The curves C1 and C2 are defined by the following parametric equations: C1: x=1+2t, y=2+3t, 2<t<5 C2: x=1/(2t-3), y=t/(2t-3), 2<t<3 Show that both curves are segments of the same straight line???

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I don't understand this at all. I've work out the cartesian equation for C1 to be y=3x/2 + 1/2 and I know that to answer the question I must show the cartesian equation for C2 to be the same. However, I am not getting the same cartesian equation for C2 and cannot figure this question out at all

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Answer 1 · 2018-05-25T08:41:11

We have:

$C_1: { (x=1+2t), (y=2+3t) :} \ \ \ \ \ 2 lt t lt 5$

$C_2: { (x=1/(2t-3)), (y=t/(2t-3)) :} \ \ \ \ \ 2 lt t lt 3$

Let us derive the cartesian equations of each curve via elimination of the parameter $t$ :

Consider $C_1$ , by rearranging, we have:

$x=1+2t => t=(x-1)/2$

So that:

$y=2+3((x-1)/2)$
$:. 2y=4+3(x-1)$
$:. 2y=4+3x-3$
$:. 2y=1+3x$

Consider $C_2$ , by rearranging, we have:

$x=1/(2t-3) => (2t-3)x=1 => t=1/2(1/x+3)$

So that:

$y=x (1/2(1/x+3))$
$:. 2y=x (1/x+3)$
$:. 2y=1+3x$

And the two cartesian equations are identical, QED

Answer 2 · 2018-05-25T08:57:48

Kindly go through the Explanation.

The curve $C_1$ is given by the parametric equations

$C_1 : x=1+2t, y=2+3t, t in (2,5)$ .

Eliminating the parameter $t$ from these eqns., we get a

relation between $x and y$ , which is the cartesian eqn. of $C_1$ .

We have,

$3x=3+6t and 2y=4+6t," so that, "3x-2y=-1$ .

Hence, the cartesian eqn. of $C_1$ is $C_1 :3x-2y+1=0$ .

Likewise, for $C_2$ , we have,

$3x-2y+1=3/(2t-3)-(2t)/(2t-3)+1=-1+1=0$ .

Comparing the cartesian eqns. of $C_1 and C_2$ , we find that

they are linear eqns. in $RR^2$ of the type,

$ax+by+c=0," where, "a^2+b^2!=0$ .

So, $C_1 and C_2$ represent the same straight line.

BONUS :

$C_1"=the segment "AB-{A(5,8), B(11,17)}$ .

$C_2"=the segment "CD-{C(1,2), D(1/3,1)}$ .

$C_1 and C_2 sub" the line "{(x,y) in RR^2 : 3x-2y+1=0}$ .

Enjoy Maths.!