Midpoint Formula - GMAT Quantitative

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Question

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

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Answer

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

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Question

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

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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?

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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 $\textup{Point }U$ ?

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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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Question

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

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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

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Question

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

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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.

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

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

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Question

Find the midpoint of the points and .

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Answer

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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Question

What is the midpoint of and ?

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Answer

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

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

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Question

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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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.

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Question

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

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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.

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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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Answer

Midpoint formula is as follows:

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

Plug in and calculate:

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

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Question

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

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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.

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

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Question

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

Tap to reveal answer

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:

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

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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 $\textup{Point }U$ ?

Tap to reveal answer

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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Question

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

Tap to reveal answer

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

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Question

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

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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.

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

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

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Question

Find the midpoint of the points and .

Tap to reveal answer

Answer

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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Question

What is the midpoint of and ?

Tap to reveal answer

Answer

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

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

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Question

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

Tap to reveal answer

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.

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