# Parametric and vectorial functions

I need some help trying to do the following:

Given $$L: (x,y)=(\sin t , 1+3\sin t)~, \quad 0

Find $$a, b, c$$ and $$d$$ so that

$$L_2: (x , y) = (-2,-5)+u(a;b)~, \quad c

represents the same points that $$L$$ does.

I don't know where I should begin. If somebody knows how to do it, please give me some hints.

I'm going to rewrite your question as $$L(x,y)=(\sin(t),1+3\sin(t))$$, for $$0.

And $$L_2(x,y)=(-2,-5)+u(a,b)$$ for $$c.

Well, if we analyze $$L$$ first, we see that its actually just a line because we have something of the form $$(x, 1+3x)$$ which just describes a line. This line segment runs from $$(0,1)$$ (at $$t=0$$) to $$(1,4)$$ (at $$t=\frac{π}{2})$$ right back to $$(0,1)$$ (at $$t=0$$).

So this graph is just the segment from $$(0,1)$$ to $$(1,4)$$.

Now, if we look at $$(-2,-5)$$, you may notice that it actually lies on the same line as $$(0,1)$$ and $$(1,4)$$.

Therefore, we can use $$L_2(x,y)=(-2,-5)+u(1,3)$$.

We find out that if we let $$u=2$$ and $$u=3$$, we get $$(0,1)$$ and $$(1,4)$$ respectively.

Therefore, we have $$L_2(x,y)=(-2,-5)+u(1,3)$$ for $$2.