For the variable traingle ABC with...

Question

For the variable triangle $ABC$ with fixed vertex at $C(1,2)$ and $A,B$ having co-ordinates $(cos t, sin t), (sin t, -cos t)$ respectively, what is the locus of its centroid?

My Approach:

The coordinate of centroid $O(x ,y )$ in triangle having $A(a,d ), B(b,e), C(c ,f)$ $$x =\frac {a +b +c}{3}, y=\frac {d + e + f}{3}$$

So coordinate of given triangle is :

$$h=\frac {cost +sint +1}{3}, k =\frac {sint - cost + 2}{3}$$

I got stuck at here. Please help me to continue.

Moreover, What is meant by 'variable triangle'?

i think since the variable $t$ in the argument of the function this must be eliminated — Dr. Sonnhard Graubner, Sep 18 '16 at 15:18
Dr. Sonnhard Graubner, How should I do that? Please give some more hints. — pi-π, Sep 18 '16 at 15:21

score 1 · Accepted Answer · answered Sep 18 '16 at 15:21

you will get the system $$3x-1=\sin(t)+\cos(t)$$ $$3y-2=\sin(t)-\cos(t)$$ adding these two equations we get $$\sin(t)=\frac{3x+3y-3}{2}$$ thus you will get $t$ and analogously $\cos(t)$ and finally you have to calculate $$\sin(t)^2+\cos(t)^2=1$$ with the tems above

score 1 · Answer 2 · answered Sep 18 '16 at 15:27

1

Alternatively, $$3x-1=\sin(t)+\cos(t)$$ $$3y-2=\sin(t)-\cos(t)$$

Squaring and adding gives $$(3x-1)^2+(3y-2)^2=2$$

btw, The triangle is variable because the coordinates change according to the value of $t$

answered Sep 18 '16 at 15:27

David Quinn

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For the variable traingle ABC with...

2 Answers2