug the x-coordinate into the line equation and solve for y.

y = 1/3x -7

Test (3,-6)

y = 1/3(3) – 7 = 1 – 7 = -6 YES!

Test (3,7)

y = 1/3(3) – 7 = 1 – 7 = -6 NO

Test (6,-12)

y = 1/3(6) – 7 = 2 – 7 = -5 NO

Test (6,5)

y = 1/3(6) – 7 = 2 – 7 = -5 NO

Test (9,5)

y = 1/3(9) – 7 = 3 – 7 = -4 NO

Since the vertical coordinates of each point are given by y, solve each equation for y and plug in 6 for x, as follows:

Taking the difference of the resulting y -values give the vertical distance between the points (6,27) and (6,48), which is 21.

Since we have –2x and +2x in the equations, it makes sense to add the equations together to give 8y = 16 yielding y = 2. Then we substitute y = 2 into one of the original equations to get x = –2. So the solution to the system of equations is (–2, 2)

When the line crosses the x-axis, the y-coordinate is 0. Substitute 0 into the equation for y and solve for x.

.25(x – 20) + 12 = 0

.25_x_ – 5 = –12

.25_x_ = –7

x = –28

The answer is the point (–28,0).

BD+AC is the length of the line, except that BC is double counted. By subtracting BC, we get the length of the line, or AD.

(2,2) when plugged in for y and x make the linear equation true, therefore those coordinates fall on that line.

y=3x-4

Because this equation is true, the point must lie on the line. The other given answer choices do not result in true equalities.

We are looking for an ordered pair that makes the given equation true. To solve, plug in the various answer choices to find the true equality.

Because this equality is true, we can conclude that the point lies on this line. None of the other given answer options will result in a true equality.

You can solve for the within these two equations by eliminating the . By doing this, you get .

Solve for to get and plug back into either equation to get the value of as 1.

Solve for by setting the two equations equal to one another:

Plugging back into either equation gives .

These are the coordinates for the intersection of the two lines.

The easy way to solve this question is to take each set of points and put it into the equation. When we do this, we find the only time the equation balances is when we use the points .

For practice, try graphing the line to see which of the points fall on it.