Calculating the endpoints of a line segment - GMAT Quantitative

Card 0 of 40

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

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

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

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

Tap the card to reveal the answer