bearings

• May 10th 2012, 11:05 AM
sarahheath1985
bearings
Not sure if I'm in the right forum as I'm new to this, but here goes.

I have a car traveling in a straight line from point C (-100,100) to point D (200,-100). I need to find the direction of travel as a bearing to the nearest degree. Any help?
• May 10th 2012, 02:28 PM
bjhopper
Re: bearings
slope diagram = -1
y-x/x+100= -1 simplifies to y= x

0,0 is the origin on a graph and the center 0f a 360 degree compass pointing true north. your course is 45 degrees W of north or 315 degrees true. Iwent the wrong way.change to 45 degrees south of east or135 degrees true
• May 10th 2012, 02:50 PM
Plato
Re: bearings
Quote:

Originally Posted by sarahheath1985
Not sure if I'm in the right forum as I'm new to this, but here goes.
I have a car traveling in a straight line from point C (-100,100) to point D (200,-100). I need to find the direction of travel as a bearing to the nearest degree. Any help?

I do not understand reply #2.
The vector of direction is $\displaystyle <3,-2>\text{ or }3i-2j~.$

Thus the direction angle is $\displaystyle \arctan\left(\frac{-2}{3}\right)$. You need to change that to degrees.
• May 10th 2012, 05:58 PM
bjhopper
Re: bearings
Yes Igoofed.the slope is -2/3 and angle = 33.7 and course is 33.7 degrees south of east or 123.7 degrees true
• May 10th 2012, 09:16 PM
sarahheath1985
Re: bearings
so,

what is the general formula? say the coordinates change to (-200,100) at C and (400,-200) at D?
• May 11th 2012, 03:16 AM
bjhopper
Re: bearings
There is no general formula. You are given two poimts on a coordinate grid. From these you determine the slope of the line connecting them.In this new case slope equals -1/2and angle equals arc tangent of -1/2= 26.6 deg. Coursefrom C to D is 26.6 deg south of east or 116.6 deg true
• May 11th 2012, 01:53 PM
HallsofIvy
Re: bearings
The "slope" of a line from $\displaystyle (x_0, y_0)$ to $\displaystyle (x_1, y_1)$ is the fraction $\displaystyle \frac{y_1- y_0}{x_1- x_}$ and the angle the line makes with the x- axis is $\displaystyle arctan\left(\frac{y_1- y_0}{x_1- x_0}\right)$. However, that is measured counterclockwise from the x-axis while the "bearing" (as on a map) is measured clockwise from the y-axis ("north"). To convert to a "bearing", subtract 90 degrees and multiply by -1:
$\displaystyle -\left(arctan\left(\frac{y_1- y_0}{x_1- x_0}\right)- 90\right)$ degrees.
• May 12th 2012, 03:09 AM
bjhopper
Re: bearings
The slope angle is always a positive acute angle. A sign is added only to show how it slants relative to the horizontal and does not apply to the angle itself.