Question

Log the coordinates of the specific points in space to which your spacecraft will travel. Please remember to include the graph of your points and the lines connecting each point along with your work.
Launch Area:___(1, 2)___
Point A:__________
Point B:__________
Point C:__________
You must show your work on each question below.

Answer 1

@agent0smith

Answer 2

Idk, are you supposed to pick points?

Answer 3

well i mean, is there certain points i have to pick? & theres more to the problem i just broke it down.

Answer 4

@ranga

Answer 5

"...which YOUR spacecraft will travel..." seems to imply there were some questions earlier to this question dealing with equation for the flight path.

Answer 6

so could i do (2,3) (4,5) & (3,2)?

Answer 7

Were there other parts to this question before this one?

Answer 8

Now it is your turn to plan the trajectories required to launch a spacecraft through a specific route in space. The launch area is identified on the map below. Select three points for your spacecraft to travel through and label them Point A, Point B, and Point C. Be sure that your points are not all in the same line.

Answer 9

Choose three points not on the same line.
(1,2) is given, You have chosen: (2,3) (4,5) & (3,2).
(1,2), (2,3) and (4,5) will all be on the same line.

Answer 10

so (3,2) (5,5) & (6,1)

Answer 11

Instead of y going up and down I will just choose three points with both x and y increasing.

Answer 12

im confused..

Answer 13

If you look at your 3 y values, it goes from2 to 5 to 1.

How about (2,3), (3,5), (4, 9)?
The numbers are easy to plot on a graph paper without using large numbers for x and y and using a scale factor.

Answer 14

how about 1 & 2?
1.Determine the equation of the line, in standard form, that will get your spacecraft from the Launch Area to Point A.

2.Determine the equation of the line, in point-slope form, that will get your spacecraft from Point A to Point B.

Answer 15

i have 8 questions. @ranga

Answer 16

@phi

Answer 17

Launch point (1,2)
A is (2,3); B is (3,5); C is (4,9)

What is the equation of the line connecting (1,2) and (2,3)?
Find the slope first.

Answer 18

Have you picked your points ?

Answer 19

2-3/1-2?

Answer 20

Yes, but use parenthesis.

Answer 21

-1/-1 = 1?

Answer 22

Yes. Can you find the equation of the line with slope 1 and passing through any one of the above two points?

Answer 23

which point do you want me to use?

Answer 24

first one is simpler. Either one will give you the same answer.

Answer 25

y-2=1(x-1)?

Answer 26

Yes. Simplify it first and then put it in the standard form: Ax + By = C

Answer 27

so would it be y=1x+1 ?

Answer 28

that is in slope-intercept form... correct, but they want standard form

Answer 29

You can drop the 1 in front of the x.
y = x + 1.
But this is the slope-intercept form. They are asking for standard form.
-x + y = 1
or x - y = -1

Answer 30

so how would i change? i forget.

Answer 31

so x-y=-1?

Answer 32

some text books prefer coefficient of x to be positive. But I think either one should be okay: -x + y = 1 or x - y = -1.

Answer 33

#2?

Answer 34

so would i use (3,5) & (4,9)?

Answer 35

Find the equation of the line passing through (2,3) and (3,5) just like before but this time leave the answer in slope-intercept form.

Answer 36

3-5/2-3?

Answer 37

once again, use parenthesis.

Answer 38

-2/-1 = 2?

Answer 39

Yes. Finish the rest.

Answer 40

y-3=2(x-2) ?

Answer 41

yes.

Answer 42

y=2x-1?

Answer 43

yes. That is the slope-intercept form of that line.

Answer 44

ok 3& 4.

Answer 45

3.Determine the equation of the line, in slope-intercept form, that will get your spacecraft from Point B to Point C.

4.In question 2, you selected one of two points (Point A or Point B) to be included in your point-slope equation. Write the point-slope form of that equation again, using the other point’s coordinates.

Answer 46

Same procedure as the previous one except here the points are (3,5) and (4,9).

Answer 47

(5-9)/(3-4)?

Answer 48

Yes.

Answer 49

-4/-1 = 4?

Answer 50

yes.

Answer 51

y-5=4(x-3)?

Answer 52

yes.

Answer 53

y=4x-7

Answer 54

yes.

Answer 55

4?

Answer 56

btw, thank you so much for your help!

Answer 57

#4 refers back to #2 where we found the equation of the line passing through (2,3 and (3,5)
You found the slope m as (3-5)/(2-3) = -2/-1 = 2.
Then you used the first point (2,3) to find the equation. Here they are asking you to use the second point.
What is the equation of the line with slope 2 passing through (3,5)?

Answer 58

y-5=2(x-3)?

Answer 59

yes. simplify.

Answer 60

y=2x-1

Answer 61

Yes. It should be the same answer as you got for #2.

Answer 62

#5&6?

Answer 63

5.Convert the equations you arrived at in question 2 and question 4 into slope-intercept form.

6.Does the point you select matter when your write a point-slope equation? Explain your reasoning using complete sentences.

Answer 64

Oops. Just realized for #2 and #4 they were asking for point-slope form and NOT slope-intercept form.

So for #2 leave the answer at: y - 3 = 2(x - 2). It was not necessary to go the next step.

Similarly for #4, leave the answer at y - 5 = 2(x - 5).

Answer 65

you mean y-5=2(x-3)?

Answer 66

for # 4?

Answer 67

yes, that was a typo. y - 5 = 2(x-3)

Answer 68

Now for #5, do the next step and put it in point-slope form:
From #2: y - 3 = 2(x - 2)
y - 3 = 2x - 4
y = 2x - 1

From #4: y - 5 = 2(x - 3)
y - 5 = 2x - 6
y = 2x - 1

Answer 69

#6?

Answer 70

#6. The point you select does NOT matter when your write a point-slope equation. You can select either point and you will get the same answer. That is because we initially used BOTH point to calculate the slope. After that it does not matter which point is chosen for the point-slope form because the slope and either of the points will ensure that the line passes through BOTH points.

Answer 71

wait im confused..

Answer 72

We started with two points and first found the slope.
Then we used the slope and one of the points to find the equation of the line in point-slope form.
They are asking if it matters which point is chosen.
That is, y - y1 = m(x-x1)
or should it be y - y2 = m(x - x2)

The answer is it does not matter. Both will simplify to the same equation.

Answer 73

You can put m = (y2-y1)/(x2-x1) into the above two equations and simplify and prove mathematically they are one and the same. But I don't think they are asking for mathematical proof. Just explanation. And that is what I did 2 replies ago.

Answer 74

7&8?

7.Reflect back on this scenario and each equation you created. Would any restrictions apply to the domain and range of those equations? Explain your reasoning using complete sentences.

8.Explain, using complete sentences, why it is important to understand any limitations on the domain and range.

Answer 75

#6 i put "It does not matter which way you write it.. y-y1=m(x-x1) or y-y2=m(x-x2) either way it comes out the same. They will both simplify to the same equation."

Answer 76

Yes.

Answer 77

#7?

Answer 78

But include that when simplified by replacing m with (y2-y1)/(x2-x1) both will yield the same result.

Answer 79

what?

Answer 80

#6 i put "It does not matter which way you write it.. y-y1=m(x-x1) or y-y2=m(x-x2) either way it comes out the same. Since m = (y2-y1)/(x2-x1) they will both simplify to the same equation."

Answer 81

#8?

Answer 82

You know the answer to #7?

Answer 83

no..

Answer 84

The equations that you found for each segment: Launch point to A, A to B and B to C are applicable only to those segments and not beyond.
The coordinates were: Launch point (1,2) A is (2,3); B is (3,5); C is (4,9)
So the equation for the launch point to point A applies only to that line segment. The equation is no longer applicable for the trajectory of the vehicle past those points. The domain and range of the first equation is: domain [1,2] and range [2,3]. Similar restrictions for the other two equations.

Answer 85

thats the answer?^

Answer 86

Yes.

Answer 87

#8?

Answer 88

Given any function, the domain is all the allowed values for x. The range is all the values y can take. The function is not defined for x values not in the domain. Therefore we cannot use the equation for x values not in the domain because the equation is no longer valid. The range gives the minimum y value to the maximum y value.

For this problem where we are dealing with 3 straight lines, I don't think we have to worry about division by 0, etc. which is one of the reasons why some x values may be excluded from the domain. Here, the restriction in the domain comes because the equation of the line is applicable only to each segment and not beyond.

Answer 89

answer^

Answer 90

?

Answer 91

Yes. But my answers to #7 & #8 are mixed in with some information to you. Hopefully you can separate them.

Answer 92

thank you so much!

Answer 93

You are welcome.

Answer 94

Once again, sorry to hear about your loss.