# Parametic equation of the straight line

• February 2nd 2010, 04:27 PM
nyasha
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 ?
• February 2nd 2010, 04:58 PM
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.
• February 2nd 2010, 06:31 PM
nyasha
Quote:

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 ?
• February 3rd 2010, 05:34 AM
Sudharaka
Dear nyasha,

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

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

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

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

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

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

Therefore, $x=t$

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

Hope this helps.
• February 3rd 2010, 07:06 AM
Soroban
Hello, nyasha!

Here's my approach . . .

Quote:

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

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

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

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

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

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

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

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