# An analytic Geometry Riddle about Locus.

• Mar 23rd 2012, 11:49 AM
serhanbener
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.
• Mar 23rd 2012, 01:38 PM
earboth
Re: An analytic Geometry Riddle about Locus.
Quote:

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:

$l:\vec c = \langle 1, \sqrt{3} \rangle \cdot t$

2. Let $\vec b = \langle b_1, b_2 \rangle$ denotes the stationary vector of point B. C is the midpoint of $\overline{AB}$.

Therefore:

$\displaystyle{\frac{\vec b + \vec a}2 = \vec c}$ with $\vec a = \langle \sqrt{3} , 1 \rangle$

3. That means:

$\left| \begin{array}{rcl}\frac{b_1+\sqrt{3}}2 &=& t \\ \frac{b_2+1}2 &=& \sqrt{3} \cdot t \end{array} \right.$ .... $\implies$ .... $\left| \begin{array}{rcl} b_1 &=& 2t - \sqrt{3} \\ b_2 &=& 2 \sqrt{3} \cdot t - 1 \end{array} \right.$

4. The stationary vector of point B is:

$\vec b = \langle 2t - \sqrt{3}, 2 \sqrt{3} \cdot t - 1 \rangle~\implies~ \vec b = 2 \langle 1, \sqrt{3} \rangle \cdot t - \langle \sqrt{3} , 1 \rangle$

This is the equation of a parallel of line c.
• Mar 23rd 2012, 02:50 PM
serhanbener
Re: An analytic Geometry Riddle about Locus.
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
• Mar 23rd 2012, 11:01 PM
earboth
Re: An analytic Geometry Riddle about Locus.
Quote:

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 (Clapping)
(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.
• Mar 23rd 2012, 11:15 PM
serhanbener
Re: An analytic Geometry Riddle about Locus.
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.