For the first part I am just unsure as to how the book has a different answer than mine. The book has the answer $y = \frac{3}{4} x - \frac{1}{4}$
but given the functions $x(t) = 3 - 4t$ and $y(t) = 2 -3t$
$x = 3 -4t$
$\frac{3}{4} x = t$
substituting t in $y = 2 - 3t$ I get:
$y = 2 - \frac{9}{4} x$
Is my answer incorrect?
For the second part I am unsure as to what to do when given the following:
$x = sin(\frac{1}{2} \Theta)$ and $y = cos(\frac{1}{2} \Theta)$, $-\pi \leq \Theta \leq \pi$
The first thing that I notice is that the values of x and y are flipped bc in the unit circle $x = cos(\Theta)$ and $y = sin(\Theta)$ so it looks as if there was some sort of transformation of rotation about the origin where x and y were reflected along the line y=0 and then rotated from $-\pi$ to $0$ and then from $0$ to $\pi$. If so this brings up another point, how could I prove this is linear? I would assume by showing that its closed under scalar addition and multiplication? I would like to see how this would look in notation if it applies here. As for trying to eliminate the parameter, Here's what I got.I tried to substitute using the half angle formula for $sin$ which gives me $sin(\frac{1}{2} \Theta) = \pm \sqrt{\frac{1-cos(\Theta)}{2}}$ But on another website I saw that the formula I used could be restated as $sin^{2} (\Theta) = \frac{1}{2} [1-cos(2 \Theta)]$ I'm so confused I don't even think this substitution would help as the book says the answer is $x^{2} + y^{2} = 1$, $y \geq 0$