Question

use the distance formula and the pythagorean theorem to find the distance, to the nearest tenth, between each pair of points. 18. u(0, 1) and v(-3, -9) 19. m(10, -1) and n(2, -5) 20. p(-10, 1) and q(5, 5) 21. consumer application televisions and computer screens are usually advertised 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 inch. 22. multi - step use the distance formula to order $overline{ab}$, $overline{cd}$, and $overline{ef}$ from shortest to longest. 23. use the pythagorean theorem to find the distance from a to e. round to the nearest hundredth. 24. x has coordinates (a, 3a), and y has coordinates (-5a, 0). find the coordinates of the midpoint of (xy). 25. describe a shortcut for finding the midpoint 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 $Q(5,5)$

Let $(x_1,y_1)=(-10,1)$ and $(x_2,y_2)=(5,5)$. Then $d=\sqrt{(5 + 10)^2+(5 - 1)^2}=\sqrt{(15)^2+(4)^2}=\sqrt{225+16}=\sqrt{241}\approx15.5$.

Step5: Solve for problem 21 using Pythagorean theorem

If height $h = 11$ in and width $w=14$ in, by the Pythagorean theorem $c=\sqrt{h^{2}+w^{2}}=\sqrt{11^{2}+14^{2}}=\sqrt{121 + 196}=\sqrt{317}\approx18$ in.

Step6: For problem 22 (assuming we find lengths of segments using distance formula)

We need to find coordinates of endpoints of $\overline{AB},\overline{CD},\overline{EF}$ from the graph (not shown here in full - detail, but the process is to apply distance formula for each pair of endpoints).

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

We need to find horizontal and vertical distances between A and E from the graph (not shown in detail here) and then apply $d=\sqrt{a^{2}+b^{2}}$.

Step8: For problem 24

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 $(x_1,y_1)=(a,3a)$ and $(x_2,y_2)=(-5a,0)$, then the mid - point is $(\frac{a-5a}{2},\frac{3a + 0}{2})=(-2a,\frac{3a}{2})$.

Step9: For problem 25

If one endpoint is $(a,b)$ and the other is $(0,0)$, the mid - point using the mid - point formula $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$ is $(\frac{a + 0}{2},\frac{b+0}{2})=(\frac{a}{2},\frac{b}{2})$.

Answer:

1. $d\approx10.4$
2. $d\approx8.9$
3. $d\approx15.5$
4. $c\approx18$ in
5. $(-2a,\frac{3a}{2})$
6. $(\frac{a}{2},\frac{b}{2})$