GRE Math : Coordinate Geometry

Study concepts, example questions & explanations for GRE Math

varsity tutors app store varsity tutors android store varsity tutors amazon store varsity tutors ibooks store

Example Questions

Example Question #1 : 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.

Possible Answers:

12

44

48

36

21

Correct answer:

21

Explanation:

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 -values give the vertical distance between the points (6,27) and (6,48), which is 21.

Example Question #1 : How To Find Out If A Point Is On A Line With An Equation

For the line

Which one of these coordinates can be found on the line?

Possible Answers:

(3, 7)

(9, 5)

(6, 12)

(3, 6)

(6, 5)

Correct answer:

(3, 6)

Explanation:

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 #91 : Lines

Solve the following system of equations:

–2x + 3y = 10

2x + 5y = 6

Possible Answers:

(–2, 2)

(3, –2)

(2, 2)

(3, 5)

(–2, –2)

Correct answer:

(–2, 2)

Explanation:

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 #1 : How To Find Out If A Point Is On A Line With An Equation

Which of the following sets of coordinates are on the line y=3x-4?

Possible Answers:

(2,-2)

(1,2)

(3,4)

(2,2)

(1,5)

Correct answer:

(2,2)

Explanation:

(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.

Example Question #98 : Coordinate Geometry

Which of the following points can be found on the line \small y=3x+2?

Possible Answers:

Correct answer:

Explanation:

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?

Possible Answers:

y = –x + 3

y = –5x + 3

y = –x – 3

y = –5x – 3

Correct answer:

y = –x + 3

Explanation:

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 (y y1)/(x– 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.

Gre_quant_179_01

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?

Possible Answers:

y = (–3/5)x + 22/5

y = (3/5)x + 22/5

y = –x + 5

y = (–3/5)x + 28/5

y = (–1/5)x + 2.75

Correct answer:

y = (–3/5)x + 22/5

Explanation:

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 #1 : How To Find The Equation Of A Line

What is the equation of a line passing through the points  and ?

Possible Answers:

Correct answer:

Explanation:

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 #4 : How To Find The Equation Of A Line

What is the equation of a line passing through the two points  and ?

Possible Answers:

Correct answer:

Explanation:

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 ?

Possible Answers:

Correct answer:

Explanation:

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 

Tired of practice problems?

Try live online GRE prep today.

1-on-1 Tutoring
Live Online Class
1-on-1 + Class
Learning Tools by Varsity Tutors