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 $x^{2} + y^{2} = 1$, $y \geq 0$

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    $\begingroup$ $t = \frac 34 - \frac 14 x, y = 2 - 3(\frac 34 - \frac 14 x)$, In the second equation. It is still a section of a circle. The rest of the changes from the standard parameters will give you different start and end points, and cause you traverse the circle counter clockwise rather than clockwise. $x = \sin t \implies x=\cos (\pi/2 - t)$ $\endgroup$ – Doug M Jun 22 '16 at 20:12
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    $\begingroup$ $x=3-4t$ does not imply that $\frac{3}{4}x=t$. $\endgroup$ – Libertron Jun 22 '16 at 20:16
  • $\begingroup$ sorry..., clockwise rather than counter-clockwise. $\endgroup$ – Doug M Jun 22 '16 at 20:19
  • $\begingroup$ your right libertron it would equal $x-3=-4t$ which then simplifies to $t = -\frac{1}{4} x + \frac{3}{4}$ $\endgroup$ – K. Gibson Jun 23 '16 at 2:53
  • $\begingroup$ plugging that into t in y gives $2 - [3(\frac{-1}{4} x + \frac{3}{4})]$ which gives $2 - \frac{3}{4} x + \frac{9}{4}$ which gives $\frac{-3}{4} x + (\frac{8}{4} + \frac{9}{4})$ whcih simplifies to $y = \frac{-3}{4} x + \frac{17}{4}$ I believe $\endgroup$ – K. Gibson Jun 23 '16 at 2:59

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