Thread: An analytic Geometry Riddle about Locus.

OC starts from (0,0) AB is a line segment that has a midpoint C(because OC is touching it at C) Slope of OC is Tan60.Coordinates of A is (SQRT3,1)Find the equation of locus of point B.

Originally Posted by serhanbener

OC starts from (0,0) AB is a line segment that has a midpoint C(because OC is touching it at C) Slope of OC is Tan60.Coordinates of A is (SQRT3,1)Find the equation of locus of point B.

1. The point C is located on the straight line l:



2. Let denotes the stationary vector of point B. C is the midpoint of .

Therefore:

with

3. That means:

.... ....

4. The stationary vector of point B is:



This is the equation of a parallel of line c.

Many Thanks for your answer. I managed to solve it like this:
A(SQRT3,1) is given. B is (x,SQRT3) and C is C(A,B).I applied the
midpoint formula.
First of all I found A as 2x- SQRT3 and B=2SQRT3x-1.I replaced A in B.
B=(A+SQRT3)SQRT3-1.Then B=SQRT3A+2.If we replace A and B with X and Y,
it becomes y= SQRT3x+2

Originally Posted by serhanbener

Many Thanks for your answer. I managed to solve it like this:
A(SQRT3,1) is given. B is (x,SQRT3) and C is C(A,B).I applied the
midpoint formula.
First of all I found A as 2x- SQRT3 and B=2SQRT3x-1.I replaced A in B.
B=(A+SQRT3)SQRT3-1.Then B=SQRT3A+2.If we replace A and B with X and Y,
it becomes y= SQRT3x+2

1. That's nearly perfect
(You must use brackets so that the equation is unambiguous: y= SQRT(3) *x + 2)

2. I was trapped by your headline: Analytic Geometry means in German vector geometry.

Vectors are a different subject of analytic geometry in Turkey. Analytic Geometry subjects are: Analytic geometry:Lines ,tangents,symmetry vs..,Analytic of the circle,Vectors,Parabolia,Hyperbolia and Elips. Thanks again for you help.