# Thread: Parametic equation of the straight line

1. ## Parametic equation of the straight line

Find the parametic equation of the line given by x+y-z=2,2x+3y+z=4

Attempt to solution:

Is the line given by this equation the intersection of these two planes ?

2. Dear nyasha,

Yes it is. I will give you a hint; the first step is to take z=0. If you need further assistance please don't hesitate to reply me.

3. Originally Posted by Sudharaka
Dear nyasha,

Yes it is. I will give you a hint; the first step is to take z=0. If you need further assistance please don't hesitate to reply me.

Can l use matrix row reduced echeloen form ?

4. Dear nyasha,

I don't know what row reduced echeloen form is. But you could easily find the line in the following method.

$\displaystyle x+y-z=2$------(1)

$\displaystyle 2x+3y+z=4$---------(2)

$\displaystyle (1)+(2)\Rightarrow{3x+4y=6}$

This is the equation of the line. Since you need to find the parametric form,

$\displaystyle x=\frac{6-4y}{3}=t$

Therefore, $\displaystyle x=t$

$\displaystyle y=\frac{6-3t}{4}$

Hope this helps.

5. Hello, nyasha!

Here's my approach . . .

Find the parametic equation of the line given by: .$\displaystyle \begin{array}{cccc} x+y-z&=&2 & [1] \\ 2x+3y+z&=&4 & [2] \end{array}$

Add [1] and [2]: .$\displaystyle 3x + 4y \:=\:6$

. . Solve for $\displaystyle x\!:\;\;x \:=\:\frac{6-4y}{3}$

Substitute into [1]: .$\displaystyle \frac{6-4y}{3} + y - z \:=\:2$

. . Solve for $\displaystyle z\!:\;\;z \:=\:-\frac{y}{3}$

We have: .$\displaystyle \begin{Bmatrix}x &=& 2 - \frac{4}{3}y \\ \\[-4mm] y &=& y \\ \\[-3mm] z &=& -\frac{1}{3}y \end{Bmatrix}$

On the right, replace $\displaystyle y$ with $\displaystyle 3t$

. . and we have: .$\displaystyle \begin{Bmatrix}x &=& 2-4t \\ y &=& 3t \\ z &=& -t \end{Bmatrix}$