Notes On Equations of Lines (Part II) - CBSE Class 11 Maths
Slope-Intercept Form

Consider a line L with slope m that intersects the Y-axis at a point A, which is at distance C from the origin. That is the given line passes through the point zero, C.
 
Point-Slope Form: yy1 = m(xx1)
 
Thus, the point-slope form of line l  is: yc = m(x – 0)
 
y = mx + c
      
This equation is the Slope-Intercept Form of the line, where:

m = Slope of the line

c = y-intercept of the line   
 

 
Consider a line L with slope m, whose x-intercept is given as D. The given line passes through the point B (d, 0).
 
Thus, the point-slope form of line l:

yy1 = m(xx1)

y – 0 = m(xd)      
 
y = m(xd)  
    
This equation is the Slope-Intercept Form, where:

m = Slope of the line

d = x-intercept of the line   
 
Intercept  Form
 
Consider a line l with x-intercept A and y-intercept B.

Let’s denote the points where the line intersects the X-axis and the Y-axis as P and Q, respectively.

Therefore, the coordinates of point ‘P’ are (a, 0) and that of point ‘Q’ are (0, b).
 
Two-Point Form: yy1 = {(y2y1)/ (x2x1)} x (xx1)

Thus, the two-point form of line l is: y -  0 = {(b – 0)/ (0 – a)} x (xa)

ay = -bx + ab    

bx + ay = ab    

Dividing both sides by ab:

 (bx + ay)/ab = ab/ab    

x/a + y/b = 1   

The above equation is the Intercept Form of equation of the straight line.
 

 
Normal Form

Consider a non-vertical line L. Let OA be the normal or perpendicular drawn to line L from the origin.

Let the perpendicular distance of line L from the origin, or the length of perpendicular OA, be P.
 
∠AOX = ω


AM ⊥ OX

Distance OM represents the x-coordinate, and distance AM represents the y-coordinate of point A.
 
From right angled triangle OMA

Cos ω = OM OA

⇒ Cos ω = OM P

⇒ OM = P Cos ω

Sin ω = AM OA

⇒ Sin ω = AM P

⇒ AM = P Sin ω
 
For the given values of P and omega, the coordinates of point A will remain the same no matter the quadrant in which line L lies.
 
The coordinates of a point on line L, we can find its equation if we know its slope, using the point-slope form. Let M be the slope of the given line L.

Given OA ⊥ l:

Slope of l x slope of OA = -1

Slope of OA = tan ω

m tan ω = -1     
                  
m = -1/ tan ω

m = -cot ω

m = -cos ω/ sin ω     
   
Point-Slope Form: yy1 = m(xx1)

Thus, the point-slope form of line l is: yp sin ω = (-cos ω/ sin ω)(xp cos ω)  

y sin ω – p sin2 ω = -x cos ω + p cos2 ω)

x cos ω + y sin ω = p(sin2 ω + cos2 ω)⇒ x cos ω + y sin ω = p     [ sin2 ω + cos2 ω = 1]

Normal Form: x cos ω + y sin ω = p    

where:

p = normal distance of the line from origin

ω = Inclination of the normal
 

Summary

Slope-Intercept Form

Consider a line L with slope m that intersects the Y-axis at a point A, which is at distance C from the origin. That is the given line passes through the point zero, C.
 
Point-Slope Form: yy1 = m(xx1)
 
Thus, the point-slope form of line l  is: yc = m(x – 0)
 
y = mx + c
      
This equation is the Slope-Intercept Form of the line, where:

m = Slope of the line

c = y-intercept of the line   
 

 
Consider a line L with slope m, whose x-intercept is given as D. The given line passes through the point B (d, 0).
 
Thus, the point-slope form of line l:

yy1 = m(xx1)

y – 0 = m(xd)      
 
y = m(xd)  
    
This equation is the Slope-Intercept Form, where:

m = Slope of the line

d = x-intercept of the line   
 
Intercept  Form
 
Consider a line l with x-intercept A and y-intercept B.

Let’s denote the points where the line intersects the X-axis and the Y-axis as P and Q, respectively.

Therefore, the coordinates of point ‘P’ are (a, 0) and that of point ‘Q’ are (0, b).
 
Two-Point Form: yy1 = {(y2y1)/ (x2x1)} x (xx1)

Thus, the two-point form of line l is: y -  0 = {(b – 0)/ (0 – a)} x (xa)

ay = -bx + ab    

bx + ay = ab    

Dividing both sides by ab:

 (bx + ay)/ab = ab/ab    

x/a + y/b = 1   

The above equation is the Intercept Form of equation of the straight line.
 

 
Normal Form

Consider a non-vertical line L. Let OA be the normal or perpendicular drawn to line L from the origin.

Let the perpendicular distance of line L from the origin, or the length of perpendicular OA, be P.
 
∠AOX = ω


AM ⊥ OX

Distance OM represents the x-coordinate, and distance AM represents the y-coordinate of point A.
 
From right angled triangle OMA

Cos ω = OM OA

⇒ Cos ω = OM P

⇒ OM = P Cos ω

Sin ω = AM OA

⇒ Sin ω = AM P

⇒ AM = P Sin ω
 
For the given values of P and omega, the coordinates of point A will remain the same no matter the quadrant in which line L lies.
 
The coordinates of a point on line L, we can find its equation if we know its slope, using the point-slope form. Let M be the slope of the given line L.

Given OA ⊥ l:

Slope of l x slope of OA = -1

Slope of OA = tan ω

m tan ω = -1     
                  
m = -1/ tan ω

m = -cot ω

m = -cos ω/ sin ω     
   
Point-Slope Form: yy1 = m(xx1)

Thus, the point-slope form of line l is: yp sin ω = (-cos ω/ sin ω)(xp cos ω)  

y sin ω – p sin2 ω = -x cos ω + p cos2 ω)

x cos ω + y sin ω = p(sin2 ω + cos2 ω)⇒ x cos ω + y sin ω = p     [ sin2 ω + cos2 ω = 1]

Normal Form: x cos ω + y sin ω = p    

where:

p = normal distance of the line from origin

ω = Inclination of the normal
 

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