# Thread: Slope (parallel and perpendicular)

1. ## Slope (parallel and perpendicular)

Hi,

I'm yet again having trouble grasping this concept. I've been given the point: (7/8, 3/4) and the equation 5x + 3y = 0. Now from this information I need to find equations of lines that are :
(a) parallel to the given line
and then
(b) perpendicular to the given line

How can one approach this problem and solve down for the equations of the lines?

Thank you yet again.

2. Originally Posted by hemi
I've been given the point: (7/8, 3/4) and the equation 5x + 3y = 0. Now from this information I need to find equations of lines that are : (a) parallel to the given line
and then
(b) perpendicular to the given line
Parallel lines have the same slope.
So any line parallel to the given line looks like $5x+3y=C$.

Perpendicular lines have slopes that multiply to $-1$
So any line perpendicular to the given line looks like $3x-5y=C$.

To find $C$ simply use the given point.

3. Originally Posted by Plato
Parallel lines have the same slope.
So any line parallel to the given line looks like $5x+3y=C$.

Perpendicular lines have slopes that multiply to $-1$
So any line perpendicular to the given line looks like $3x-5y=C$.

To find $C$ simply use the given point.

So:
$5(7/8)+3(3/4)=6.625$

and:
$3(7/8)-5(3/4)=-1.125$

But this answer does not give me an equation of a line (if I did it correctly)? How can I write this as such as an equation of a line. Thanks a lot for your help Plato.

4. Originally Posted by hemi
So:
$5(7/8)+3(3/4)=6.625$

and:
$3(7/8)-5(3/4)=-1.125$

But this answer does not give me an equation of a line (if I did it correctly)? How can I write this as such as an equation of a line.
$5x+3y=\frac{53}{8}$

and:
$3x-5y=\frac{-9}{8}$

5. Thank you Plato.