How to find the endpoints of a line segment - PSAT Math

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Question

The midpoint of line segment AB is (2, -5). If the coordinates of point A are (4, 4), what are the coordinates of B?

Answer

The fastest way to find the missing endpoint is to determine the distance from the known endpoint to the midpoint and then performing the same transformation on the midpoint. In this case, the x-coordinate moves from 4 to 2, or down by 2, so the new x-coordinate must be 2-2 = 0. The y-coordinate moves from 4 to -5, or down by 9, so the new y-coordinate must be -5-9 = -14.

An alternate solution would be to substitute (4,4) for (x1,y1) and (2,-5) for (x,y) into the midpoint formula:

x=(x1+x2)/2

y=(y1+y2)/2

Solving each equation for (x2,y2) yields the solution (0,-14).

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Question

Line segment AB has an endpoint, A, located at , and a midpoint at . What are the coordinates for point B of segment AB?

Answer

With an endpoint A located at (10,-1), and a midpoint at (10,0), we want to add the length from A to the midpoint onto the other side of the segment to find point B. The total length of the segment must be twice the distance from A to the midpoint.

A is located exactly one unit below the midpoint along the y-axis, for a total displacement of (0,1). To find point B, we add (10+0, 0+1), and get the coordinates for B: (10,1).

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Question

Point A is (5, 7). Point B is (x, y). The midpoint of AB is (17, –4). What is the value of B?

Answer

Point A is (5, 7). Point B is (x, y). The midpoint of AB is (17, –4). What is the value of B?

We need to use our generalized midpoint formula:

MP = ( (5 + x)/2, (7 + y)/2 )

Solve each separately:

(5 + x)/2 = 17 → 5 + x = 34 → x = 29

(7 + y)/2 = –4 → 7 + y = –8 → y = –15

Therefore, B is (29, –15).

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Question

One endpoint of a line segment on the coordinate plane is . The segment has length 10; the other endpoint has -coordinate 10 and is in Quadrant I. Give the -coordinate of the other endpoint. (Nearest tenth if applicable).

Answer

Let be the -coordinate of the other endpoint.

The segment has endpoints and .

Apply the distance formula

$d = \sqrt{(x_{2}-x_{1})^{2}+ (y_{2}-y_{1})^{2} }$

setting $d = 10, x_{1}= 3, x_{2} = 10, y_{1} = 7, y_{2}= Y$ ,

$10 = \sqrt{(10-3)^{2}+ (Y-7)^{2} }$

$100 =7^{2}+ (Y-7)^{2}$

$49+ (Y-7)^{2} = 100$

$(Y-7)^{2} = 51$

$Y-7 = \pm \sqrt{51}$

$Y = 7 \pm \sqrt{51}$

Therefore, there are two values of which fit the distance criterion.

One is $Y = 7 - \sqrt{51} \approx 7 - 7.1 \approx -0.1$ .

However, since the endpoint is in Quadrant I, it must have a positive coordinate, so this is eliminated as a choice.

The other is $Y = 7 + \sqrt{51} \approx 7 + 7.1 \approx 14.1$ , which is correct since it is positive.

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Question

The midpoint of a line segment is the point . One endpoint is ; give the -coordinate of the other.

Answer

Let the endpoints of a line segment be

$\left ( x_{1}, y_{1}\right ), \left ( x_{2}, y_{2}\right )$

Then the midpoint of the segment will be

$\left ( \frac{x_{1}+x_{2} }{2}, \frac{y_{1}+y_{2} }{2} \right )$

The -coordinate of the midpoint is $x_{m}= 1.4$ , and the -coordinate of the known endpoint is $x_{1} = 5.1$ , so we can solve for $x_{2}$ , the -coordinate of the unknown endpoint, in the following equation:

$\frac{x_{1}+x_{2} }{2} = x_{m}$

$\frac{5.1+x_{2} }{2} =1.4$

$5.1+x_{2} = 2.8$

$x_{2} = -2.3$

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Question

The midpoint of a line segment is the point . One endpoint is ; give the -coordinate of the other.

Answer

Let the endpoints of a line segment be

$\left ( x_{1}, y_{1}\right ), \left ( x_{2}, y_{2}\right )$

Then the midpoint of the segment will be

$\left ( \frac{x_{1}+x_{2} }{2}, \frac{y_{1}+y_{2} }{2} \right )$

The -coordinate of the midpoint is $y_{m}= -4.3$ , and the -coordinate of the known endpoint is $y_{1} = 3.1$ , so we can solve for $y_{2}$ , the -coordinate of the unknown endpoint, in the following equation:

$\frac{y_{1}+y_{2} }{2} = y_{m}$

$\frac{3.1+y_{2} }{2} =-4.3$

$3.1+x_{2} = -8.6$

$x_{2} = -11.7$

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Question

The midpoint of a line segment is the point $\left (4 \frac{3}{5},- 3 \frac{1 }{3} \right )$ . One endpoint is $\left (1 \frac{2}{3}, 1 \frac{2 }{5} \right )$ ; give the -coordinate of the other.

Answer

Let the endpoints of a line segment be

$\left ( x_{1}, y_{1}\right ), \left ( x_{2}, y_{2}\right )$

Then the midpoint of the segment will be

$\left ( \frac{x_{1}+x_{2} }{2}, \frac{y_{1}+y_{2} }{2} \right )$

The -coordinate of the midpoint is $y_{m}=- 3 \frac{1 }{3} = - \frac{10 }{3}$ , and the -coordinate of the known endpoint is $y_{1} = 1 \frac{2 }{5} = \frac{7}{5}$ ,

so we can solve for $y_{2}$ , the -coordinate of the unknown endpoint, in the following equation:

$\frac{ \frac{7}{5}+y_{2} }{2} = - \frac{10 }{3}$

$\frac{7}{5}+y_{2} = - \frac{20}{3}$

$y_{2} = - \frac{20}{3} - \frac{7}{5}$

$y_{2} = - \frac{100}{15} - \frac{21}{15}$

$y_{2} = - \frac{121}{15}$

$y_{2} = - 8\frac{1}{15}$

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Question

The midpoint of a line segment is the point $\left (4 \frac{3}{5},- 3 \frac{1 }{3} \right )$ . One endpoint is $\left (1 \frac{2}{3}, 1 \frac{2 }{5} \right )$ ; give the -coordinate of the other.

Answer

Let the endpoints of a line segment be

$\left ( x_{1}, y_{1}\right ), \left ( x_{2}, y_{2}\right )$

Then the midpoint of the segment will be

$\left ( \frac{x_{1}+x_{2} }{2}, \frac{y_{1}+y_{2} }{2} \right )$

The -coordinate of the midpoint is $x_{m}= 4 \frac{3}{5} = \frac{23}{5}$ , and the -coordinate of the known endpoint is $x_{1} = 1 \frac{2}{3} = \frac{5}{3}$ , so we can solve for $x_{2}$ , the -coordinate of the unknown endpoint, in the following equation:

$\frac{x_{1}+x_{2} }{2} = x_{m}$

$\frac{\frac{5}{3}+x_{2} }{2} = \frac{23}{5}$

$\frac{5}{3}+x_{2} = \frac{46}{5}$

$x_{2} = \frac{46}{5} - \frac{5}{3}$

$x_{2} = \frac{138}{15} - \frac{25}{15}$

$x_{2} = \frac{113}{15}$

$x_{2} = 7 \frac{8}{15}$

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Question

The midpoint of line segment AB is (2, -5). If the coordinates of point A are (4, 4), what are the coordinates of B?

Answer

The fastest way to find the missing endpoint is to determine the distance from the known endpoint to the midpoint and then performing the same transformation on the midpoint. In this case, the x-coordinate moves from 4 to 2, or down by 2, so the new x-coordinate must be 2-2 = 0. The y-coordinate moves from 4 to -5, or down by 9, so the new y-coordinate must be -5-9 = -14.

An alternate solution would be to substitute (4,4) for (x1,y1) and (2,-5) for (x,y) into the midpoint formula:

x=(x1+x2)/2

y=(y1+y2)/2

Solving each equation for (x2,y2) yields the solution (0,-14).

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Question

Line segment AB has an endpoint, A, located at , and a midpoint at . What are the coordinates for point B of segment AB?

Answer

With an endpoint A located at (10,-1), and a midpoint at (10,0), we want to add the length from A to the midpoint onto the other side of the segment to find point B. The total length of the segment must be twice the distance from A to the midpoint.

A is located exactly one unit below the midpoint along the y-axis, for a total displacement of (0,1). To find point B, we add (10+0, 0+1), and get the coordinates for B: (10,1).

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Question

Point A is (5, 7). Point B is (x, y). The midpoint of AB is (17, –4). What is the value of B?

Answer

Point A is (5, 7). Point B is (x, y). The midpoint of AB is (17, –4). What is the value of B?

We need to use our generalized midpoint formula:

MP = ( (5 + x)/2, (7 + y)/2 )

Solve each separately:

(5 + x)/2 = 17 → 5 + x = 34 → x = 29

(7 + y)/2 = –4 → 7 + y = –8 → y = –15

Therefore, B is (29, –15).

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Question

One endpoint of a line segment on the coordinate plane is . The segment has length 10; the other endpoint has -coordinate 10 and is in Quadrant I. Give the -coordinate of the other endpoint. (Nearest tenth if applicable).

Answer

Let be the -coordinate of the other endpoint.

The segment has endpoints and .

Apply the distance formula

$d = \sqrt{(x_{2}-x_{1})^{2}+ (y_{2}-y_{1})^{2} }$

setting $d = 10, x_{1}= 3, x_{2} = 10, y_{1} = 7, y_{2}= Y$ ,

$10 = \sqrt{(10-3)^{2}+ (Y-7)^{2} }$

$100 =7^{2}+ (Y-7)^{2}$

$49+ (Y-7)^{2} = 100$

$(Y-7)^{2} = 51$

$Y-7 = \pm \sqrt{51}$

$Y = 7 \pm \sqrt{51}$

Therefore, there are two values of which fit the distance criterion.

One is $Y = 7 - \sqrt{51} \approx 7 - 7.1 \approx -0.1$ .

However, since the endpoint is in Quadrant I, it must have a positive coordinate, so this is eliminated as a choice.

The other is $Y = 7 + \sqrt{51} \approx 7 + 7.1 \approx 14.1$ , which is correct since it is positive.

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Question

The midpoint of a line segment is the point . One endpoint is ; give the -coordinate of the other.

Answer

Let the endpoints of a line segment be

$\left ( x_{1}, y_{1}\right ), \left ( x_{2}, y_{2}\right )$

Then the midpoint of the segment will be

$\left ( \frac{x_{1}+x_{2} }{2}, \frac{y_{1}+y_{2} }{2} \right )$

The -coordinate of the midpoint is $x_{m}= 1.4$ , and the -coordinate of the known endpoint is $x_{1} = 5.1$ , so we can solve for $x_{2}$ , the -coordinate of the unknown endpoint, in the following equation:

$\frac{x_{1}+x_{2} }{2} = x_{m}$

$\frac{5.1+x_{2} }{2} =1.4$

$5.1+x_{2} = 2.8$

$x_{2} = -2.3$

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Question

The midpoint of a line segment is the point . One endpoint is ; give the -coordinate of the other.

Answer

Let the endpoints of a line segment be

$\left ( x_{1}, y_{1}\right ), \left ( x_{2}, y_{2}\right )$

Then the midpoint of the segment will be

$\left ( \frac{x_{1}+x_{2} }{2}, \frac{y_{1}+y_{2} }{2} \right )$

The -coordinate of the midpoint is $y_{m}= -4.3$ , and the -coordinate of the known endpoint is $y_{1} = 3.1$ , so we can solve for $y_{2}$ , the -coordinate of the unknown endpoint, in the following equation:

$\frac{y_{1}+y_{2} }{2} = y_{m}$

$\frac{3.1+y_{2} }{2} =-4.3$

$3.1+x_{2} = -8.6$

$x_{2} = -11.7$

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Question

The midpoint of a line segment is the point $\left (4 \frac{3}{5},- 3 \frac{1 }{3} \right )$ . One endpoint is $\left (1 \frac{2}{3}, 1 \frac{2 }{5} \right )$ ; give the -coordinate of the other.

Answer

Let the endpoints of a line segment be

$\left ( x_{1}, y_{1}\right ), \left ( x_{2}, y_{2}\right )$

Then the midpoint of the segment will be

$\left ( \frac{x_{1}+x_{2} }{2}, \frac{y_{1}+y_{2} }{2} \right )$

The -coordinate of the midpoint is $y_{m}=- 3 \frac{1 }{3} = - \frac{10 }{3}$ , and the -coordinate of the known endpoint is $y_{1} = 1 \frac{2 }{5} = \frac{7}{5}$ ,

so we can solve for $y_{2}$ , the -coordinate of the unknown endpoint, in the following equation:

$\frac{ \frac{7}{5}+y_{2} }{2} = - \frac{10 }{3}$

$\frac{7}{5}+y_{2} = - \frac{20}{3}$

$y_{2} = - \frac{20}{3} - \frac{7}{5}$

$y_{2} = - \frac{100}{15} - \frac{21}{15}$

$y_{2} = - \frac{121}{15}$

$y_{2} = - 8\frac{1}{15}$

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Question

The midpoint of a line segment is the point $\left (4 \frac{3}{5},- 3 \frac{1 }{3} \right )$ . One endpoint is $\left (1 \frac{2}{3}, 1 \frac{2 }{5} \right )$ ; give the -coordinate of the other.

Answer

Let the endpoints of a line segment be

$\left ( x_{1}, y_{1}\right ), \left ( x_{2}, y_{2}\right )$

Then the midpoint of the segment will be

$\left ( \frac{x_{1}+x_{2} }{2}, \frac{y_{1}+y_{2} }{2} \right )$

The -coordinate of the midpoint is $x_{m}= 4 \frac{3}{5} = \frac{23}{5}$ , and the -coordinate of the known endpoint is $x_{1} = 1 \frac{2}{3} = \frac{5}{3}$ , so we can solve for $x_{2}$ , the -coordinate of the unknown endpoint, in the following equation:

$\frac{x_{1}+x_{2} }{2} = x_{m}$

$\frac{\frac{5}{3}+x_{2} }{2} = \frac{23}{5}$

$\frac{5}{3}+x_{2} = \frac{46}{5}$

$x_{2} = \frac{46}{5} - \frac{5}{3}$

$x_{2} = \frac{138}{15} - \frac{25}{15}$

$x_{2} = \frac{113}{15}$

$x_{2} = 7 \frac{8}{15}$

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Question

The midpoint of line segment AB is (2, -5). If the coordinates of point A are (4, 4), what are the coordinates of B?

Answer

The fastest way to find the missing endpoint is to determine the distance from the known endpoint to the midpoint and then performing the same transformation on the midpoint. In this case, the x-coordinate moves from 4 to 2, or down by 2, so the new x-coordinate must be 2-2 = 0. The y-coordinate moves from 4 to -5, or down by 9, so the new y-coordinate must be -5-9 = -14.

An alternate solution would be to substitute (4,4) for (x1,y1) and (2,-5) for (x,y) into the midpoint formula:

x=(x1+x2)/2

y=(y1+y2)/2

Solving each equation for (x2,y2) yields the solution (0,-14).

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Question

Line segment AB has an endpoint, A, located at , and a midpoint at . What are the coordinates for point B of segment AB?

Answer

With an endpoint A located at (10,-1), and a midpoint at (10,0), we want to add the length from A to the midpoint onto the other side of the segment to find point B. The total length of the segment must be twice the distance from A to the midpoint.

A is located exactly one unit below the midpoint along the y-axis, for a total displacement of (0,1). To find point B, we add (10+0, 0+1), and get the coordinates for B: (10,1).

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Question

Point A is (5, 7). Point B is (x, y). The midpoint of AB is (17, –4). What is the value of B?

Answer

Point A is (5, 7). Point B is (x, y). The midpoint of AB is (17, –4). What is the value of B?

We need to use our generalized midpoint formula:

MP = ( (5 + x)/2, (7 + y)/2 )

Solve each separately:

(5 + x)/2 = 17 → 5 + x = 34 → x = 29

(7 + y)/2 = –4 → 7 + y = –8 → y = –15

Therefore, B is (29, –15).

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Question

One endpoint of a line segment on the coordinate plane is . The segment has length 10; the other endpoint has -coordinate 10 and is in Quadrant I. Give the -coordinate of the other endpoint. (Nearest tenth if applicable).

Answer

Let be the -coordinate of the other endpoint.

The segment has endpoints and .

Apply the distance formula

$d = \sqrt{(x_{2}-x_{1})^{2}+ (y_{2}-y_{1})^{2} }$

setting $d = 10, x_{1}= 3, x_{2} = 10, y_{1} = 7, y_{2}= Y$ ,

$10 = \sqrt{(10-3)^{2}+ (Y-7)^{2} }$

$100 =7^{2}+ (Y-7)^{2}$

$49+ (Y-7)^{2} = 100$

$(Y-7)^{2} = 51$

$Y-7 = \pm \sqrt{51}$

$Y = 7 \pm \sqrt{51}$

Therefore, there are two values of which fit the distance criterion.

One is $Y = 7 - \sqrt{51} \approx 7 - 7.1 \approx -0.1$ .

However, since the endpoint is in Quadrant I, it must have a positive coordinate, so this is eliminated as a choice.

The other is $Y = 7 + \sqrt{51} \approx 7 + 7.1 \approx 14.1$ , which is correct since it is positive.

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