Straight line question ( K method) (1 Viewer)

[Twickel]

A ship sails from point A on a bearing of 236 to point B a distance of 425km. The ship then turns and sails south to point C. The bearing of A from C is 42

draw a diagram. Thats easy but how do they figure out that at B the angle is 124?
Once I draw the triangle I did not have the angle at B, how would I have gotten 124?

OH alt angles. DW

 

[me121]

The K method is used when you want to find the equation of the line that goes through the point of intersection of two lines, and another point.

If the two lines are a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂, and the point is (m,n), you write it as

a₁x + b₁y + c₁ + k(a₂x + b₂y + c₂) = 0
then substitute in m and n for x and y to solve for k, then expand and simplify and you get the equation of the line you wanted.

ahh.. you see i would have just found the point of intersection, then write down the equation joining the two points.

 

[12o9]

I don't think so...the rules regarding general form (I think), are Ax+By+C = 0, where A, B and C are integers, and A is positive.

I never knew that such a rule existed =/.

 

[me121]

I never knew that such a rule existed =/.

yeh.. its seems very trivial.. its all a matter of the human interaction of the maths, the actual maths is all the same.

A ship sails from point A on a bearing of 236 to point B a distance of 425km. The ship then turns and sails south to point C. The bearing of A from C is 42

draw a diagram. Thats easy but how do they figure out that at B the angle is 124?
Once I draw the triangle I did not have the angle at B, how would I have gotten 124?

OH alt angles. DW

yeh. i don't want to say don't post, but try to think about it first.. because quite often you can pick it up with a little more thinking, and its better that way. though i think you did this anyway.

 

[Iruka]

I don't think that rule actually is a rule.

How would you write x + sqrt(2)y - 6 = 0 in general form, if all the coefficients had to be integers?

The so-called general form wouldn't be very general then, would it, if there were an infinite number of lines that can't be written that way.

 

[Aerath]

I never knew that such a rule existed =/.

Look at your Fitzpatrick textbook.
Noone ever reads the information bits of their textbook, they always get stuck into the questions.

 

[Twickel]

A car is north of two towns. The car is 39km from twon A and 52km from twon B the distance between AB is 68km (A is west of B) what it the bearing of the car from town A and B.

THIS IS SO EASY WHY AM I GETTING TI WRONG. FOR CAR FROM A it should be like this
39^2+68^2-52^2/ 2 x 39 x 68. Or am I doing ti wrong

 

[bored of sc]

A car is north of two towns. The car is 39km from twon A and 52km from twon B the distance between AB is 68km (A is west of B) what it the bearing of the car from town A and B.

THIS IS SO EASY WHY AM I GETTING TI WRONG. FOR CAR FROM A it should be like this
39^2+68^2-52^2/ 2 x 39 x 68. Or am I doing ti wrong

What answer did you get?

I did it and got 49.55^o.

Oh, don't forget to do the cos^-1 of it all.

 

[Twickel]

The same as you but the book says 40. Can you try doing the bearing of the car from B

 

[bored of sc]

The same as you but the book says 40. Can you try doing the bearing of the car from B

Oh!!! It's asking for bearing. So from the diagram you have you can see we have the angle that is COMPLIMENTARY to the bearing. Therefore,

bearing = 90 - ans
= 90 - 49.55
= 40.45
= 40^o to nearest degree
= 040

 

[jordankuai22]

Wow this thread has so much activity.

It must be a really hard question :/

 

[jordankuai22]

wow, you are a failure

 

[Aerath]

Maybe cause people are asking more than one question?

 

[Twickel]

Thankyou.

 

[Twickel]

(0,-2) (3,4) find the equation of the line passing through those points. Please use the two point formula.
y-y1/x-x1= y-y1/x2-x1
y+2/x-0=6/3

I get 3y+6=3x.

So what am I doing wrong. There are two points so im using the the two point formula.

 

[bored of sc]

(0,-2) (3,4) find the equation of the line passing through those points. Please use the two point formula.
y-y1/x-x1= y-y1/x2-x1
y+2/x-0=6/3

I get 3y+6=3x.

So what am I doing wrong. There are two points so im using the the two point formula.

Factorise (common factor of 3) and rearrange equation and you get =

y = x - 2 (what's the answer?)

 

[Twickel]

0=2x-y-4.


Pretend I wrote nothing did you try and do it? Please be an error.

 

[bored of sc]

You've done the cross-multiplication wrong; that's all. It should be 3y + 6 = 6x --> 0 = 2x -y -2... Oh wait, that's still the wrong answer.

 

[Twickel]

*crys* NNNNNOOOOOOOOOOOOOOOOOOOOOOOO. They use the one point method. y-y1=m(x-x1) but because there are two points the two point method should work aswell correct?

 

[Twickel]

teacher did a mistake look. y--2=2(x-0) next line she put y+4 instead of y+2.