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Flashcards: Midpoint of line segments

Give the coordinates of the midpoint of the line segment whose endpoints are the intercepts of the line of the equation

$\displaystyle 2x+ 5y = 40$.

$\displaystyle (-4, -14)$

(4, 14) $\displaystyle (4, 14)$

$\displaystyle (-14, -4)$

(14, 4) $\displaystyle (14, 4)$

None of the other choices gives the correct response.

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Flashcards: Midpoint of line segments

Give the coordinates of the midpoint of the line segment whose endpoints are the intercepts of the line of the equation

$\displaystyle 2x+ 5y = 40$.

$\displaystyle (-4, -14)$

(4, 14) $\displaystyle (4, 14)$

$\displaystyle (-14, -4)$

(14, 4) $\displaystyle (14, 4)$

None of the other choices gives the correct response.

Report an Error Copyright Notice

Flashcards: Midpoint of line segments

Give the coordinates of the midpoint of the line segment whose endpoints are the intercepts of the line of the equation

$\displaystyle 2x+ 5y = 40$.

Your answer:

$\displaystyle (-4, -14)$

Your answer:

(4, 14) $\displaystyle (4, 14)$

Your answer:

$\displaystyle (-14, -4)$

Your answer:

None of the other choices gives the correct response.

Correct answer:

(14, 4) $\displaystyle (14, 4)$

Explanation:

The x-intercept and y-intercept of the line of an equation can be found by substituting 0 for $\displaystyle y$ and $\displaystyle x$, respectively, as follows:

x-intercept:

$\displaystyle 2x+ 5y = 40$

Substitute and simplify:

$\displaystyle 2x+ 5(0)= 40$

$\displaystyle 2x+ 0 = 40$

2x= 40 $\displaystyle 2x= 40$

Solve for $\displaystyle x$ by dividing by 2 on both sides:

$2x \div 2 = 40 \div 2$ $\displaystyle 2x \div 2 = 40 \div 2$

x = 20 $\displaystyle x = 20$

The x-intercept of the line is located at (20,0) $\displaystyle (20,0)$

The y-intercept can be found by setting x= 0 $\displaystyle x= 0$ and solving for $\displaystyle y$ in a similar fashion:

$\displaystyle 2x+ 5y = 40$

$\displaystyle 2 (0)+ 5y = 40$

$\displaystyle 0 + 5y = 40$

5y = 40 $\displaystyle 5y = 40$

$5y \div 5 = 40 \div 5$ $\displaystyle 5y \div 5 = 40 \div 5$

y = 8 $\displaystyle y = 8$

The y-intercept is located at (0,8) $\displaystyle (0,8)$.

The midpoint of a line segment, given its endpoints $(x_{1}, y_{1})$ $\displaystyle (x_{1}, y_{1})$ and $(x_{2}, y_{2})$ $\displaystyle (x_{2}, y_{2})$, is located at

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

substituting accordingly,

$\frac{x_{1}+x_{2}}{2} =\frac{20+8}{2} = \frac{28}{2} = 14$ $\displaystyle \frac{x_{1}+x_{2}}{2} =\frac{20+8}{2} = \frac{28}{2} = 14$

$\frac{y_{1}+y_{2}}{2} = \frac{ 0+8}{2} = \frac{8}{2} = 4$ $\displaystyle \frac{y_{1}+y_{2}}{2} = \frac{ 0+8}{2} = \frac{8}{2} = 4$

The midpoint is located at (14, 4) $\displaystyle (14, 4)$.

Flashcards: Midpoint of line segments

What is the midpoint of the line segment created by the points (-2,16) $\displaystyle (-2,16)$ and (3,7) $\displaystyle (3,7)$?

$(-\frac{5}{2},-\frac{13}{2})$ $\displaystyle (-\frac{5}{2},-\frac{13}{2})$

$(\frac{9}{2},-\frac{17}{2})$ $\displaystyle (\frac{9}{2},-\frac{17}{2})$

$(\frac{23}{2},\frac{3}{2})$ $\displaystyle (\frac{23}{2},\frac{3}{2})$

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

$(\frac{1}{2},\frac{23}{2})$ $\displaystyle (\frac{1}{2},\frac{23}{2})$

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Flashcards: Midpoint of line segments

What is the midpoint of the line segment created by the points (-2,16) $\displaystyle (-2,16)$ and (3,7) $\displaystyle (3,7)$?

Your answer:

$(-\frac{5}{2},-\frac{13}{2})$ $\displaystyle (-\frac{5}{2},-\frac{13}{2})$

Your answer:

$(\frac{9}{2},-\frac{17}{2})$ $\displaystyle (\frac{9}{2},-\frac{17}{2})$

Your answer:

$(\frac{23}{2},\frac{3}{2})$ $\displaystyle (\frac{23}{2},\frac{3}{2})$

Your answer:

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

Correct answer:

$(\frac{1}{2},\frac{23}{2})$ $\displaystyle (\frac{1}{2},\frac{23}{2})$

Explanation:

The midpoint of the line segment between two points

$(x_{0},y_{0})$ $\displaystyle (x_{0},y_{0})$ and $(x_{1},y_{1})$ $\displaystyle (x_{1},y_{1})$

is given by the formula

$(\frac{x_{0}+x_{1}}{2},\frac{y_{0}+y_{1}}{2})$ $\displaystyle (\frac{x_{0}+x_{1}}{2},\frac{y_{0}+y_{1}}{2})$.

Thus, since the line segment is generated by

(-2,16) $\displaystyle (-2,16)$ and (3,7) $\displaystyle (3,7)$

applying the midpoint formula we have

$(\frac{-2+3}{2},\frac{16+7}{2})= (\frac{1}{2},\frac{23}{2})$ $\displaystyle (\frac{-2+3}{2},\frac{16+7}{2})= (\frac{1}{2},\frac{23}{2})$

as the midpoint.

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Flashcards: Midpoint of line segments

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