This page will take a look at a few of the various examples that utilize the concept of slopes associated with parallel and perpendicular lines.

 Questions Asking Directly for Equations of Parallel or Perpendicular Lines:

 1. What is the equation of a line parallel to y = -4x + 5 and passing through the point (6,-3)?

Solution: The slope of the given line, y = -4x + 5 is -4 (remember y = mx + b).
Since parallel lines have = slopes, our line has a slope of -4.
Use the point -slope form of the equation of a line: y - y1 = m(x - x1).

m = -4      (x1, y1) = (6,-3)
ANSWER:    y - (-3) = -4(x - 6)
y + 3 = -4x + 24
y = -4x + 21

 2. What is the equation of a line perpendicular to 2y = x - 4 and passing through the point (-4,1)?

Solution: The slope of the given line, 2y = x - 4 is ½ (remember to solve for y first).
Since perpendicular lines have negative reciprocal slopes, our line has a slope of -2.
Use the point-slope form of the equation of a line: y - y1 = m(x - x1).

m = -2      (x1, y1) = (-4,1)
ANSWER:    y - 1 = -2(x - (-4))
y -1 = -2x - 8
y = -2x - 7

 Questions Asking Indirectly for Equations of Perpendicular Lines:

 3. What is the equation of a line tangent to a circle whose equation is (x - 5)2 + (y + 2)2 = 25 at the point (8,-4), which lies on the circle ?

Solution: The circle (x - 5)2 + (y + 2)2 = 25 has a center at (5,-2) and a radius of 5.
A tangent to a circle is perpendicular to the radius at the point of tangency.
Find the slope of the radius:
The slope of the tangent will be
.