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???

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

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.!