Calculating the midpoint of a line segment

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GMAT Quantitative › Calculating the midpoint of a line segment

Questions 1 - 7

Find the midpoint of the points and .

Explanation

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

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

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 $\overline{JK}$ ?

$\small (74,40)$

$\small \small (70,35)$

$\small \small (74,35)$

$\small \small (75,45)$

$\small \small (73,42)$

Explanation

Midpoint formula is as follows:

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

Plug in and calculate:

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

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

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.

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

What is the midpoint of and ?

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

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

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

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

Explanation

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

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

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

Explanation

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

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

Quadrants II and IV

Quadrants II and III

Quadrants III and IV

Quadrants I and III

Quadrants I and IV

Explanation

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.

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

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:

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

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

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

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

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.

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