GMAT Math : Lines

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Example Questions

Example Question #1 : Lines

Find the equation of the line that is perpendicular to the line connecting the points $\dpi{100} \small (0,-4)\ and\ (-1,-7)$ .

Possible Answers:

the line between points $\dpi{100} \small (0,0)\ and\ (2,2)$

the line between the points $\dpi{100} \small (3,0)\ and\ (-3,2)$

$\dpi{100} \small y=\frac{x}{3}+1$

$\dpi{100} \small y=3x-1$

$\dpi{100} \small y=-4x+8$

Correct answer:

the line between the points $\dpi{100} \small (3,0)\ and\ (-3,2)$

Explanation:

Lines are perpendicular if their slopes are negative reciprocals of each other. First we need to find the slope of the line in the question stem.

$slope = \frac{rise}{run} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{-7 + 4}{-1 - 0} = \frac{-3}{-1} = 3$

The negative reciprocal of 3 is $\dpi{100} \small -\frac{1}{3}$ , so our answer will have a slope of $\dpi{100} \small -\frac{1}{3}$ . Let's go through the answer choices and see.

$\dpi{100} \small y=3x-1$ : This line is of the form $\dpi{100} \small y=mx+b$ , where $\dpi{100} \small m$ is the slope. The slope is 3, so this line is parallel, not perpendicular, to our line in question.

$\dpi{100} \small y=-4x+8$ : The slope here is $\dpi{100} \small -4$ , also wrong.

$\dpi{100} \small y=\frac{x}{3}+1$ : The slope of this line is $\dpi{100} \small \frac{1}{3}$ . This is the reciprocal, but not the negative reciprocal, so this is also incorrect.

The line between the points $\dpi{100} \small (3,0)\ and\ (-3,2)$ : $\dpi{100} \small slope = \frac{2}{(-3-3)}=\frac{2}{-6}=-\frac{1}{3}$ .

This is the correct answer! Let's check the last answer choice as well.

The line between points $\dpi{100} \small (0,0)\ and\ (2,2)$ : $\dpi{100} \small slope = \frac{2}{2}=1$ , which is incorrect.

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Example Question #2 : Lines

Determine whether the lines with equations 3y+x=2 and 4y-2x=3 are perpendicular.

Possible Answers:

There is not enough information to determine the answer

They are not perpendicular

They are perpendicular

Correct answer:

They are not perpendicular

Explanation:

If two equations are perpendicular, then they will have inverse negative slopes of each other. So if we compare the slopes of the two equations, then we can find the answer. For the first equation we have $3y+x=2\Rightarrow y=\frac{-1}{3}x+\frac{2}{3}$

so the slope is $-\frac{1}{3}$ .

So for the equations to be perpendicular, the other equation needs to have a slope of 3. For the second equation, we have

$4y-2x=3\Rightarrow y=\frac{1}{2}x+\frac{3}{4}$

so the slope is $\frac{1}{2}$ .

Since the slope of the second equation is not equal to 3, then the lines are not perpendicular.

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Example Question #2 : Lines

Transversal

Refer to the above figure. l || m . True or false: $m \perp t$

Statement 1: $\angle 3 \cong \angle 5$

Statement 2: $\angle 2$ and $\angle 6$ are supplementary.

Possible Answers:

BOTH statements TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient to answer the question.

EITHER statement ALONE is sufficient to answer the question.

BOTH statements TOGETHER are insufficient to answer the question.

Statement 1 ALONE is sufficient to answer the question, but Statement 2 ALONE is NOT sufficient to answer the question.

Statement 2 ALONE is sufficient to answer the question, but Statement 1 ALONE is NOT sufficient to answer the question.

Correct answer:

EITHER statement ALONE is sufficient to answer the question.

Explanation:

If transversal crosses two parallel lines and , then same-side interior angles are supplementary, so $\angle 3$ and $\angle 5$ are supplementary angles. Also, corresponding angles are congruent, so $\angle 2 \cong \angle 6$ .

By Statement 1 alone, angles $\angle 3$ and $\angle 5$ are congruent as well as supplementary; by Statement 2 alone, $\angle 2$ and $\angle 6$ are also supplementary as well as congruent. Two angles that are both supplementary and congruent are both right angles, so from either statement alone, and intersect at right angles, so, consequently, $m \perp t$ .

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Example Question #383 : Geometry

Transversal

Figure NOT drawn to scale.

Refer to the above figure.

True or false: $m \perp t$

Statement 1: $\angle 4$ is a right angle.

Statement 2: $\angle 3$ and $\angle 5$ are supplementary.

Possible Answers:

BOTH statements TOGETHER are insufficient to answer the question.

EITHER statement ALONE is sufficient to answer the question.

Statement 1 ALONE is sufficient to answer the question, but Statement 2 ALONE is NOT sufficient to answer the question.

Statement 2 ALONE is sufficient to answer the question, but Statement 1 ALONE is NOT sufficient to answer the question.

BOTH statements TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient to answer the question.

Correct answer:

BOTH statements TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient to answer the question.

Explanation:

Statement 1 alone establishes by definition that $l\perp t$ , but does not establish any relationship between and .

By Statement 2 alone, since same-side interior angles are supplementary, l || m , but no conclusion can be drawn about the relationship of , since the actual measures of the angles are not given.

Assume both statements are true. If two lines are parallel, then any line in their plane perpendicular to one must be perpendicular to the other. l || m and $l\perp t$ , so it can be established that $m \perp t$ .

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Example Question #3 : Lines

Find the equation of the line that is perpendicular to the following equation and passes through the point (4,7) .

$y=-\frac{x}{2}+14$

Possible Answers:

y=4x-8

y=2x+1

$y=\frac{1}{2}x-1$

y=2x-1

y=2x-14

Correct answer:

y=2x-1

Explanation:

To solve this equation, we want to begin by recalling how to find the slope of a perpendicular line. In this case, our original line is modeled by the following:

$y=-\frac{x}{2}+14$

To find the slope of any line perpendicular to the above equation, we simply need to take the reciprocal of the first slope, and then change its sign. Our original slope is $-\frac{1}{2}$ , so

$m=-\frac{1}{2}$

becomes

{m}'=2 .

If we flip $\frac{1}{2}$ , we get , and the opposite sign of a negative is a positive; hence, our slope is positive .

So, we know our perpendicular line should look something like this:

y=2x+b

However, we need to find out what (our -intercept) is in order to complete our equation. To do so, we need to plug in the ordered pair we received in the question, (4,7) , and solve for :

7=2*4+b

7=8+b

7-8=b

b=-1

So, by putting everything together, we get our final equation:

y=2x-1

This equation satisfies the conditions of being perpendicular to our initial equation and passing through (4,7) .

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Example Question #2 : Lines

Which of the following lines is perpendicular to $y=-\frac{3}{4}x+7$ ?

Possible Answers:

$y=\frac{3}{4}x+7$

$y=-\frac{4}{3}x+7$

$y=-\frac{3}{4}x-7$

Two of the equations are perpendicular to the given line.

$y=\frac{4}{3}x+7$

Correct answer:

$y=\frac{4}{3}x+7$

Explanation:

In order for a line to be perpendicular to another line defined by the equation y=mx+b , the slope of line must be a negative reciprocal of the slope of line . Since line 's slope is in the slope-intercept equation above, line 's slope would therefore be $-\frac{1}{m}$ .

In this instance, $m=-\frac{3}{4}$ , so $-\frac{1}{m}=-\frac{1}{-\frac{3}{4}}=\frac{4}{3}$ . Therefore, the correct solution is $y=\frac{4}{3}x+7$ .

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Example Question #2 : Calculating Whether Lines Are Perpendicular

A given line has a slope of m=6 . What is the slope of any line perpendicular to ?

Possible Answers:

Not enough information provided

$-\frac{1}{6}$

Correct answer:

$-\frac{1}{6}$

Explanation:

Given that we have a line with a slope m=6 , we can therefore conclude that any perpendicular line would have a slope $-\frac{1}{m}=-\frac{1}{6}$ .

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Example Question #3 : Calculating Whether Lines Are Perpendicular

Which of the following lines are perpendicular to $y=\frac{5}{6}x+10$ ?

Possible Answers:

$y=-\frac{6}{5}x-7$

Two answers are perpendicular to the given line.

$y=-\frac{5}{6}x-2$

$y=\frac{5}{6}x-7$

$y=-\frac{6}{5}x+10$

Correct answer:

Two answers are perpendicular to the given line.

Explanation:

Since in this instance the slope $m=\frac{5}{6}$ , $-\frac{1}{m}=-\frac{1}{\frac{5}{6}}=-\frac{6}{5}$ . Two of the above answers have this as their slope, so therefore that is the answer to our question.

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Example Question #4 : Calculating Whether Lines Are Perpendicular

Do the functions f(x) and g(x) intersect at a ninety-degree angle, and how can you tell?

$\small f(x)=3x+7$

$\small g(x)=-\frac{1}{3}x+7$

Possible Answers:

It is impossible to determine from the information provided.

Yes, because f(x) and g(x) have the same y-intercept.

No, because f(x) and g(x) never intersect.

Yes, because the slope of g(x) is the reciprocal of the slope of f(x) and it has the opposite sign.

No, because f(x) and g(x) have different slopes.

Correct answer:

Yes, because the slope of g(x) is the reciprocal of the slope of f(x) and it has the opposite sign.

Explanation:

If two lines intersect at a ninety-degree angle, they are said to be perpendicular. Two lines are perpendicular if their slopes are opposite reciprocals. In this case:

$f(x)\:slope=\frac{3}{1}$

$g(x)\:slope=-\frac{1}{3}$

The two lines' slopes are reciprocals with opposing signs, so the answer is yes. Of our two yes answers, only one has the right explanation. Eliminate the option dealing with -intercepts.

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Example Question #5 : Calculating Whether Lines Are Perpendicular

Find the slope of a line that is perpendicular to the line running through the points (7,4) and (5,8) .

Possible Answers:

$\frac{1}{2}$

$-\frac{1}{2}$

Not enough information provided.

Correct answer:

$\frac{1}{2}$

Explanation:

To find the slope of the line running through (7,4) and (5,8) , we use the following equation:

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{8-4}{5-7}=\frac{4}{-2}=-2$

The slope of any line perpendicular to the given line would have a slope that is the negative reciprocal of , or $-\frac{1}{m}$ . Therefore, $-\frac{1}{-2}=\frac{1}{2}$

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GMAT Math : Lines

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #1 : Lines

Example Question #2 : Lines

Example Question #2 : Lines

Example Question #383 : Geometry

Example Question #3 : Lines

Example Question #2 : Lines

Example Question #2 : Calculating Whether Lines Are Perpendicular

Example Question #3 : Calculating Whether Lines Are Perpendicular

Example Question #4 : Calculating Whether Lines Are Perpendicular

Example Question #5 : Calculating Whether Lines Are Perpendicular

All GMAT Math Resources

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