# Thread: Find equation of the line

1. ## Find equation of the line

Hi everyone,

I would to check with someone if I didn't make a mistake with the problem below

Positions are given with reference to a Cartesian coordinate system whose x- and y-axes point due East and due North, respectively. Distance is measured in kilometres.

An aeroplane flies in a straight line from city A, at (300,-200), to city B, at (-100,600).

((i) Find the equation of the line of flight of the aeroplane.
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A(300,-200) B(-100,600)

First find the slope

Run is: -100-300 = -400
Rise is: 600-(-200) = 800

The slope is

M = rise/run = 800/-400 = -2

We can apply equation y-y1 = m(x-x1) with m= -2 and either (x1,y1) = (300,-200) or (x1,y1)= (-100,600).

First possibility:
y(-200)= -2(x-1)
y= -2x+202
[HTML][/HTML]

Thanks again for helping me

2. Originally Posted by valerie-petit
Hi everyone,

I would to check with someone if I didn't make a mistake with the problem below

Positions are given with reference to a Cartesian coordinate system whose x- and y-axes point due East and due North, respectively. Distance is measured in kilometres.

An aeroplane flies in a straight line from city A, at (300,-200), to city B, at (-100,600).

((i) Find the equation of the line of flight of the aeroplane.

A(300,-200) B(-100,600)

First find the slope

Run is: -100-300 = -400
Rise is: 600-(-200) = 800

The slope is

M = rise/run = 800/-400 = -2 Mr F says: Correct.

We can apply equation y-y1 = m(x-x1) with m= -2 and either (x1,y1) = (300,-200) or (x1,y1)= (-100,600). Mr F says: Correct.

First possibility:
y(-200)= -2(x-1) Mr F says: Wrong! How on Earth did you get it? y - (-200) = -2 (x - 300) => y = .....

y= -2x+202

Thanks again for helping me
..

3. ## Thanks Mr Fantastic

y-200= -2(x - 300)
y= -2x+302

Thanks again for helping me out

4. ## Direction of travel

The second question of that exercise I have no idea what is the steps to achieve it.

''Find the direction of travel of the aeroplane, as a bearing, with the angle in degrees correct to one decimal place.''

Can someone Explain me what they looking for, It's the degrees part which confuse me. How do I find the angle if I don't have the distance between A and B.(distance between A & B is asked in the next question)

Thanks again

5. Originally Posted by valerie-petit
The second question of that exercise I have no idea what is the steps to achieve it.

''Find the direction of travel of the aeroplane, as a bearing, with the angle in degrees correct to one decimal place.''

Can someone Explain me what they looking for, It's the degrees part which confuse me. How do I find the angle if I don't have the distance between A and B.(distance between A & B is asked in the next question)

Thanks again
Plot the two points. Draw the obvious big right-angle triangle (it has a base length of 400 and a height of 800). Use tan to get one of the angles inside this triangle. Use this angle to get the bearing of B from A.

6. ## Re: direction of travel

If I didn't do a big mess, I find the angle of B is 30.9 degree. So that should be S 30.9 W.

7. Originally Posted by valerie-petit
If I didn't do a big mess, I find the angle of B is 30.9 degree. So that should be S 30.9 W.
1. $\tan^{-1} 2 = 63.4^0$.
2. A is in the fourth quadrant and B is in the second quadrant so the bearing of B from A will definitely not be S 30.9 W.

Did you draw a set of xy-axes, plot the points and draw the triangle?

8. ## Re: direction of aeroplane

Sorry I did tan B= b/a instead of a/b, and I didn't draw the triangle with xy-axes which made my triangle look completely wrong.

So the direction from B to A is S63.4E degree.

I'm really bad in maths, the course I have give a book the size of a magazine, no much exercises and examples, but this is my last exam and I can't back up

9. ## Re: direction of an aeroplane

I start again from the begining, attached I put my drawing of the triangle.

10. ## Thanks again

Sorry about that completely confuse myself. I got it right this time.

Thanks again Mr Fantastic

11. Originally Posted by valerie-petit
y-200= -2(x - 300)
y= -2x+302

Thanks again for helping me out
Not sure I agree with the final line equation. I get y = -2x + 400. What do you think?

http://www.mathcelebrity.com/slope.php

12. [quote=valerie-petit;180131]Hi everyone,

I would to check with someone if I didn't make a mistake with the problem below

Positions are given with reference to a Cartesian coordinate system whose x- and y-axes point due East and due North, respectively. Distance is measured in kilometres.

An aeroplane flies in a straight line from city A, at (300,-200), to city B, at (-100,600).

((i) Find the equation of the line of flight of the aeroplane.
A(300,-200) B(-100,600)

First find the slope

Run is: -100-300 = -400
Rise is: 600-(-200) = 800

The slope is

M = rise/run = 800/-400 = -2

We can apply equation y-y1 = m(x-x1) with m= -2 and either (x1,y1) = (300,-200) or (x1,y1)= (-100,600).

First possibility:

You are right upto here. Now see, the equation:

$y - y_1 = m(x - x_1)$

y - (- 200) = - 2 (x - 300)

y + 200 = - 2x + 600

y = - 2x + 600 - 200

y = - 2x + 400.

13. ## Thanks again

Thanks mathceleb

14. ## Thanks again

Yes, at the end I did find my mistake
Sad to know that question was just the beginning of a long exercise

15. Originally Posted by valerie-petit
Yes, at the end I did find my mistake
Sad to know that question was just the beginning of a long exercise
No worries. Keep working at it and you will do great.