Parametric equation and distance formula Help with this assignment! For the past three weeks now, I've been battling with mathematics assignment. I've successfully solved some but these ones posed a great challenge to me


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*Eliminate the parameter and identify the graph of the resulting equation


$x = 3v+7, y = v-1$
$x = \sin t, y = \cos t, 0\leq t \leq \pi$
$x = 3\sin t, y = 2\cos t, 0\leq t \leq \pi$
$x = 7u^2-3, y = 4-2u$
$x = 2+\cos t, y = 1+\sin t$


*Prove that the distance between line $y = mx+c$ and the origin is given by $d = \frac{|c|}{\sqrt{1+m^2}}$.

*Use the above equation again to show that the distance between the line and the point $(x,y)$ is $d = \frac{|y-mx-c|}{\sqrt{1+m^2}}$.
For question 1, I just need a hint to go about solving them since they are kinda similar. I seriously need help in question 2 and 3 because I'm lacking when its comes to proving or showing relation between one thing and another.
 A: $x = \sin t, y = \cos t, 0\leq t \leq \pi$
I would really hope that you just recognize this as a circle, as this is close to the unit circle definition of cosine and sine.  
You could say:
$x^2 = \sin^2 t\\ y^2= \cos^2 t\\
x^2 + y^2 = \sin^2 t + \cos^2 t = 1$
Considering the limits on $t$ once that is applied, you will get a semi-circle.
Moving on to,
$x = 3\sin t, y = 2\cos t, 0\leq t \leq \pi$
That is just a circle with a dilation applied.
And using the same logic for the algebra.
$\frac {x}{3} = \sin t, \frac {y}{2} = \cos t\\
\frac {x^2}{9} + \frac {y^2}{4} = 1$
Regarding:
$x = 7u^2-3, y = 4-2u$ 
Isolate $u,$ and substitute.
$u = \frac {4-y}{2}\\
x = 7\frac {(4-y)^2}{4}-3$
And simplify.
As for $2, 3$ the distance between a point and a line, is the shortest distance.
I would start at the point (the origin in problem 2) and move along a line perpendicular to the targeted line, until I found the point of intersection.
$y = -\frac {1}{m} x$ is perpendicular to $y= mx + c$ and goes through the origin.
Now use substitution or elimination.
$(1+m^2) x + mc = 0\\
x = \frac {-mc}{1+m^2}\\
y = \frac {c}{1+m^2}$
$d((0,0),(x,y)) = \sqrt {x^2 + y^2} = \sqrt {\frac {c^2 + c^2m^2}{(1+m^2)^2}} = \frac {|c|}{\sqrt{1+m^2}}$ 
A: Hint for #2 and #3:
The distance between the line $y = mx+c$ and a point $(w,z)$ is defined as the minimum distance between any point $(x,y)$ on the line and the point $(w,z)$.
Using Euclidean distance (which I am assuming the question asks for), the distance between a point on the line and the point $(w,z)$ is:
$$d((x,y),(w,z)) = \sqrt{(x-w)^2 + (y-z)^2} = \sqrt{(x-w)^2 + (mx+c-z)^2}$$
So, you need to find the minimum value of that distance.
Hint for #1 (I'll just go ahead and do the first, which is the simplest):
Start by isolating the parameter $v$ in the first equation:
$$ v = \frac{x-7}{3}$$
Then, substitute into the second equation:
$$ y(x) = v - 1 = \frac{x-7}{3} - 1$$
The graph is the graph of y(x).
