# 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

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

##### 2 Answers

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

Consider

# 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

#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

Kindly go through the **Explanation.**

#### Explanation:

The **curve** **parametric equations**

**Eliminating** the **parameter**

**relation** between **cartesian eqn.** of

We have,

Hence, the **cartesian eqn.** of

Likewise, for

**Comparing** the cartesian eqns. of

they are **linear eqns.** in

So, **represent** the **same straight line.**

**BONUS :**

**Enjoy Maths.!**