All GRE Math Resources
Example Questions
Example Question #2 : How To Find Out If A Point Is On A Line With An Equation
Consider the lines described by the following two equations:
4y = 3x2
3y = 4x2
Find the vertical distance between the two lines at the points where x = 6.
44
48
12
21
36
21
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.
Example Question #3 : Other Lines
For the line
Which one of these coordinates can be found on the line?
(3, –6)
(9, 5)
(6, –12)
(6, 5)
(3, 7)
(3, –6)
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
Example Question #52 : Lines
Solve the following system of equations:
–2x + 3y = 10
2x + 5y = 6
(3, 5)
(2, 2)
(3, –2)
(–2, –2)
(–2, 2)
(–2, 2)
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)
Example Question #51 : Coordinate Geometry
Which of the following sets of coordinates are on the line ?
when plugged in for and make the linear equation true, therefore those coordinates fall on that line.
Because this equation is true, the point must lie on the line. The other given answer choices do not result in true equalities.
Example Question #52 : Coordinate Geometry
Which of the following points can be found on the line ?
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.
Example Question #1 : How To Find The Equation Of A Line
What is the equation of the straight line passing through (–2, 5) with an x-intercept of 3?
y = –x – 3
y = –x + 3
y = –5x – 3
y = –5x + 3
y = –x + 3
First you must figure out what point has an x-intercept of 3. This means the line crosses the x-axis at 3 and has no rise or fall on the y-axis which is equivalent to (3, 0). Now you use the formula (y2 – y1)/(x2 – x1) to determine the slope of the line which is (5 – 0)/(–2 – 3) or –1. Now substitute a point known on the line (such as (–2, 5) or (3, 0)) to determine the y-intercept of the equation y = –x + b. b = 3 so the entire equation is y = –x + 3.
Example Question #2 : How To Find The Equation Of A Line
What is the equation of the line passing through (–1,5) and the upper-right corner of a square with a center at the origin and a perimeter of 22?
y = (3/5)x + 22/5
y = (–1/5)x + 2.75
y = –x + 5
y = (–3/5)x + 22/5
y = (–3/5)x + 28/5
y = (–3/5)x + 22/5
If the square has a perimeter of 22, each side is 22/4 or 5.5. This means that the upper-right corner is (2.75, 2.75)—remember that each side will be "split in half" by the x and y axes.
Using the two points we have, we can ascertain our line's equation by using the point-slope formula. Let us first get our slope:
m = rise/run = (2.75 – 5)/(2.75 + 1) = –2.25/3.75 = –(9/4)/(15/4) = –9/15 = –3/5.
The point-slope form is: y – y0 = m(x – x0). Based on our data this is: y – 5 = (–3/5)(x + 1); Simplifying, we get: y = (–3/5)x – (3/5) + 5; y = (–3/5)x + 22/5
Example Question #243 : Geometry
What is the equation of a line passing through the points and ?
Based on the information provided, you can find the slope of this line easily. From that, you can use the point-slope form of the equation of a line to compute the line's full equation. The slope is merely:
Now, for a point and a slope , the point-slope form of a line is:
Let's use for our point
This gives us:
Now, distribute and solve for :
Example Question #2 : How To Find The Equation Of A Line
What is the equation of a line passing through the two points and ?
Based on the information provided, you can find the slope of this line easily. From that, you can use the point-slope form of the equation of a line to compute the line's full equation. The slope is merely:
Now, for a point and a slope , the point-slope form of a line is:
Let's use for our point
This gives us:
Now, distribute and solve for :
Example Question #5 : How To Find The Equation Of A Line
What is the equation of a line passing through with a -intercept of ?
Based on the information that you have been provided, you can quickly find the slope of your line. Since the y-intercept is , you know that the line contains the point . Therefore, the slope of the line is found:
Based on this information, you can use the standard slope-intercept form to find your equation:
, where and