In this post we will try to understand different form of equation of straight line. These topics are important part of Class 11 Mathematics and its questions are regularly asked in CBSE/NCERT Math Exam

- General Equation of a Straight Line
- General Equation of a Straight Line to Different Standard forms

**General Equation of a Straight Line**

The General Equation of a straight line is ** Ax + By + C = 0**

where both A and B are not zero simultaneously.

⟹ Here A, B and C are constants

⟹The graph of Ax + By + C = 0 is always a straight line.

**General Equation of a Straight Line to Different Standard forms**

**1. Slope-Intercept Form of Straight Line**

We know that equation of a straight line in slope-intercept form is

⟹ **y = mx + c**

and the general equation of the straight line is

⟹ Ax + By + C = 0

Reducing the general equation in the slope-intercept form

⟹ By = – Ax – C

⟹ y\ =\ -\ \frac{A}{B} x\ -\ \frac{C}{B} – – – – – eq(1)

Eq.(1) is in the form of y = mx + c

Comparing the values

**Points to be noted **

a) **Slope of the line Ax + By + C = 0** is equal to

=\ -\frac{A}{B\ } \ =\ -\ \frac {coefficient\ of\ x}{coefficient\ of\ y}

b) **Intercept of the line Ax + By + C = 0** is equal to

= \ -\frac{C}{B\ } \ =\ -\ \frac{constant\ term\ C}{coefficient\ of\ y}

**How to solve general equation to slope intercept questions**

**Step 01**: Write the equation of the straight line in the form of general equation as shown below –

Ax + By + C = 0**Step 02**: Find slope (m) \ =\ -\ \frac{A}{B}

and intercept (c) \ =\ -\frac{C}{B}

**Step 03**: Put the values of ‘m’ and ‘c’ in the slope-intercept form equation –

y = mx + c

You will get the equation of the straight line in slope-intercept form

**Slope Intercept Equation Questions**

**(Q1) Reduce the equation into slope-intercept form: 6x + 3y – 5 =0**

Given – the equation of the straight line in the general form

6x + 3y – 5 = 0

Comparing it with the general equation: Ax + By + C = 0

We get,

A = 6, B = 3 and C =-5

therefore,

slope (m) = -\frac{A}{B} \ =\ -\frac{6}{3} \ =\ -2

and intercept(c) = -\frac{C}{B} \ =\ -\frac{( -5)}{3} \ =\ \frac{5}{3}

Putting it in the slope-intercept form equation –

y = mx + c

y = -2x + \frac{5}{3}

This is the required equation in slope intercept form {Ans.}

**(Q2) Reduce the equation into slope-intercept form:****x +7y = 0**

Given – the equation of the straight line in the general form

x + 7y = 0

Comparing it with the general equation: Ax + By + C = 0

We get,

A = 1, B = 7 and C = 0

therefore, slope (m) = -\frac{A}{B} \ =\ -\frac{1}{7}

and intercept (c) = -\frac{C}{B} \ = -\frac{0}{7} \ =\ 0

Putting it in the slope-intercept form equation –

y = mx + c

y = \left( -\frac{1}{7}\right) x\ +\ 0\\\ \\

y = -\frac{1}{7} x\ +\ 0 \\

This is the required equation in slope intercept form.

**(Q3) Transform the equation** -3x\ +\ \sqrt{3} y=\ 3\ **into slope-intercept form** and also find the angle which straight line makes with the x-axis.

Sol.

Given – the equation of the straight line in the general form

Comparing it with the general equation: Ax + By + C = 0

We get,

A =-3,

B = \sqrt{3}

and C = -3

therefore,

slope(m) = -\frac{A}{B} \ =\ -\frac{( -3)}{\sqrt{3}} \ =\ \frac{3}{\sqrt{3}} \ =\ \sqrt{3}

and intercept (c) = -\frac{C}{B} \ =\ -\frac{( -3)}{\sqrt{3}} \ =\ \frac{3}{\sqrt{3}} \ =\ \sqrt{3}

Putting it in the slope-intercept form equation –

y = mx + c

y = \left(\sqrt{3}\right) x\ +\ \sqrt{3}\\

y = \sqrt{3} x\ +\ \sqrt{3}\\

This is the required equation in slope intercept form.{Ans.}

Now, since, slope (m) = \sqrt{3}\\ \\ tan\ \theta \ =\ \sqrt{3}\\\ \\ tan\ \theta \ =\ tan\ 60\degree \\\ \\ therefore,\ \theta \ =\ 60\degree \\\ \\ Hence,\ the\ line\ makes\ an\ angle\ of\ 60\degree \\ \\ with\ the\ positive\ direction\ of\ x-axis.\

**2. Intercept Form Equation of Straight Line**

We know that **equation of a straight line in intercept form** is:

⟹ \frac{x}{a} \ +\ \frac{y}{b} \ =\ 1

and the general equation of the straight line is:

⟹ Ax + By + C = 0

Reducing it in the slope-intercept form

Ax + By + C = 0

Ax + By = – C

**Points to be noted **

a) **x-intercept of the line** Ax + By + C = 0 is :

= -\frac{C}{A} = -\frac{constant\ term\ C}{coefficient\ of\ x}

b) **y-intercept of the line** Ax + By + C = 0 is:

= -\frac{C}{B} = -\frac{constant\ term\ C}{coefficient\ of\ y}

**How to reduce General Equation of a straight line to Intercept form**

Way to approach such questions**Step 1**: Write the equation of the straight line in the form of general equation as shown below –

Ax + By + C = 0

**Step 2**: Find x-intercept = \ -\ \frac{C}{A} and

y-intercept =\ -\frac{C}{B}

**Step 3**: Put the values of x-intercept and y-intercept in the intercept form equation –

\frac{x}{a} \ +\ \frac{y}{b} \ =\ 1

You will get the equation of the straight line in intercept form.

**Questions on Intercept form of Straight Line**

**(Q1) Reduce the equation into slope-intercept form:**

4x – 3y – 6 = 0

Sol.

Given – the equation of the straight line in the general form

4x – 3y – 6 = 0

Comparing it with the general equation: Ax + By + C = 0

We get,

A = 4, B = -3 and C = -6

therefore, x-intercept (a) = \ -\ \frac{C}{A} \ =\ -\frac{( -6)}{4} \ =\ \frac{3}{2}

and y-intercept (b) \ = \ -\frac{C}{B} \ =\ -\frac{( -6)}{( -3)} \ =\ -2

Putting it in the slope-intercept form equation –

\frac{x}{a} \ +\ \frac{y}{b} \ =\ 1\\\ \\ \frac{x}{\frac{3}{2}} \ +\ \frac{y}{-2} \ =\ 1\\

This is the required equation in intercept form.

**(Q2) Reduce the equation into slope-intercept form:****⟹ 3x + 2y – 12 = 0**

Sol. Given – the equation of the straight line in the general form

3x + 2y – 12 = 0

Comparing it with the general equation:

Ax + By + C = 0

We get,

A = 3, B = 2 and C = -12

therefore,

x-intercept (a) = - \frac{C}{A} \ =\ -\frac{( -12)}{3} \ =\ \frac{12}{3} \ =4\

and y-intercept (b) = \ -\frac{C}{B} \ =\ -\frac{( -12)}{2} \ =\ \frac{12}{2} =6

Putting it in the slope-intercept form equation –

\frac{x}{a} \ +\ \frac{y}{b} \ =\ 1 \\\ \\ \frac{x}{4} \ +\ \frac{y}{6} \ =\ 1\\

This is the required equation in intercept form.{Ans.}

**(Q3)** \ Reduce\ the\ equation\ into\ slope-intercept\ form:\\\ \\ \sqrt{3} y\ -3x=\ 3. \ \

Sol. Given – the equation of the straight line in the general form

\sqrt{3} y\ -3x=\ 3\\\ \\ 3x\ -\ \sqrt{3} y\ +\ 3\ =\ 0 \\\ \\ \\\ \\ Comparing\ it\ with\ the\ general\ equation:\\ \\ \ Ax\ +\ By\ +\ C\ =\ 0 \\\ \\ We\ get,\\ \\ A\ =\ 3,\ B\ =\ -\sqrt{3} \ and\ C\ =\ 3 \\\ \\therefore, x-intercept (a) = \ -\ \frac{C}{A} \ =\ -\frac{3}{3} \ =\ -1\\\ \\ and\ y-intercept\ ( b) \ =\ -\frac{C}{B} \ =\ -\frac{3}{( -\sqrt{3)}} \ =\ \frac{3}{\sqrt{3}} =\sqrt{3} \\\ \\ Putting\ it\ in\ the\ slope-intercept\ form\ equation\ - \\\ \\ \frac{x}{a} \ +\ \frac{y}{b} \ =\ 1\\\ \\ \frac{x}{-1} \ +\ \frac{y}{\sqrt{3}} \ =\ 1\\\ \\ This\ is\ the\ required\ equation\ in\ intercept\ form.

**(Q4) ** Equation\ of\ a\ straight\ line\ is\ 3x\ -\ 4y\ +\ 10\ =\ 0.\ Find\ its\ \\\ \\ (i) \ slope \\ \\ ( ii) \ x-intercept\ and\ y-intercept

Sol. Given – the equation of the straight line in the general form

3x – 4y + 10 = 0

Comparing it with the general equation: Ax + By + C = 0

We get,

A = 3, B = -4 and C = 10

(i) slope\ ( m) \ =\ -\frac{A}{B} \\ \\ ⟹ \ -\frac{3}{( -4)} \ =\ \frac{3}{4} \\\ \\ (ii) x\ intercept\ (a) =\ -\ \frac{C}{A} \\ \\ =\ -\frac{10}{3} \ \

and y intercept (b) \ =\ -\frac{C}{B} \ =\ -\frac{10}{( -4)} \ =\ \frac{5}{2}

Hence,

(i) slope = \frac{3}{4}

(ii) x-intercept = -\frac{10}{3}

(iii) y-intercept = \frac{5}{2}

**3. Normal Form of Straight Line**

We know that **equation of a straight line in normal form is:**

⟹ x\ cos\ \alpha \ +\ y\ sin\ \alpha \ =\ p

and the **general equation** of the straight line is

⟹ Ax + By + C = 0

Reducing it in the slope-intercept form

⟹ Ax + By + C = 0

⟹ Ax + By =- C**{Case 1} : When C < 0 i.e., -C > 0 **

Dividing both sides of equation (1) by \sqrt{A^{2} \ +\ B^{2}} , we get:

**{Case II} : When C >0 i.e., -C < 0**

⟹ – Ax – By = C

Dividing\ both\ sides\ of\ equation\ ( 1) \ by\\ \\ \sqrt{A^{2} \ +\ B^{2}} ,\ we\ get \\\ \\ -\frac{A}{\ \sqrt{A^{2} \ +\ B^{2}}} x\ -\ \frac{B}{\ \sqrt{A^{2} \ +\ B^{2}}} \ =\ \frac{C}{\ \sqrt{A^{2} \ +\ B^{2}}}\\\ \\ comparing\ eq.( 2) \ with\ normal\ form\ equation\\ \\ x\ cos\ \alpha \ +\ y\ sin\ \alpha \ =\ p,\ we\ get \\\ \\ cos\ \alpha \ =\ -\frac{A}{\ \sqrt{A^{2} \ +\ B^{2}}} ,\ sin\ \alpha \ =-\ \frac{B}{\ \sqrt{A^{2} \ +\ B^{2}}} \ and\ p\ =\ \frac{C}{\ \sqrt{A^{2} \ +\ B^{2}}} \\\ \\Note: In the normal form:

⟹ x\ cos\ \alpha \ +\ y\ sin\ \alpha \ =\ p ,

p is always taken as positive.

**How to reduce General Equation of a straight line to Normal form**

Way to approach such questions

**Step 01**: Write the equation of the straight line in the form of general equation as shown below –

Ax + By + C = 0

**Step 02**: Transpose the constant term to R.H.S.

After transportation, if the constant term on R.H.S. is (-)ve, make it positive by changing the sign throughout.

**Step 03**: Divide both sides by \sqrt{( coefficient\ of\ x)^{2\ } \ +\ ( coefficient\ of\ y)^{2}}

Thus, the normal form of line Ax + By + C = 0 is –

-\frac{A}{\ \sqrt{A^{2} \ +\ B^{2}}} x\ -\ \frac{B}{\ \sqrt{A^{2} \ +\ B^{2}}} \ =\ \frac{C}{\ \sqrt{A^{2} \ +\ B^{2}}} ,\\ \\ if\ c >\ 0 \\\ \\ \\\ \\
\frac{A}{\ \sqrt{A^{2} \ +\ B^{2}}} x\ +\ \frac{B}{\ \sqrt{A^{2} \ +\ B^{2}}} \ =\ -\frac{C}{\ \sqrt{A^{2} \ +\ B^{2}}} ,\\ \\ if\ c\ < \ 0

You will get the equation of the straight line in normal form.

**Questions on Normal form of Straight Line**

Q1) Reduce\ the\ equation\ \sqrt{3} x\ +\ y\ -\ 8\ =\ 0\ to\ the\ normal\ form\ and\ find\ the\ length\ of \\ \\ the\ perpendicular\ from\ the\ origin\ to\ the\ line

Sol. the equation of the straight line in the general form

\sqrt{3} x\ +\ y\ -\ 8\ =\ 0 \\\ \\ \sqrt{3} x\ +\ y\ =\ 8\ \\\ \\ Dividing\ both\ sides\ of\ eq.( 1) \ by \\ \\ \sqrt{( coefficient\ of\ x)^{2\ } \ +\ ( coefficient\ of\ y)^{2}} \\\ \\ \sqrt{\left( -\sqrt{3}\right)^{2} +\ ( 1)^{2}} \ =\ \sqrt{3\ +\ 1} \ =\ \sqrt{4} \ =\ 2 \\\ \\ we\ get,\ \frac{\sqrt{3} x}{2} \ +\frac{y}{2} \ =\ \frac{8}{2} \\\ \\ \frac{\sqrt{3}}{2} x\ +\frac{1}{2} y\ =\ 4 \\\ \\ Here,\ cos\ \alpha \ =\frac{\sqrt{3}}{2} \\ \\ and\ sin\ \alpha \ =\ \frac{1}{2} \\ \\ and\ p\ =\ 4 \\\ \\

here, both cos\ \alpha \ and\ sin\ \alpha \ are\ positive\\ \\ hence,\ \alpha \ lies\ in\ 1st\ quadrant \\\ \\ sin\ \alpha \ =\ \frac{1}{2} \ =\ sin\ 30\degree \\\ \\ therefore,\ \alpha \ =\ 30\degree \\\ \\ hence,\ eq.( 2) \ becomes\\\ \\ x\ cos\ 30\degree \ +\ y\ sin\ 30\degree \ =\ 4

(Q2) Reduce\ x\ +\ \sqrt{3} y\ +\ 4\ =\ 0\ to\ the\ normal\ form\\ \\ and\ the\ find\ the\ values\ of\ 'p'\ and\ '\alpha

Sol. the equation of the straight line in the general form

x\ +\ \sqrt{3} y\ +\ 4\ =\ 0 \\\ \\ x\ +\ \sqrt{3} y\ =-\ 4 \\\ \\ Multiplying\ eq.( 1) \ with\ '-1'\ to\ make\ the\ R.H.S.\ positive. \\\ \\ -\ x\ -\sqrt{3} y\ =\ 4\ \\\ \\ Dividing\ both\ sides\ of\ eq.( 2) \ by\\ \\ \sqrt{( coefficient\ of\ x)^{2\ } \ +\ ( coefficient\ of\ y)^{2}} \\\ \\ \sqrt{( -1)^{2\ } \ +\ \left( -\sqrt{3}\right)^{2}} \ =\ \sqrt{1\ +\ 3} \ =\ \sqrt{4} \ =\ 2 \\\ \\ we\ get,\ -\frac{x}{2} \ -\frac{\sqrt{3} y}{2} \ =\ \frac{4}{2} \\\ \\ -\frac{1}{2} x\ -\frac{\sqrt{3}}{2} y\ =\ 2 \\\ \\ comparing\ it\ with\ x\ cos\ \alpha \ +\ y\ sin\ \alpha \ =\ p,\ we\ get \\\ \\ cos\ \alpha \ =\ -\frac{1}{2} \ ;\ sin\ \alpha \ =\ -\frac{\sqrt{3}}{2} \ and\ p\ =\ 2 \\\ \\ As\ cos\ \alpha \ and\ sin\ \alpha \ are\ both\ negative,\\ \\ therefore,\ \alpha \ lies\ in\ the\ third\ quadrant \\\ \\ therefore,\ cos\ \alpha \ =\ -\frac{1}{2} \\ \\ =\ -\ cos\ 60\degree \ =\ cos\ ( 180\degree \ +\ 60\degree ) \\\ \\ cos\ \alpha \ =\ cos\ 240\degree \\\ \\ or\ \alpha \ =\ 240\degree \\\ \\ Hence,\ p\ =\ 2\ and\ \alpha \ =\ 240\degree .</p>
<p>and\ the\ equation\ in\ normal\ form\ becomes:\\ \\ \ x\ cos\ 240\degree \ +\ y\ sin\ 240\degree \ =\ 2

(Q3) Reduce\ the\ equation\ x\ -\ y\ =\ 4\ to\ the\ normal\ form\ and\ find\ the\ length\ of \\ \\ \ the\ perpendicular\ from\ the\ origin\ to\ the\ line.

The equation of the straight line in the general form

⟹ x – y = 4 – – – eq(1)

Dividing both sides of eq.( 1) by \sqrt{( coefficient\ of\ x)^{2\ } \ +\ ( coefficient\ of\ y)^{2}} \\\ \\ \sqrt{( 1)^{2} +\ ( -1)^{2}} \ =\ \sqrt{1\ +\ 1} \ =\ \sqrt{2} \\\ \\ we\ get,\ \frac{x}{\sqrt{2}} \ -\frac{y}{\sqrt{2}} \ =\ \frac{4}{\sqrt{2}} \\\ \\ \frac{1}{\sqrt{2}} x\ -\frac{1}{\sqrt{2}} y\ =\ 2\sqrt{2}\\\ \\ Here,\ cos\ \alpha \ =\frac{1}{\sqrt{2}} \\ \\ and\ sin\ \alpha \ =\ -\frac{1}{\sqrt{2}} \\ \\ and\ p\ =\ 2\sqrt{2}

here, both cos\ \alpha \ is\ positive\ and\ sin\ \alpha \ is\ negative\\ \\ hence,\ \alpha \ lies\ in\ 4th\ quadrant \\\ \\ sin\ \alpha \ =\ -\frac{1}{\sqrt{2}} \\ \\ =\ -sin\ 45\degree \ =\ sin\ ( 270\degree +45\degree ) \\ \\ =\ sin\ 315\degree \\\ \\ therefore,\ \alpha \ =\ 315\degree \\\ \\ hence,\ eq.( 2) \ becomes\\\ \\ x\ cos\ 315\degree \ +\ y\ sin\ 315\degree \ =\ 2\sqrt{2}