Question

GEOMETRY: Help asap pls - show all work (please only answer if you know the answer)

Find the distance between the points to the nearest tenth.
1. C(-1, -1), D(6, 2)
2. E(-7, 0), F(5 ,8)

Line AB has endpoints A(-3, 2) and B(3, -2).
3. find the coordinates of the midpoint of Line AB
4. Find AB to the nearest tenth

Answer 1

distance formula

distance between (x1,y1) and (x2,y2) is

`d=√((x_1-x_2)^2+(y_1-y_2)^2)`

1. (-1,-1) and (6,2)

`d=√((-1-6)^2+(-1-2)^2)`

`d=√((-7)^2+(-3)^2)`

`d=√(49+9)`

`d=√(58)`

d≈7.6

2. (-7,0) and (5,8)

`d=√((-7-5)^2+(0-8)^2)`

`d=√((-12)^2+(-8)^2)`

`d=√(144+64)`

`d=√(208)`

d≈14.4

midpoint of (x1,y1) and (x2,y2) is

`((x1+x2)/(2),(y1+y2)/(2))`

midpoint of (-3,2) and (3,-2) is

`((-3+3)/(2),(2-2)/(2))`

`((0)/(2),(0)/(2))`

`(0,0)`

distance between them

(-3,2) and (3,-2)

`d=√((-3-3)^2+(2-(-2))^2)`

`d=√((-6)^2+(4)^2)`

`d=√(36+16)`

`d=√(52)`

d≈7.2

Answer 2

Find the distance between the points to the nearest tenth.

1. C(-1, -1), D(6, 2) Think of a right triangle whose hypotenuse connects. C and D; the length of this hypotenuse is the distance between C and D. As we go from C to D, x increases by 7 and y increases by 3. These distances represent the lengths of the shorter two legs of the right triangle. Then, according to the distance formula (which is essentially the same thing as the Pythagorean Theorem): 3^2 + 7^2 = distance^2. This becomes distance^2 = 9 + 49, or distance^2 = sqrt(58). The distance from C to D is sqrt(58) units.

2. E(-7, 0), F(5 ,8) Following the same reasoning and same approach, the distance between E and F is sqrt(12^2 + 8^2), or sqrt(144+64), or sqrt(208). This reduces to sqrt (4*52), or 2sqrt(52). The distance from E to F is 2sqrt(52).

Line AB has endpoints A(-3, 2) and B(3, -2).

3. find the coordinates of the midpoint of Line AB. Use the midpoint formulas to find the coordinates of the midpoint of the line connecting A and B. The x-coordinate is (-3+3)/2, or 0. The y-coordinate is (2-2)/2, or 0. Thus, the midpoint of AB is (0,0).

4. Find AB to the nearest tenth. Follow the same reasoning and approach used in (1) and (2) above: change in x = 6; change in y = -4. Then the distance between A and B (that is, the length of AB) is: sqrt(36+16) =sqrt(52), or, to the nearest tenth, 7.2.