Rotation and Transformation of Axes

Hi,I have a question that I have problems with.It's my homework due Thursday.

M1(9,-3);M2(-6,5).Bring the origin to M2 and rotate the system so that M1,M2 agrees with the direction of the new x axis.(tanθ=-8/15).

I can't understand what I'm supposed to find as a solution.Why do I need to change the origin?Should I only show how I did it?Thanks in advance.(I'm not asking for having my hw done by sb else.I studied all weekend on this subject but I can't understand the question.)

Rectangular to Polar Co-ordinates

[quote=truevein;474315]Hi,I have a question that I have problems with.It's my homework due Thursday.

M1(9,-3);M2(-6,5).Bring the origin to M2 and rotate the system so that M1,M2 agrees with the direction of the new x axis.(tanθ=-8/15).

[This is simply a variation on pythagorus theorum. Firstly, by bringing the origin to M2 (that is to say applying +6 and -5 respectively to the co-rdinates of both M1 & M2) the resultant values for M1 are now also the difference values as follows;

XM1-XM2=15 and YM1-YM2=-8

From these you can define a bearing (or azimuth) from M2 to M1;

TANθ=15/-8 so ATAN(15/-8)= -61.9275 (Decimal Deg)

-68.8083 (Grads)

Note now that your new bearing has simply swaped the X and Y difference values which means that the axis has rotated thro 90Deg (100Grad).

TANθ=-8/15 so ATAN(-8/15)= -28.0725 (Decimal Deg)

-31.1917 (Grads)

Simply add the two calculated bearings and your result will be;

-90(Degrees) or -100(Grads)

And if you require a distance;

SQRT(15^2 + (-8)^2) = 17

which is the same as

SQRT((-8)^2 + 15^2) = 17]