# Find Equation of Parallel and Perpendicular Lines

• Jun 2nd 2010, 11:35 PM
larry21
Find Equation of Parallel and Perpendicular Lines
Find an equation of the following parallel and perpendicular lines.

A.) The line parallel to \$\displaystyle x+3=0\$ and passing through \$\displaystyle (-6, -7)\$

B.) The line perpendicular to \$\displaystyle y-4=0\$ passing through \$\displaystyle (-1, 6)\$

Is there a specific formula I can use to to solve these?
• Jun 2nd 2010, 11:43 PM
harish21
Quote:

Originally Posted by larry21
Find an equation of the following parallel and perpendicular lines.

A.) The line parallel to \$\displaystyle x+3=0\$ and passing through \$\displaystyle (-6, -7)\$

B.) The line perpendicular to \$\displaystyle y-4=0\$ passing through \$\displaystyle (-1, 6)\$

Is there a specific formula I can use to to solve these?

The equation of a line passing through a point \$\displaystyle (x_1, y_1)\$ is given by:

\$\displaystyle y-y_1 = m(x-x_1)\$

where m is the slope.

Now, two lines are parallel if their slopes are equal.

and two lines are perpendicular if the product of their slopes is -1.
• Jun 2nd 2010, 11:47 PM
undefined
Quote:

Originally Posted by harish21
\$\displaystyle y-y_1 = m(x-x_1)\$

For reference, this equation is known as the point-slope equation.
• Jun 3rd 2010, 12:14 AM
nikhil
1)required line is parallel to x+3=0 and passes through (-6,-7)
equation of such line will be x=c. since it passes through (-6,-7)
-6=c or
x=-6
x+6=0
[NOTE: line x+k=0 is perpendicular to x axis so any line parallel to it must also be perpendicular to x axis and in the form x+k=0]

2) line perpendicular to y-4=0 will be in the form x=c.
since it passes through(-1,6)
we have -1=c
or c=-1
x=-1
x+1=0