# Standard form of Equation of a Straight line

## Definition

Expressing a straight line in standard form equation is called general form of equation of a straight line.

A straight line can be expressed in a standard equation form to study the properties of the line, understanding behaviour of line and also solving the problems in mathematics. The standard form equation denotes a line. So, it is also known a linear equation. The linear equation of a straight line plays vital role in developing the fundamentals in mathematics and it is given here.

$ax+by+c=0$

Many of you donâ€™t know how this expression is developed mathematically but you will study how it actually represents a straight line from a mathematical proof. The expression can be developed geometrically and also based on a trigonometric principle.

#### Proof

Consider two points on a plane and named A and B. Point A and B are located at $\left({x}_{1}$, ${y}_{1}\right)$ and $\left({x}_{2}$, ${y}_{2}\right)$ respectively.

Draw a straight line which passes through both points. Consider a point on the straight line and name it C and assume it is located at $\left(x,y\right)$.

Now construct two different triangles as shown in the below picture for developing the expression using both geometry and trigonometry. A vertical line is drawn through point B. Similarly, two horizontal straight lines are drawn and these two are passed through the point A and C. The vertical line meets the horizontal lines at two points, named them point E and D.

Now we have two right angle triangles ΔBAD and ΔBCE. Assume, the angle made by the lines $\stackrel{↔}{\mathrm{AB}}$ and $\stackrel{↔}{\mathrm{AD}}$ is θ in degrees. As you know, both are right angle triangles. Therefore, the angle at $\angle$B and $\angle$ D is 90°.

According to trigonometric properties, if angle $\angle$A is theta and Angle $\angle$ D is 90°, then angle B should be (90-θ)°. Similarly, if angle at point E is 90° and angle at B is (90-θ)°, then the angle at point C will be θ°. Therefore, the angle of triangle ΔBAD is equal to the angle of another ΔBCD.

The above picture explains the positions and lengths of vertices of both triangles. As per this, the length of the each side is given here.

AD $={x}_{2}-{x}_{1}$

BD $={y}_{2}-{y}_{1}$

CE $={x}_{2}-x$

BE $={y}_{2}-y$

According to triangle ΔBAD,

$\mathrm{tan}\theta =\frac{\mathrm{BD}}{\mathrm{AD}}$

$⇒\mathrm{tan}\theta =\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$

Similarly, according to triangle ΔBCE,

$\mathrm{tan}\theta =\frac{\mathrm{BE}}{\mathrm{CD}}=\frac{{y}_{2}-y}{{x}_{2}-x}$

According to above both two trigonometric ratio expressions, the both expressions of trigonometric ratio tangent can be equated because they both are same.

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

Apply principle of cross multiplication for performing the multiplication mathematically.

$⇒\left({y}_{2}-{y}_{1}\right).\left({x}_{2}-x\right)=\left({y}_{2}-y\right).\left({x}_{2}-{x}_{1}\right)$

$⇒{y}_{2}.{x}_{2}-{y}_{2}.x-{y}_{1}.{x}_{2}+{y}_{1}.x={y}_{2}.{x}_{2}-{y}_{2}.{x}_{1}-y.{x}_{2}+y.{x}_{1}$

Bring all terms of right hand side to left hand side.

$⇒{y}_{2}.{x}_{2}-{y}_{2}.x-{y}_{1}.{x}_{2}+{y}_{1}.x-{y}_{2}.{x}_{2}+{y}_{2}.{x}_{1}+y.{x}_{2}-y.{x}_{1}=0$

It can be written as follows.

$-{y}_{2}.x+{y}_{1}.x+y.{x}_{2}-y.{x}_{1}-{y}_{1}.{x}_{2}+{y}_{2}.{x}_{1}+{y}_{2}.{x}_{2}-{y}_{2}.{x}_{2}=0$

Two ${y}_{2}.{x}_{2}$ terms are appearing in the expression with opposite signs. So, they both get cancelled.

$-{y}_{2}.x+{y}_{1}.x+y.{x}_{2}-y.{x}_{1}-{y}_{1}.{x}_{2}+{y}_{2}.{x}_{1}=0$

Now, take common x and y terms for writing the expression in simple form.

$\left(-{y}_{2}+{y}_{1}\right)x+\left({x}_{2}-{x}_{1}\right)y+\left(-{y}_{1}.{x}_{2}+{y}_{2}.{x}_{1}\right)=0$

The coefficients of variables $x$ and $y$ and remaining part are constants in value. So, we can represent them by some assumed constants.

Assume,

$a=-{y}_{2}+{y}_{1},b={x}_{2}-{x}_{1}$ and $c=-{y}_{1}.{x}_{2}+{y}_{2}.{x}_{1}$

Now, the linear expression can be written in a simple form.

$ax+by+c=0$

The straight line equations are given directly in mathematical problems by displaying various values to a, b and c in the expression.