Equation of a Line

The standard forms of the equation of a line are

  • Slope-intercept form
  • Intercept form
  • Normal form

General Equation of a Line

The general equation in two variables of the first degree is represented as

Ax + By +C = 0,

A, B ≠ 0 where A, B and C are constants which belong to real numbers.

When we represent the equation geometrically, we always get a straight line.

Below is a representation of straight line formulas Ax + By + C = 0 in different forms:

Slope-intercept Form

We know that the equation of a straight line in slope-intercept form is given as:

y = mx + c

Where m indicates the slope of the line and c is the y-intercept
When B ≠ 0 then, the standard equation of first degree Ax + By + C = 0 can be rewritten in slope-intercept form as:

y = − AB x − CB

Thus m= –AB and c = –CB .

Intercept Form

The equation of a line making intercepts equal to a and b on the x-axis and the y-axis respectively is given by:

ax + by =c

Now if, C ≠ 0 then the general equation of line Ax + By + C = 0 can be obtained by replacing the x-intercept by –CA and y-intercept by –CB

x−CA + y−CB = 1Equation of a LineIf C = 0 then Ax + By = 0 represent the equation of a line passing through the origin.

Normal Form

The equation of the line whose length of the perpendicular from the origin is p and the angle made by the perpendicular with the positive x-axis is given by α is given by:

x cos α+y sin α = p

This is known as the normal form of the line.
The general form of the line Ax + By + C = 0 can be represented in normal form as:

A cos α = B sin α = – p

From this we can say that \(\cos \alpha = -\frac{p}{A}\) and \(\sin \alpha = -\frac{p}{B}\).
Also it can be inferred that,

\(\cos^{2} \alpha + \sin^{2} \alpha = \left (\frac{p}{A} \right )^{2} + \left (\frac{p}{B} \right )^{2}\)

\(\Rightarrow 1 = p^{2} \left (\frac{A^{2} + B^{2}}{A^{2}B^{2}} \right )\)

\(\Rightarrow p = \left (\frac{AB}{\sqrt {A^{2} + B^{2}}} \right )\)

From the general equation of a straight line Ax + By + C = 0 , we can conclude the following:

i) The slope is given by -A/B, given that B ≠ 0.

ii) The x-intercept is given by \(- \frac{C}{A}\) and the y-intercept is given by \(- \frac{C}{B}\).

iii) It can be seen from the above discussion that:
\(p = \pm\left (\frac{AB}{\sqrt {A^{2} + B^{2}}} \right )\) , \(\cos \alpha = \pm \left (\frac{B}{\sqrt {A^{2} + B^{2}}} \right )\)  , \(\sin \alpha = \pm \left (\frac{A}{\sqrt {A^{2} + B^{2}}} \right )\)

iv) If two points (x1,y1) and(x2,y2)are said to lie on the same side of the line Ax + By + C = 0, then the expressions \(Ax_{1} + By_{1} + C\) and \(Ax_{2} + By_{2} + C\) will have the same sign or else these points would lie on the opposite sides of the line.

Equations of a Line Examples

To understand this concept better let us take an example:

(i)The equation of a line is given by, 2x – 6y +3 = 0. Find the slope and both the intercepts.
Solution: The given equation 2x – 6y + 3 = 0 can be represented in slope intercept form as:

y = 13x + 12
Comparing it with y = mx + c, we can say that,
m = 13
Also the above equation can be re-framed in intercept form and be written as;
x−32 + y- 12 = 1
Thus x-intercept is given as a = −32 and y-intercept as b = 12.
To learn more about concepts of line, visit our site BYJU’S.
This is known as the normal form of a line.
The general form of the line Ax + By + C = 0 can be represented in normal form as:

A cos α = B sin α = – p

From this we can say that \(\cos \alpha = -\frac{p}{A}\) and \(\sin \alpha = -\frac{p}{B}\).
Also it can be inferred that,

\(\cos^{2} \alpha + \sin^{2} \alpha = \left (\frac{p}{A} \right )^{2} + \left (\frac{p}{B} \right )^{2}\)

\(\Rightarrow 1 = p^{2} \left (\frac{A^{2} + B^{2}}{A^{2}B^{2}} \right )\)

\(\Rightarrow p = \left (\frac{AB}{\sqrt {A^{2} + B^{2}}} \right )\)

From the general equation of a straight line Ax + By + C = 0 , we can conclude the following:
i) The slope is given by -A/B, given that B ≠ 0.

ii) The x-intercept is given by \(- \frac{C}{A}\) and the y-intercept is given by \(- \frac{C}{B}\).

iii) It can be seen from the above discussion that:
\(p = \pm\left (\frac{AB}{\sqrt {A^{2} + B^{2}}} \right )\) , \(\cos \alpha = \pm \left (\frac{B}{\sqrt {A^{2} + B^{2}}} \right )\)  , \(\sin \alpha = \pm \left (\frac{A}{\sqrt {A^{2} + B^{2}}} \right )\)

iv) If two points (x1,y1) and(x2,y2)are said to lie on the same side of the line Ax + By + C = 0, then the expressions \(Ax_{1} + By_{1} + C\) and \(Ax_{2} + By_{2} + C\) will have the same sign or else these points would lie on the opposite sides of the line.

(ii)The equation of a line is given by, 2x – 6y +3 = 0. Find the slope and both the intercepts.

Solution: The given equation 2x – 6y + 3 = 0 can be represented in slope intercept form as:

y = 13x + 12
Comparing it with y = mx + c, we can say that,
m = 13
Also the above equation can be re-framed in intercept form and be written as;
x−32 + y- 12 = 1
Thus x-intercept is given as a = −32 and y-intercept as b = 12.

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