# General Form of a Straight Line Equation

Go back to  'Straight Lines'

Thus, we now know of three forms in which the equation of an arbitrary straight line can be written.

From those three forms, you might be able to deduce that the most general form for the equation of an arbitrary straight line is $$Ax + By + C = 0$$ . Let us try to prove this assertion, that is, let us try to show that $$Ax + By + C = 0$$ represents the equation of a straight line.

For this purpose, it will suffice to show that if we take any three arbitrary points $$\left( {{x_1},\,\,{y_1}} \right),\,\,\left( {{x_2},\,\,{y_2}} \right)$$ and  $$\left( {{x_3},\,\,{y_3}} \right)$$ on the curve $$Ax + By + C = 0$$  these three points will turn out to be collinear. Equivalently, the area of the triangle with the vertices as these three points will turn out to be zero.

Since all the three points satisfy the equation $$Ax + By + C = 0$$ ,we have

\begin{align}& A{{x}_{1}}+B{{y}_{1}}+C=0 \\ & A{{x}_{2}}+B{{y}_{2}}+C=0 \\ & A{{x}_{3}}+B{{y}_{3}}+C=0 \end{align}

We can eliminate A, B and C from these three equations simultaneously to obtain a relation involving only the co-ordinates of the three points. A basic knowledge of elimination in determinant form will tell you that the relation we’ll get after elimination is

$\left| \begin{matrix}{{x}_{1}} & {{y}_{1}} & 1 \\{{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\ \end{matrix} \right|=0$

which means that the area of the triangle formed by these three points as vertices is zero! Hence, the assertion is true.

With this discussion in mind, you should be able to write the equation for any arbitrary straight line. We will encounter the use of all these forms in the coming examples.

Before concluding this article, do this as a simple exercise based on the discussion we've already done:

(a) Show that the equation of the straight line with slope m and passing through the fixed point (x1, y1) is  $$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$$

(b) Show that the equation of the straight line passing through the two fixed points $$\left( {{x}_{1}},\,\,{{y}_{1}} \right)\,\,\text{and}\,\,\left( {{x}_{2}},\,\,{{y}_{2}} \right)$$ is

$\frac{y-{{y}_{1}}}{x-{{x}_{1}}}=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$

The following table summarizes the various forms of the straight line that we've encountered.

 Known parameters about the line Equation Name of this form 1. \begin{align} & \text{Slope}\,\,\,\,\,\,\,\,\,\,\,\,\,\,~~~~m \\ & y\text{-intercept}~~~\,\,\,\,~c \\ \end{align} $y=mx+c$ Slope-intercept form 2. \begin{align} & x-intercept~~~~a \\ & y-intercept~~~~b \\ \end{align} $\frac{x}{a}+\frac{y}{b}=1$ Intercept form 3. \begin{align} & \begin{array}{*{35}{l}} \text{Length of perpendicular} \\ \text{from origin to the line } \;\qquad\qquad\text{ :} p \end{array} \\ & \text{Inclination of perpendicular}\qquad:\alpha \end{align} $x\cos \alpha +y\sin \alpha =p$ Normal form 4. \begin{align}& \begin{array}{*{35}{l}} \,\,\,\,\,\,\,\,\text{Slope}\quad\text{ : } \;m \\ \text{Any point through} \end{array} \\ & \text{which the line passes }\!\!~\!\!\text{ }~~~~~~~~~:\left( {{x}_{1}},{{y}_{1}} \right) \end{align} $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ Point-slope form 5. \begin{align} & \text{Any two points through }\!\!~\!\!\text{ }~~~~~~~~~~~:\left( {{x}_{1}},{{y}_{1}} \right) \\ & \text{which the line passes }~~~~~~~~~~~~~~~:\left( {{x}_{2}},{{y}_{2}} \right) \\ \end{align} $\frac{y-{{y}_{1}}}{x-{{x}_{1}}}=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ Two point form \begin{align}& \text{Most general form }:\quad Ax+By+C=0 \quad \quad \text{where }A,B,C\in \mathbb{R} \\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \qquad \,\,\,\text{and at least one of }A,B\;\text{is non-zero} \end{align}

Note that each of the five specific forms mentioned in the table above can be converted easily to the most general form of the equation of a line. You are urged to do this as an exercise.

Also, the five forms are inter convertible among themselves in most cases too. For example, $$y = mx + c$$ can be written in intercept form as \begin{align}\frac{x}{-\left( c/m \right)}+\frac{y}{c}=1\end{align} so that the x-intercept of this line is  \begin{align}a=-\frac{c}{m}\end{align} and the y-intercept is b = c. You are urged to try out all the (possible) conversions from one form to another.

You should now be able to understand that to determine a straight line uniquely, we must have two quantities given. Thus, two points could uniquely fix a line, or a point on the line and its slope could do so too, and so on. Notice that the general equation of the line also in fact contains only two arbitrary constants:

\begin{align}&\qquad\quad Ax+By+C=0 \\ & \Rightarrow \quad\left( \frac{A}{C} \right)x+\left( \frac{B}{C} \right)y+1=0 \\ & \Rightarrow \quad Px+Qy+1=0\qquad\qquad\left\{ \begin{gathered} \text{contains only two} \\ \text{arbitrary constants} \end{gathered} \right\} \end{align}

Straight Lines
Straight Lines
grade 11 | Questions Set 1
Straight Lines
Straight Lines
grade 11 | Questions Set 2