Coordinate Geometry - GMAT Quantitative

Card 0 of 2080

Question

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

Answer

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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Question

Determine whether the lines with equations and are perpendicular.

Answer

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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Question

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.

Answer

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, , 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. and $l\perp t$ , so it can be established that $m \perp t$ .

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Question

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

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

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

Answer

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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Question

Find the equation of the line that is perpendicular to the following equation and passes through the point .

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

Answer

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

.

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:

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, , and solve for :

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

This equation satisfies the conditions of being perpendicular to our initial equation and passing through .

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Question

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

Answer

In order for a line to be perpendicular to another line defined by the equation , 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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Question

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

Answer

In order for a line to be perpendicular to another line defined by the equation , 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 $-\frac{1}{m}$ .

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

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Question

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

Answer

In order for a line to be perpendicular to another line defined by the equation , 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 $-\frac{1}{m}$ .

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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Question

Find the slope of a line that is perpendicular to the line running through the points and .

Answer

To find the slope of the line running through and , 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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Question

Which of the following lines is perpendicular to $y=\frac{3}{4}x+16$ ?

Answer

Given a line defined by the equation with a slope of , any line perpendicular to would have a slope that is the negative reciprocal of , $-\frac{1}{m}$ . Given our equation $y=\frac{3}{4}x+16$ , we know that $m=\frac{3}{4}$ and that $-\frac{1}{m}=-\frac{1}{\frac{3}{4}}=-\frac{4}{3}$ .

The only answer choice with this slope is $y=-\frac{4}{3}x+12$ .

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Question

Which of the following lines is perpendicular to

Answer

Given a line defined by the equation with a slope of , any line perpendicular to would have a slope that is the negative reciprocal of , $-\frac{1}{m}$ . Given our equation , we know that and that $-\frac{1}{m}=-\frac{1}{5}$ .

There are two answer choices with this slope, $y=-\frac{1}{5}x-4$ and $y=-\frac{1}{5}x+7$ .

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Question

Do the functions and intersect at a ninety-degree angle, and how can you tell?

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

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

Answer

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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Question

A given line is defined by the equation . Which of the following lines would be perpendicular to line ?

Answer

For any line with an equation and slope , a line that is perpendicular to must have a slope of $-\frac{1}{m}$ , or the negative reciprocal of . Given , we know that and therefore know that $-\frac{1}{m}=-\frac{1}{2}$ .

Only one equation above has a slope of $-\frac{1}{2}$ : $y=-\frac{1}{2}x+7$ .

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Question

What is the slope of a line that is perpendicular to $y=-\frac{1}{7}x+9$

Answer

For any line with an equation and slope , a line that is perpendicular to must have a slope of $-\frac{1}{m}$ , or the negative reciprocal of . Given the equation $y=-\frac{1}{7}x+9$ , we know that $m=-\frac{1}{7}$ and therefore know that $-\frac{1}{m}=-\frac{1}{-\frac{1}{7}}=7$ .

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Question

Which of the following lines is perpendicular to ?

Answer

For any line with an equation and slope , a line that is perpendicular to must have a slope of $-\frac{1}{m}$ , or the negative reciprocal of . Given the equation , we know that and therefore know that $-\frac{1}{m}=-\frac{1}{4}$ .

Given a slope of $-\frac{1}{4}$ , we know that there are two solutions provided: $y=-\frac{1}{4}x+5$ and $y=-\frac{1}{4}x-4$ .

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Question

What is the slope of a line perpendicular to that of

Answer

First, we need to rearrange the equation into slope-intercept form. .

$6y-3x=2->y=\frac{3x+2}{6}.$ Therefore, the slope of this line equals $\frac{1}{2}.$ Perpendicular lines have slope that are the opposite reciprocal, or

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Question

If is defined as follows, is the point $\left ( -2,5\right )$ on ?

$\small f(x)=4x+13$

Answer

To find out if (-2,5) is on f(x), simply plug the point into f(x)

$\small f(x)=4x+13$

Becomes,

$\small \small 5=4*-2+13=-8+13=5$

So yes, it does!

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Question

The midpoint of a line segment with endpoints and is . What is ?

Answer

If the midpoint of a line segment with endpoints and is , then by the midpoint formula,

$\frac{2A + (B + 7)}{2} = 9$

and

$\frac{B + (3A - 2)}{2} = 6$ .

The first equation can be simplified as follows:

or

The second can be simplified as follows:

or

This is a system of linear equations. can be calculated by subtracting:

$\underline{2A + B = 11 }$

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Question

The quadrilateral with vertices is a trapezoid. What are the endpoints of its midsegment?

Answer

The midsegment of a trapezoid is the segment whose endpoints are the midpoints of its legs - its nonparallel opposite sides. These two sides are the ones with endpoints and . The midpoint of each can be found by taking the means of the - and -coordinates:

$(0,0), (4,4): \left ( \frac{0+4}{2},\frac{0+4}{2} \right ) \Rightarrow (2,2)$

$(22,0), (16,4): \left ( \frac{22+16}{2},\frac{0+4}{2} \right ) \Rightarrow (19,2)$

The midsegment is the segment that has endpoints (2,2) and (19,2)

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Question

If the midpoint of $\overline{VU}$ is $\left ( 6,7\right )$ and $\textup{Point }V$ is at $\left (2,9\right )$ , what are the coordinates of $\textup{Point }U$ ?

Answer

Midpoint formula is as follows:

$\small \small m(x,y)=(\frac{x+x'}{2},\frac{y+y'}{2})$

In this case, we have x,y and the value of the midpoint. We need to findx' and y'

V is at (2,9) and the midpoint is at (6,7)

$\small 6=\frac{2+x'}{2}$ and $\small 7=\frac{9+y'}{2}$

$\small x'=62-2=10$ $\small y'=72-9=14-9=5$

So we have (10,5) as point U

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