Question

example: find the midpoint of the line segment with the endpoints (4, 1) and (2, -5). midpoint = (\\(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\\)) = (\\(\frac{4 + 2}{2}, \frac{1 - 5}{2}\\)) = (3, -2) find the midpoint of the line segment with the given endpoints. 1) (5, 0), (1, 4) 2) (-9, 3), (7, -8) 3) (-2, 9), (-7, 7) 4) (5, 10), (-3, 6) 5) (-1, -6), (3, 0) 6) (8, 1), (-2, -5) 7) (-6, -10), (-2, -8) 8) (4, -1), (-5, 9) 9) (2, 3), (4, -7) 10) (-9, -4), (-3, 6)

Explanation:

Step1: Recall mid - point formula

The mid - point formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$.

Step2: Solve for each pair of points

1) For points $(5,0)$ and $(1,4)$

$(\frac{5 + 1}{2},\frac{0+4}{2})=(\frac{6}{2},\frac{4}{2})=(3,2)$

2) For points $(-9,3)$ and $(7,-8)$

$(\frac{-9 + 7}{2},\frac{3+( - 8)}{2})=(\frac{-2}{2},\frac{3 - 8}{2})=(-1,-\frac{5}{2})$

3) For points $(-2,9)$ and $(-7,7)$

$(\frac{-2+( - 7)}{2},\frac{9 + 7}{2})=(\frac{-2-7}{2},\frac{16}{2})=(-\frac{9}{2},8)$

4) For points $(5,10)$ and $(-3,6)$

$(\frac{5+( - 3)}{2},\frac{10 + 6}{2})=(\frac{5 - 3}{2},\frac{16}{2})=(1,8)$

5) For points $(-1,-6)$ and $(3,0)$

$(\frac{-1 + 3}{2},\frac{-6+0}{2})=(\frac{2}{2},\frac{-6}{2})=(1,-3)$

6) For points $(8,1)$ and $(-2,-5)$

$(\frac{8+( - 2)}{2},\frac{1+( - 5)}{2})=(\frac{8 - 2}{2},\frac{1 - 5}{2})=(3,-2)$

7) For points $(-6,-10)$ and $(-2,-8)$

$(\frac{-6+( - 2)}{2},\frac{-10+( - 8)}{2})=(\frac{-6-2}{2},\frac{-10 - 8}{2})=(-4,-9)$

8) For points $(4,-1)$ and $(-5,9)$

$(\frac{4+( - 5)}{2},\frac{-1 + 9}{2})=(\frac{4 - 5}{2},\frac{8}{2})=(-\frac{1}{2},4)$

9) For points $(2,3)$ and $(4,-7)$

$(\frac{2 + 4}{2},\frac{3+( - 7)}{2})=(\frac{6}{2},\frac{3 - 7}{2})=(3,-2)$

10) For points $(-9,-4)$ and $(-3,6)$

$(\frac{-9+( - 3)}{2},\frac{-4 + 6}{2})=(\frac{-9-3}{2},\frac{2}{2})=(-6,1)$

Answer:

1. $(3,2)$
2. $(-1,-\frac{5}{2})$
3. $(-\frac{9}{2},8)$
4. $(1,8)$
5. $(1,-3)$
6. $(3,-2)$
7. $(-4,-9)$
8. $(-\frac{1}{2},4)$
9. $(3,-2)$
10. $(-6,1)$