Question

find the midpoint of the two given points (1 point each): 1) a: (4, 4) & b:(6, 6) 2) a: (-6, 1) & b:(-8, 3) 3) a: (-5, 1) & b:(7, 9) 4) a: (10, -5) & b:(12, 3) 5) a: (6, 2) & b:(8, 0) 6) a: (4, -1) & b:(-5, 9) 7) a: (-2, 9) & b:(-7, 7) 8) a: (-9, 3) & b:(7, -8) 9) a: (-1, -2) & b:(-6, -4) 10) a: (1, -2) & b:(-5, -1) 11) a: (5, 5) & b:(-3, -3) 12) a: (-5, -5) & b:(3, -3)

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 problem 1

For points $A(4,4)$ and $B(6,6)$, $x_1 = 4$, $y_1 = 4$, $x_2 = 6$, $y_2 = 6$.
$x=\frac{4 + 6}{2}=\frac{10}{2}=5$
$y=\frac{4+6}{2}=\frac{10}{2}=5$
The mid - point is $(5,5)$.

Step3: Solve for problem 2

For points $A(-6,1)$ and $B(-8,3)$, $x_1=-6$, $y_1 = 1$, $x_2=-8$, $y_2 = 3$.
$x=\frac{-6+( - 8)}{2}=\frac{-6 - 8}{2}=\frac{-14}{2}=-7$
$y=\frac{1 + 3}{2}=\frac{4}{2}=2$
The mid - point is $(-7,2)$.

Step4: Solve for problem 3

For points $A(-5,1)$ and $B(7,9)$, $x_1=-5$, $y_1 = 1$, $x_2 = 7$, $y_2 = 9$.
$x=\frac{-5 + 7}{2}=\frac{2}{2}=1$
$y=\frac{1+9}{2}=\frac{10}{2}=5$
The mid - point is $(1,5)$.

Step5: Solve for problem 4

For points $A(10,-5)$ and $B(12,3)$, $x_1 = 10$, $y_1=-5$, $x_2 = 12$, $y_2 = 3$.
$x=\frac{10 + 12}{2}=\frac{22}{2}=11$
$y=\frac{-5 + 3}{2}=\frac{-2}{2}=-1$
The mid - point is $(11,-1)$.

Step6: Solve for problem 5

For points $A(6,2)$ and $B(8,0)$, $x_1 = 6$, $y_1 = 2$, $x_2 = 8$, $y_2 = 0$.
$x=\frac{6 + 8}{2}=\frac{14}{2}=7$
$y=\frac{2+0}{2}=\frac{2}{2}=1$
The mid - point is $(7,1)$.

Step7: Solve for problem 6

For points $A(4,-1)$ and $B(-5,9)$, $x_1 = 4$, $y_1=-1$, $x_2=-5$, $y_2 = 9$.
$x=\frac{4+( - 5)}{2}=\frac{4 - 5}{2}=-\frac{1}{2}$
$y=\frac{-1 + 9}{2}=\frac{8}{2}=4$
The mid - point is $(-\frac{1}{2},4)$.

Step8: Solve for problem 7

For points $A(-2,9)$ and $B(-7,7)$, $x_1=-2$, $y_1 = 9$, $x_2=-7$, $y_2 = 7$.
$x=\frac{-2+( - 7)}{2}=\frac{-2-7}{2}=-\frac{9}{2}$
$y=\frac{9 + 7}{2}=\frac{16}{2}=8$
The mid - point is $(-\frac{9}{2},8)$.

Step9: Solve for problem 8

For points $A(-9,3)$ and $B(7,-8)$, $x_1=-9$, $y_1 = 3$, $x_2 = 7$, $y_2=-8$.
$x=\frac{-9 + 7}{2}=\frac{-2}{2}=-1$
$y=\frac{3+( - 8)}{2}=\frac{3 - 8}{2}=-\frac{5}{2}$
The mid - point is $(-1,-\frac{5}{2})$.

Step10: Solve for problem 9

For points $A(-1,-2)$ and $B(-6,-4)$, $x_1=-1$, $y_1=-2$, $x_2=-6$, $y_2=-4$.
$x=\frac{-1+( - 6)}{2}=\frac{-1-6}{2}=-\frac{7}{2}$
$y=\frac{-2+( - 4)}{2}=\frac{-2-4}{2}=-3$
The mid - point is $(-\frac{7}{2},-3)$.

Step11: Solve for problem 10

For points $A(1,-2)$ and $B(-5,-1)$, $x_1 = 1$, $y_1=-2$, $x_2=-5$, $y_2=-1$.
$x=\frac{1+( - 5)}{2}=\frac{1 - 5}{2}=\frac{-4}{2}=-2$
$y=\frac{-2+( - 1)}{2}=\frac{-2-1}{2}=-\frac{3}{2}$
The mid - point is $(-2,-\frac{3}{2})$.

Step12: Solve for problem 11

For points $A(5,5)$ and $B(-3,-3)$, $x_1 = 5$, $y_1 = 5$, $x_2=-3$, $y_2=-3$.
$x=\frac{5+( - 3)}{2}=\frac{5 - 3}{2}=1$
$y=\frac{5+( - 3)}{2}=\frac{5 - 3}{2}=1$
The mid - point is $(1,1)$.

Step13: Solve for problem 12

For points $A(-5,-5)$ and $B(3,-3)$, $x_1=-5$, $y_1=-5$, $x_2 = 3$, $y_2=-3$.
$x=\frac{-5 + 3}{2}=\frac{-2}{2}=-1$
$y=\frac{-5+( - 3)}{2}=\frac{-5-3}{2}=-4$
The mid - point is $(-1,-4)$.

Answer:

1. $(5,5)$
2. $(-7,2)$
3. $(1,5)$
4. $(11,-1)$
5. $(7,1)$
6. $(-\frac{1}{2},4)$
7. $(-\frac{9}{2},8)$
8. $(-1,-\frac{5}{2})$
9. $(-\frac{7}{2},-3)$
10. $(-2,-\frac{3}{2})$
11. $(1,1)$
12. $(-1,-4)$