Question

use the distance formula and the pythagorean theorem to find the distance, to the nearest tenth, between each pair of points.

1. (u(0,1)) and (v(-3,-9)) 19. (m(10, - 1)) and (n(2,-5)) 20. (p(-10,1)) and (q(5,1))
2. consumer application televisions and computer screens are usually measured based on the length of their diagonals. if the height of a computer screen is 11 in. and the width is 14 in., what is the length of the diagonal? round to the nearest tenth.
3. multi - step use the distance formula to order (ab), (cd), and (ef) from shortest to longest.
4. use the pythagorean theorem to find the distance from (a) to (e). round to the nearest hundredth.
5. (x) has coordinates ((a,3a)), and (y) has coordinates ((-5a,0)). find the coordinates of the mid - point of ((xy)).
6. describe a shortcut for finding the mid - point of a segment when one of its endpoints has coordinates ((a,b)) and the other endpoint is the origin.

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 problem 18 with points $U(0,1)$ and $V(-3,-9)$

Let $(x_1,y_1)=(0,1)$ and $(x_2,y_2)=(-3,-9)$. Then $d=\sqrt{(-3 - 0)^2+(-9 - 1)^2}=\sqrt{(-3)^2+(- 10)^2}=\sqrt{9 + 100}=\sqrt{109}\approx10.4$.

Step3: Solve for problem 19 with points $M(10,-1)$ and $N(2,-5)$

Let $(x_1,y_1)=(10,-1)$ and $(x_2,y_2)=(2,-5)$. Then $d=\sqrt{(2 - 10)^2+(-5+1)^2}=\sqrt{(-8)^2+(-4)^2}=\sqrt{64 + 16}=\sqrt{80}\approx8.9$.

Step4: Solve for problem 20 with points $P(-10,1)$ and $Q(5,6)$

Let $(x_1,y_1)=(-10,1)$ and $(x_2,y_2)=(5,6)$. Then $d=\sqrt{(5 + 10)^2+(6 - 1)^2}=\sqrt{(15)^2+(5)^2}=\sqrt{225+25}=\sqrt{250}\approx15.8$.

Step5: Solve for problem 21 (using Pythagorean theorem for computer - screen diagonal)

If the height $h = 11$ in and width $w = 14$ in, by the Pythagorean theorem $c=\sqrt{h^{2}+w^{2}}$, where $c$ is the diagonal. So $c=\sqrt{11^{2}+14^{2}}=\sqrt{121 + 196}=\sqrt{317}\approx17.8$ in.

Step6: For problem 22 (assuming we have coordinates of points for $AB$, $CD$, and $EF$)

First find the distances of $AB$, $CD$, and $EF$ using the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$ for each pair of endpoints of the line - segments, then order them from shortest to longest.

Step7: For problem 23 (using Pythagorean theorem to find distance from $A$ to $E$)

Find the horizontal and vertical displacements between $A$ and $E$, then use $d=\sqrt{\text{horizontal}^2+\text{vertical}^2}$ and round to the nearest hundredth.

Step8: For problem 24 (finding mid - point of $XY$)

The mid - point formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. If $X=(a,3a)$ and $Y=(-5a,0)$, then the mid - point is $(\frac{a-5a}{2},\frac{3a + 0}{2})=(-2a,\frac{3a}{2})$.

Step9: For problem 25 (shortcut for mid - point when one endpoint is the origin)

If one endpoint is $(0,0)$ and the other is $(a,b)$, the mid - point is $(\frac{a + 0}{2},\frac{b+0}{2})=(\frac{a}{2},\frac{b}{2})$, which means we just take half of the non - origin coordinates.

Answer:

1. Approximately $10.4$; 19. Approximately $8.9$; 20. Approximately $15.8$; 21. Approximately $17.8$ in; 22. Order after calculating distances; 23. Calculate using Pythagorean theorem and round; 24. $(-2a,\frac{3a}{2})$; 25. Take half of non - origin coordinates.