GRE Math : Lines

Study concepts, example questions & explanations for GRE Math

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

Example Questions

Example Question #1 : Other Lines

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:

21

44

48

12

36

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

For the line

\displaystyle y=\frac{1}{3}x-7

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

Possible Answers:

(9, 5)

(6, 12)

(6, 5)

(3, 6)

(3, 7)

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 #21 : Other Lines

Solve the following system of equations:

–2x + 3y = 10

2x + 5y = 6

Possible Answers:

(–2, 2)

(–2, –2)

(3, 5)

(2, 2)

(3, –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 #481 : Geometry

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

Possible Answers:

(2,-2)\displaystyle (2,-2)

(2,2)\displaystyle (2,2)

(3,4)\displaystyle (3,4)

(1,2)\displaystyle (1,2)

(1,5)\displaystyle (1,5)

Correct answer:

(2,2)\displaystyle (2,2)

Explanation:

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

y=3x-4\displaystyle y=3x-4

\displaystyle 2=3(2)-4

\displaystyle 2=6-4

\displaystyle 2=2

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

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

Possible Answers:

\displaystyle (2, 7)

\displaystyle (0, 1)

\displaystyle (-1,2)

\displaystyle (1, 5)

\displaystyle (1, 0)

Correct answer:

\displaystyle (1, 5)

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.

\displaystyle y=3x+2

\displaystyle 5=3(1)+2

\displaystyle 5=3+2

\displaystyle 5=5

Because this equality is true, we can conclude that the point \displaystyle (1,5) 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 = –5x – 3

y = –x – 3

y = –5x + 3

y = –x + 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 #1 : 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 = (–1/5)x + 2.75

y = (–3/5)x + 22/5

y = (–3/5)x + 28/5

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

y = –x + 5

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 \displaystyle (-2,17) and \displaystyle (5,-11)?

Possible Answers:

\displaystyle y=\frac{12}{5}x-20

\displaystyle y=-27x-15

\displaystyle y=-4x+9

\displaystyle y=4x+17

\displaystyle y=\frac{14}{3}x+17

Correct answer:

\displaystyle y=-4x+9

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:

\displaystyle \frac{rise}{run}=\frac{17-(-11)}{-2-5}=\frac{17+11}{-7}=\frac{-28}{7}=-4

Now, for a point \displaystyle (a,b) and a slope \displaystyle m, the point-slope form of a line is:

\displaystyle y-b=m(x-a)

Let's use \displaystyle (-2,17) for our point \displaystyle (a,b)

This gives us:

\displaystyle y-17=-4(x+2)

Now, distribute and solve for \displaystyle y:

\displaystyle y-17=-4x-8

\displaystyle y=-4x+9

 

 

Example Question #1 : How To Find The Equation Of A Line

What is the equation of a line passing through the two points \displaystyle (41,11) and \displaystyle (4,-9)?

Possible Answers:

\displaystyle y=14x-18

\displaystyle y=\frac{20}{27}x-\frac{14}{15}

\displaystyle y=\frac{20}{37}x-\frac{413}{37}

\displaystyle y=\frac{17}{14}x-\frac{148}{25}

\displaystyle y=\frac{7}{2}x-\frac{85}{3}

Correct answer:

\displaystyle y=\frac{20}{37}x-\frac{413}{37}

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:

\displaystyle \frac{rise}{run}=\frac{11-(-9)}{41-4}=\frac{11+9}{37}=\frac{20}{37}

Now, for a point \displaystyle (a,b) and a slope \displaystyle m, the point-slope form of a line is:

\displaystyle y-b=m(x-a)

Let's use \displaystyle (4,-9) for our point \displaystyle (a,b)

This gives us:

\displaystyle y+9=\frac{20}{37}(x-4)

Now, distribute and solve for \displaystyle y:

\displaystyle y+9=\frac{20}{37}x-\frac{80}{37}

\displaystyle y=\frac{20}{37}x-\frac{80}{37}-9

\displaystyle y=\frac{20}{37}x-\frac{80}{37}-\frac{333}{37}

\displaystyle y=\frac{20}{37}x-\frac{413}{37}

 

Example Question #243 : Geometry

What is the equation of a line passing through \displaystyle (4,-12) with a \displaystyle y-intercept of \displaystyle 9?

Possible Answers:

\displaystyle y=-5x-9

\displaystyle y=\frac{-15}{4}x+12

\displaystyle y=\frac{27}{5}x+9

\displaystyle y=-22x+15

\displaystyle y=\frac{-21}{4}x+9

Correct answer:

\displaystyle y=\frac{-21}{4}x+9

Explanation:

Based on the information that you have been provided, you can quickly find the slope of your line.  Since the y-intercept is \displaystyle 9, you know that the line contains the point \displaystyle (0,9).  Therefore, the slope of the line is found:

\displaystyle \frac{-12-9}{4-0}=\frac{-21}{4}

Based on this information, you can use the standard slope-intercept form to find your equation:

\displaystyle y=mx+b, where \displaystyle m=\frac{-21}{4} and \displaystyle b=9

\displaystyle y=\frac{-21}{4}x+9

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