Midpoint Formula - GMAT Quantitative

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Find the midpoint of the points and .

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

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

What is the midpoint of and ?

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

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

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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 .

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The midpoint of a line segment with endpoints 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.

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 $\overline{JK}$ ?

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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 ?

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

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The midpoint formula is

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

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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:

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

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

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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)

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 $\textup{Point }U$ ?

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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

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

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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

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

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Recall midpoint formula:

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$

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

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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:

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

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

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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)

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 $\textup{Point }U$ ?

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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

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

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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

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

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Recall midpoint formula:

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$

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

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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:

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

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

Tap to see back →

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)

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 $\textup{Point }U$ ?

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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