GMAT Math : Lines

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Example Questions

Example Question #1 : Midpoint Formula

Which of the following quadrants can contain the midpoint of a line segment with endpoints (0,A) and (-A,0) for some nonzero value of ?

Possible Answers:

Quadrants I and III

Quadrants II and III

Quadrants I and IV

Quadrants III and IV

Quadrants II and IV

Correct answer:

Quadrants II and IV

Explanation:

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

If A > 0 , then the -coordinate is negative and the -coordinate is positive, so the midpoint is in Quadrant II. If A < 0 , the reverse is true, so the midpoint is in Quadrant IV.

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Example Question #2 : Midpoint Formula

The midpoint of a line segment with endpoints (A + 2, 2B-1) and (B+4, 2A+1) is (8,10) . Sove for .

Possible Answers:

It cannot be determined from the information given.

Correct answer:

It cannot be determined from the information given.

Explanation:

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:

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

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

A + B + 6 = 16

A + B = 10

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

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

A + B = 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.

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Example Question #1 : Lines

Find the midpoint of the points (2, 9) and (4, 3) .

Possible Answers:

(-3,6)

(3,6)

(2.5,5.5)

(6,12)

(2,5)

Correct answer:

(3,6)

Explanation:

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

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

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Example Question #2 : Lines

What is the midpoint of (-5,1) and (1,4) ?

Possible Answers:

(-3,6)

$(-2,\frac{5}{2})$

$(3,\frac{3}{2})$

$(3,\frac{5}{2})$

$(2,\frac{5}{2})$

Correct answer:

$(-2,\frac{5}{2})$

Explanation:

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

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

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Example Question #4 : Coordinate Geometry

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 $\overline{JK}$ ?

Possible Answers:

$\small (74,40)$

$\small \small (74,35)$

$\small \small (75,45)$

$\small \small (73,42)$

$\small \small (70,35)$

Correct answer:

$\small (74,40)$

Explanation:

Midpoint formula is as follows:

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

Plug in and calculate:

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

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Example Question #4 : Coordinate Geometry

Segment has endpoints of (-6,8) and (4,26) . If the midpoint of is given by point , what are the coordinates of point ?

Possible Answers:

(-1,17)

(17,-1)

(5,17)

(-5,-17)

(17,5)

Correct answer:

(-1,17)

Explanation:

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.

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

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Example Question #5 : Calculating The Midpoint Of A Line Segment

What are the coordinates of the mipdpoint of the line segment if A=(2,3) and B=(2,-7)?

Possible Answers:

(0,-2)

(2,4)

(2,-2)

(0,-10)

(0,4)

Correct answer:

(2,-2)

Explanation:

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

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Example Question #5 : Midpoint Formula

The quadrilateral with vertices (0,0),(22,0), (16,4), (4,4) is a trapezoid. What are the endpoints of its midsegment?

Possible Answers:

(0,0), (16,4)

(2,2),(19,2 )

(22,0), (4,4)

(0,2),(23,2 )

(11,0),(10,4)

Correct answer:

(2,2),(19,2 )

Explanation:

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 (0,0), (4,4) and (22,0), (16,4) . 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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Example Question #3 : Lines

The midpoint of a line segment with endpoints (2A, B) and (B + 7, 3A - 2) is (9, 6) . What is ?

Possible Answers:

It cannot be determined from the information given.

Correct answer:

Explanation:

If the midpoint of a line segment with endpoints (2A, B) and (B + 7, 3A - 2) is (9, 6) , 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:

2A + B + 7 = 18

2A + B = 11

The second can be simplified as follows:

3A + B - 2 = 12

3A + B = 14

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

3A + B = 14

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

$A \; \; \; \; \; \; \; \; = 3$

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Example Question #7 : Midpoint Formula

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 $\textup{Point }U$ ?

Possible Answers:

$\left ( 14,23\right )$

$\left ( 23,14\right )$

$\left ( -5,10\right )$

$\left ( 10,5\right )$

$\left ( 5,10\right )$

Correct answer:

$\left ( 10,5\right )$

Explanation:

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

$\small x'=6*2-2=10$ $\small y'=7*2-9=14-9=5$

So we have (10,5) as point U

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GMAT Math : Lines

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #1 : Midpoint Formula

Example Question #2 : Midpoint Formula

Example Question #1 : Lines

Example Question #2 : Lines

Example Question #4 : Coordinate Geometry

Example Question #4 : Coordinate Geometry

Example Question #5 : Calculating The Midpoint Of A Line Segment

Example Question #5 : Midpoint Formula

Example Question #3 : Lines

Example Question #7 : Midpoint Formula

All GMAT Math Resources

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