General Equation of a Line (C)

Every line in the plane is described by an equation of the form

Ax + By = C

where at least one of A and B is not zero. Here we are assuming x denotes the first coordinate of a point P in the plane, and y denotes the second coordinate of this point.

Each line has many such equations. In fact, if

Ax + By = C

is an equation of a line, so is

ADx + BDy = CD

for any non-zero real number D. To effect a standardization, one can require that A and B are always chosen so that

A² + B² = 1

and that A is not negative. If A is zero, make B equal to 1. If we make this choice, then there is a geometric interpretation which makes it easy to see why all lines can be described by such equations. Suppose that $${P=(x_0, y_0)}$$is a point on the given line. Draw a circle of radius 1 about about P, and then draw the diameter of this circle which is perpendicular to the given line. Then one of the endpoints of this diameter has coordinates $${(A + x_0, B + y_0)}$$ where A and B meet the requirements outlined above. From the Pythagorean Theorem, if $${(x,y)}$$ is another point on the line we must have

$$\left(\sqrt{A^2 + B^2}\right)^2 + \left(\sqrt{(x-x_0)^2 + (y... ...^2}\right)^2 = \left(\sqrt{(A+x_0-x)^2 + (B+y_0-y)^2}\right)^2$$

which simplifies to

0 = A(x-x₀) + B(y-y₀)

or

Ax + By = Ax₀ + By₀

so $${ C = Ax_0 + By_0}$$.

To find a form of the general equation of a line given two points we proceed as in the following example:

Given: The line passes through (8, 4) and (3,11).

Find: A, B, and C so that that $${Ax +By = C}$$ describes this line.

Solution: We must have

8A + 4B = C = 3A + 11B

or

5A = 7B

We also want $${A^2 + B^2 = 1}$$, or $${49A^2 + 49B^2 = 49}$$. Since $${25A^2 = 49B^2}$$ we have $${49A^2 + 25A^2 = 49}$$ or $${74A^2 = 49}$$.From here we can find A, then B, and finally C, since $${C = 8A + 4B}$$.