Question

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. o(1, -5), p(9, 1)
5. understand if m is the mid - point of st, write an equation that describes the relationship between sm and mt.
6. apply the area 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 for 11.

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

Step3: Solve for 12.

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

Step4: Solve for 13.

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

Step5: Solve for 14.

For points $O(1,-5)$ and $P(9,1)$, $x_1 = 1,y_1=-5,x_2 = 9,y_2 = 1$.
$d=\sqrt{(9 - 1)^2+(1-(-5))^2}=\sqrt{8^2+6^2}=\sqrt{64 + 36}=\sqrt{100}=10$.

Step6: Solve for 15.

If $M$ is the mid - point of $ST$, then $SM=MT$ and $ST=SM + MT = 2MT=2SM$.

Step7: Solve for 16.

Assume the coordinates of the farthest corner of the bed from the origin. Let's say the coordinates of the farthest corner of the bed are $(8,3)$ (from the grid). Using the distance formula with $(x_1,y_1)=(0,0)$ and $(x_2,y_2)=(8,3)$.
$d=\sqrt{(8 - 0)^2+(3 - 0)^2}=\sqrt{64+9}=\sqrt{73}\approx8.5$.

Answer:

1. 7
2. 12
3. 13
4. 10
5. $ST = 2MT$
6. 8.5