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Equation of a line

created by Enzondio

(thing) by hobyrne (2.8 y) (print)   ?   I like it! Wed Feb 12 2003 at 8:08:38

There are two common forms for the equation of a (straight) line.

The standard form is ax + by = c. The slope of the line is -a/b, the x-intercept is c/a, the y-intercept is c/b.

Another common form is the slope-intercept form, it looks like this: y = mx + b. (Note that this b is different from the one in the standard form.) This form has the advantage that the slope can be easily recognised - it is m. The y-intercept is b, the x-intercept is -b/m. It has the disadvantage that vertical lines (lines parallel to the y-axis) cannot be represented in this form.

If you are given a point on a line (x0, y0), and its slope (m), the equation of the line can be found: y - y0 = m * (x - x0). If you are given two points on the line, the equation of the line is (y - y0) / (y1 - y0) = (x - x0) / (x1 - x0). If you are given the x-intercept (v) and y-intercept (w), the equation of the line is (x / v) + (y / w) = 1


(idea) by srkorn (2.2 y) (print)   ?   I like it! Tue Aug 10 2004 at 19:22:18

So, you want to convey the idea of a straight line, but don't want to deal with the horrible imprecision of ASCII art?

   y
   | / 
   |/
   /
  /|
-/-------- x
/  |
   |
Rejoice! You can use math in a variety of flavors to express this concept abstractly.

In elementary algebra, we use the slope-intercept form of the equation of a line in the plane: y = mx+b. Pity the fool who does not recognize this formula.

In complex analysis, we can interpret x and y as linearly independent components of the complex variable z = x+iy, and reinterpret the slope-intercept form as:


0 = -y+mx+b
  = Re((m+i)(x+iy)+b)
  = Re((m+i)z+b)

If you want to generalize this to include vertical lines, you can use


0 = Re(az+b)
which more closely resembles the "standard" equation of a line from high school:

a1x - a2y + Re(b) = 0
with the slope being a2/a1.

In the wide world of Euclidean n-space, given two points in , or equivalently, the endpoints of two vectors emanating from the origin, we can describe the line passing through them in terms of vector addition and scalar multiplication. Say the vectors in question are u and v. Then, the line passing through them is the locus of points given by


u + t(v-u)
for all real t. Equivalently, of course, we can also write

v + t(u-v).

(thing) by redgirlie (2.3 mon) (print)   ?   1 C! I like it! Tue Oct 16 2001 at 4:23:46

There are basically three types of line equations:

1. y=mx+b

x and y represent a point on the line (x,y)
m represents the slope of the line
b represents the y-intercept
to find the y-intercept let x=0
to find the x-intercept let y=0

2. ax+by=c

x and y represent a point on the line (x,y)
-a/b represents the slope of the line
a, b, and c are constants
the y-intercept is c/b when x=0
the x-intercept is c/a when y=0

3. (y-y1)=m(x-x1)

x1 and y1 represent a point(x1,y1) on the line
m represents the slope of the line
*usually the preferred formula for lines because if you know a point on the line and the slope you know the equation

Parallel lines have the same slope

Perpedicular lines have slopes that are negative reciprocals
e.g. if m is the slope of one line then the slope of the other line is -1/m

Slope can be calculated by using this formula:
(y2-y1)/(x2-x1)
*or is more easily known as rise/run


printable version
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