Question

for exercises 6 - 9, find the coordinates of each point described in relation to line segment cd.

1. 1/3 of the way from c to d
2. 2/3 of the way from d to c
3. 2/3 of the way from c to d
4. 1/3 of the way from d to c
5. what is the distance formula?

for exercises 11 - 14, find the distance between each pair of points.

1. a(6, 8), b(-1, 8)
2. c(5, -6), d(5, 6)
3. e(-2, 0), f(11, 0)
4. q(1, -5), t(9, 1)
5. understand if m is the mid - point of st, write an equation that describes the relationship between st and mt.
6. apply the axes in the coordinate grid at the right represent the walls of a bedroom. one corner of the room is at the origin. what is the distance from that corner of the room to the corner of the bed that is farthest away? if necessary, round to the nearest tenth of a foot.

Explanation:

Step1: Recall distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.

Step2: Solve Exercise 11

For points $A(6,8)$ and $B(-1,8)$, $x_1 = 6,y_1 = 8,x_2=-1,y_2 = 8$. Substitute into the distance formula:
$d=\sqrt{(-1 - 6)^2+(8 - 8)^2}=\sqrt{(-7)^2+0^2}=\sqrt{49}=7$.

Step3: Solve Exercise 12

For points $C(5,-6)$ and $D(5,6)$, $x_1 = 5,y_1=-6,x_2 = 5,y_2 = 6$. Substitute into the distance formula:
$d=\sqrt{(5 - 5)^2+(6-(-6))^2}=\sqrt{0+(12)^2}=\sqrt{144}=12$.

Step4: Solve Exercise 13

For points $E(-2,0)$ and $F(11,0)$, $x_1=-2,y_1 = 0,x_2 = 11,y_2 = 0$. Substitute into the distance formula:
$d=\sqrt{(11-(-2))^2+(0 - 0)^2}=\sqrt{(13)^2+0^2}=\sqrt{169}=13$.

Step5: Solve Exercise 14

For points $Q(1,-5)$ and $T(9,1)$, $x_1 = 1,y_1=-5,x_2 = 9,y_2 = 1$. Substitute into the distance formula:
$d=\sqrt{(9 - 1)^2+(1-(-5))^2}=\sqrt{8^2+6^2}=\sqrt{64 + 36}=\sqrt{100}=10$.

Step6: Solve Exercise 15

If $M$ is the mid - point of $\overline{ST}$, then $ST = 2MT$.

Answer:

1. 7
2. 12
3. 13
4. 10
5. $ST = 2MT$