Lines - GMAT Quantitative

Card 0 of 912

Question

Determine whether and $y+8=-\frac{2}{3}x +2$ are parallel lines.

Answer

Parallel lines have the same slope. Therefore, we need to find the slope once both equations are in slope intercept form :

$y=-\frac{2}{3}x+\frac{10}{3}$

$m=-\frac{2}{3}$

$y+8=-\frac{2}{3}x +2$

$y=-\frac{2}{3}x+2-8$

$y=-\frac{2}{3}x-6$

$m=-\frac{2}{3}$

The lines are parallel because the slopes are the same.

Compare your answer with the correct one above

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:

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

Compare your answer with the correct one above

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)

Compare your answer with the correct one above

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

Compare your answer with the correct one above

Question

A line segment has its midpoint at and an endpoint at . What are the coordinates of the other endpoint?

Answer

Because we are given the midpoint and one of the endpoints, we know the x coordinate of the other endpoint will be the same distance away from the midpoint in the x direction, and the y coordinate of the other endpoint will be the same distance away from the midpoint in the y direction. Given two endpoints of the form:

The midpoint of these two endpoints has the coordinates:

$(x_m,y_m)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Plugging in values for the given midpoint and one of the endpoints, which we can see is because it lies to the right of the midpoint, we can solve for the other endpoint as follows:

$(x_m,y_m)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$(-1,3)=(\frac{x_1+3}{2},\frac{y_1+8}{2})$

$\frac{x_1+3}{2}=-1\rightarrow x_1+3=-2\rightarrow x_1=-5$

$\frac{y_1+8}{2}=3\rightarrow y_1+8=6\rightarrow y_1=-2$

So the other endpoint has the coordinates

Compare your answer with the correct one above

Question

Consider segment with endpoint at . If the midpoint of can be found at , what are the coordinates of point ?

Answer

Recall midpoint formula:

$(x',y')=\left(\frac{x^1+x^2}{2}, \frac{y^1+y^2}{2}\right)$

In this case we have (x'y') and one of our other (x,y) points.

Plug and chug:

$(-15,-67)=\left(\frac{8+x}{2}, \frac{9+y}{2}\right)$

If you make this into two equations and solve you get the following.

$\small -15=\frac{8+x}{2}, x=-38$

$\small -67=\frac{9+y}{2}, y=-143$

Compare your answer with the correct one above

Question

Find the midpoint of the points and .

Answer

Add the corresponding points together and divide both values by 2:

$(\frac{2+4}{2},\frac{9+3}{2}) = (3, 6)$

Compare your answer with the correct one above

Question

What is the midpoint of and ?

Answer

Add the x-values and divide by 2, and then add the y-values and divide by 2. Be careful of the negatives!

$(\frac{-5+1}{2},\frac{1+4}{2}) = (-2, \frac{5}{2})$

Compare your answer with the correct one above

Question

Which of the following quadrants can contain the midpoint of a line segment with endpoints and for some nonzero value of ?

Answer

The midpoint of the line segment with endpoints and is $\left ( \frac{0+ (-A) }{2}, \frac{A+ 0 }{2}, \right )$ , or $\left (- \frac{A }{2}, \frac{A }{2}, \right )$

If , then the -coordinate is negative and the -coordinate is positive, so the midpoint is in Quadrant II. If , the reverse is true, so the midpoint is in Quadrant IV.

Compare your answer with the correct one above

Question

The midpoint of a line segment with endpoints and is . Sove for .

Answer

The midpoint of a line segment with endpoints $(x_{1}, y_{1}), (x_{2}, y_{2})$ is

$\left ( \frac{x_{1}+x_{2} }{2},\frac{y_{1}+y_{2} }{2} \right )$ .

Substitute the coordinates of the endpoints, then set each equation to the appropriate midpoint coordinate.

-coordinate: $\frac{(A + 2)+ (B + 4) }{2} = 8$

-coordinate: $\frac{(2B-1)+(2A +1) }{2} = 10$

Simplify each, then solve the system of linear equations in two variables:

$\frac{(A + 2)+ (B + 4) }{2} = 8$

$\frac{(A + B + 6) }{2} = 8$

$\frac{(2B-1)+(2A +1) }{2} = 10$

$\frac{2A + 2B }{2} = 10$

The two linear equations turn out to be equivalent, meaning that there are infinitely many solutions to the system. Therefore, insufficient information is given to answer the question.

Compare your answer with the correct one above

Question

Consider segment $\overline{JK}$ which passes through the points $\left ( 4,5\right )$ and $\left ( 144,75\right )$ .

What are the correct coordinates for the midpoint of $\overline{JK}$ ?

Answer

Midpoint formula is as follows:

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

Plug in and calculate:

$\small \small \small m=(\frac{4+144}{2},\frac{75+5}{2})=(74,40)$

Compare your answer with the correct one above

Question

Segment has endpoints of and . If the midpoint of is given by point , what are the coordinates of point ?

Answer

Midpoints can be found using the following:

$midpoint(x,y)=\left(\frac{x+x'}{2},\frac{y+y'}{2}\right)$

Plug in our points (-6,8) and (4,26) to find the midpoint.

$\small \small midpoint(x,y)=\left(\frac{-6+4}{2},\frac{8+26}{2}\right)=\left(\frac{-2}{2},\frac{34}{2}\right)=(-1,17)$

Compare your answer with the correct one above

Question

What are the coordinates of the mipdpoint of the line segment if and

Answer

The midpoint formula is $(\frac{x_{2}+x_{1}}{2},\frac{y_{2}+y_{1}}{2}) = (\frac{2+2}{2},\frac{3+-7}{2})=(2,-2).$

Compare your answer with the correct one above

Question

What is the distance between the points and ?

Answer

Let's plug our coordinates into the distance formula.

$\sqrt{(2-7)^{2}+(5-17)^{2}}= \sqrt{(-5)^{2}+(-12)^{2}} = \sqrt{25+144}= \sqrt{169} = 13$

Compare your answer with the correct one above

Question

What is the distance between the points and ?

Answer

We need to use the distance formula to calculate the distance between these two points.

$\sqrt{(1-5)^{2}+(4-2)^{2}} = \sqrt{(-4)^{2}+(2)^{2}} =\sqrt{20}=\sqrt{4}\sqrt{5}=2\sqrt{5}$

Compare your answer with the correct one above

Question

A line segement on the coordinate plane has endpoints and . Which of the following expressions is equal to the length of the segment?

Answer

Apply the distance formula, setting

$x_{ 1} = A + 4, x_{2} = A ,y_{ 1} = B, y_{2} = B+3$ :

$d = \sqrt{(x_{2}- x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$d = \sqrt{[A-(A+4)]^{2}+[(B+3)-B)]^{2}}$

$d = \sqrt{(-4)^{2}+3^{2}} = \sqrt{16+9} = \sqrt{25} = 5$

Compare your answer with the correct one above

Question

What is distance between and ?

Answer

$distance= \sqrt{(x_{2}-x_{1})^2+y_{2}-y_{1})^2}$

$= \sqrt{(6-1)^2+(7-5)^2}$

$= \sqrt{(5)^2+(2)^2}$

$= \sqrt{(25+4)}$

$\sqrt{29}$

Compare your answer with the correct one above

Question

Consider segment $\overline{JK}$ which passes through the points $\left ( 4,5\right )$ and $\left ( 144,75\right )$ .

Find the length of segment $\overline{JK}$ .

Answer

This question requires careful application of distance formula, which is really a modified form of Pythagorean theorem.

$\small d=\sqrt{(x-x')^2+(y-y')^2}$

Plug in everthing and solve:

$\small \small d=\sqrt{(144-4)^2+(75-7)^2}=\sqrt{24500}\approx156.5$

So our answer is 156.6

Compare your answer with the correct one above

Question

What is the length of a line segment that starts at the point and ends at the point ?

Answer

Using the distance formula for the length of a line between two points, we can plug in the given values and determine the length of the line segment by calculating the distance between the two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$d=\sqrt{(4-1)^2+(1-(-3))^2}$

$d=\sqrt{(3)^2+(4)^2}$

$d=\sqrt{25}$

Compare your answer with the correct one above

Question

What is the equation of the line that is perpendicular to and goes through point ?

Answer

Perpendicular lines have slopes that are negative reciprocals of each other.

The slope for the given line is , from , where is the slope. Therefore, the negative reciprocal is $-\frac{1}{2}$ .

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

$y-y_{1}=m(x-x_{1})$

$y-1=-\frac{1}{2}(x-5)$

$y-1=-\frac{1}{2}x+\frac{5}{2}$

$y=-\frac{1}{2}x+\frac{5}{2}+1$

$y=-\frac{1}{2}x+\frac{5}{2}+\frac{2}{2}$

$y=-\frac{1}{2}x+\frac{7}{2}$

Compare your answer with the correct one above

Tap the card to reveal the answer