Question

lesson 5-2

1. determine the coordinates of the midpoint of the segment with endpoints r(3, 16) and s(7, -6).
2. determine the coordinates of the midpoint of the segment with endpoints w(-5, 10.2) and x(12, 4.5).
3. point c is the midpoint of $overline{ab}$. point a has coordinates (2, 4), and point c has coordinates (5, 0).

a. what are the coordinates of point b?
b. what is ab?
c. what is bc?

1. $overline{jl}$ has endpoints j(8, 10) and l(20, 5). point k has coordinates (13, 9).

a. is point k the midpoint of $overline{jl}$? explain how you know.
b. how could you check that your answer to part a is reasonable?

1. find the coordinates of the midpoint of $overline{de}$.
2. what are the coordinates of the midpoint of the segment with endpoints at (-3, -4) and (5, 8)?

a. (1, 2)
b. (2, 4)
c. (4, 6)
d. (8, 12)

Explanation:

Step1: Recall mid - point formula

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

Step2: Solve problem 15

For points $R(3,16)$ and $S(7, - 6)$, $x_1=3,y_1 = 16,x_2=7,y_2=-6$.
$x_m=\frac{3 + 7}{2}=\frac{10}{2}=5$
$y_m=\frac{16+( - 6)}{2}=\frac{16 - 6}{2}=5$
The mid - point is $(5,5)$.

Step3: Solve problem 16

For points $W(-5,10.2)$ and $X(12,4.5)$, $x_1=-5,y_1 = 10.2,x_2=12,y_2 = 4.5$.
$x_m=\frac{-5 + 12}{2}=\frac{7}{2}=3.5$
$y_m=\frac{10.2+4.5}{2}=\frac{14.7}{2}=7.35$
The mid - point is $(3.5,7.35)$.

Step4: Solve problem 17a

Let point $A(x_1,y_1)=(2,4)$ and point $C(x_m,y_m)=(5,0)$. Let point $B(x_2,y_2)$.
Using the mid - point formula $x_m=\frac{x_1 + x_2}{2}$ and $y_m=\frac{y_1 + y_2}{2}$.
For $x$: $5=\frac{2 + x_2}{2}$, then $10=2 + x_2$, so $x_2=8$.
For $y$: $0=\frac{4 + y_2}{2}$, then $0 = 4 + y_2$, so $y_2=-4$.
Point $B$ has coordinates $(8,-4)$.

Step5: Solve problem 17b

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.
For $A(2,4)$ and $B(8,-4)$, $x_1 = 2,y_1 = 4,x_2=8,y_2=-4$.
$AB=\sqrt{(8 - 2)^2+(-4 - 4)^2}=\sqrt{6^2+( - 8)^2}=\sqrt{36 + 64}=\sqrt{100}=10$.

Step6: Solve problem 17c

Since $C$ is the mid - point of $AB$, $BC=\frac{AB}{2}=5$.

Step7: Solve problem 18a

For points $J(8,10)$ and $L(20,5)$, using the mid - point formula, $x_m=\frac{8 + 20}{2}=\frac{28}{2}=14$ and $y_m=\frac{10 + 5}{2}=\frac{15}{2}=7.5$.
Since the coordinates of $K$ are $(13,9)$ and the mid - point coordinates are $(14,7.5)$, point $K$ is not the mid - point of $\overline{JL}$.

Step8: Solve problem 18b

We can check the distance between $J$ and $K$ and $K$ and $L$. If $K$ is the mid - point, $JK=KL$.
$JK=\sqrt{(13 - 8)^2+(9 - 10)^2}=\sqrt{25 + 1}=\sqrt{26}$
$KL=\sqrt{(20 - 13)^2+(5 - 9)^2}=\sqrt{49+16}=\sqrt{65}$
Since $JK
eq KL$, our answer in part a is reasonable.

Step9: Solve problem 19

From the graph, $D(-3,3)$ and $E(3,-1)$.
$x_m=\frac{-3 + 3}{2}=0$
$y_m=\frac{3+( - 1)}{2}=\frac{2}{2}=1$
The mid - point is $(0,1)$.

Step10: Solve problem 20

For points $(-3,-4)$ and $(5,8)$, $x_1=-3,y_1=-4,x_2=5,y_2 = 8$.
$x_m=\frac{-3 + 5}{2}=\frac{2}{2}=1$
$y_m=\frac{-4 + 8}{2}=\frac{4}{2}=2$
The mid - point is $(1,2)$, so the answer is A. $(1,2)$.

Answer:

1. $(5,5)$
2. $(3.5,7.35)$

17a. $(8,-4)$
17b. $10$
17c. $5$
18a. No, because the mid - point of $\overline{JL}$ is $(14,7.5)$ and the coordinates of $K$ are $(13,9)$.
18b. Check if $JK = KL$ using the distance formula. Since $JK=\sqrt{26}$ and $KL=\sqrt{65}$, $JK
eq KL$.

1. $(0,1)$
2. A. $(1,2)$