Question 153309
can you please help me indicate if the line is parallel, perpendicular, or neither.
<pre>
{{{system (3x+y=5, -x+3y=6)}}}

Solve them both for y and look at their respective coefficients of x.
These coefficients are the slopes of the lines. 

1. If the are the same, the lines are parallel.
2. If they are not parallel, then find the reciprocal of one of those
   coefficients of x, then change the sign.  If that is the same as the 
   other coefficient of x, then they are perpendicular.

If neither, then neither!

Solve the first for y:

{{{3x+y=5}}}

Add -3x to both sides:

{{{3x+y-3x=5-3x}}}

{{{cross(3x)+y-cross(3x)=5-3x}}}

{{{y=5-3x}}}  

The coefficient of {{{x}}} is {{{-3}}}

Solve the second for y:

{{{-x+3y=6}}}

Add +x to both sides:

{{{-x+3y+x=6+x}}}

{{{cross(-x)+3y-cross(x)=6+x}}}

{{{3y=6+x}}}  

Now we must multiply both sides by {{{1/3}}}

{{{(1/3)3y=(1/3)6+(1/3)x}}}

                2
{{{(1/cross(3))cross(3)y=(1/cross(3))cross(6)+1/3}}}{{{x}}}

{{{y = 2+1/3}}}{{{x}}}


The coefficient of {{{x}}} is {{{1/3}}}

------

So the two coefficients are {{{-3}}} and {{{1/3}}}

They certainly are not the same, so they are not parallel.

So let's find the reciprocal of the {{{1/3}}} which is
{{{3/1}}} which equals {{{3}}} and then change the sign to 
{{{-3}}}.

We notice that we get the other.  So therefore they are perpendicular.

Edwin</pre>