Equations of Lines (Part II)

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: y – y₁ = m(x – x₁)
 
Thus, the point-slope form of line l  is: y – c = 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:

y – y₁ = m(x – x₁)

⇒y – 0 = m(x –d)      
 
⇒ y = m(x – d)  
    
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: y -  y₁ = {(y₂ – y₁)/ (x₂ – x₁)} x (x – x₁)

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

⇒ 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 = $\frac{\text{OM}}{\text{OA}}$

⇒ Cos ω = $\frac{\text{OM}}{\text{P}}$

⇒ OM = P Cos ω

Sin ω = $\frac{\text{AM}}{\text{OA}}$

⇒ Sin ω = $\frac{\text{AM}}{\text{P}}$

⇒ 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: y – y₁ = m(x – x₁)

Thus, the point-slope form of line l is: y – p sin ω = (-cos ω/ sin ω)(x – p 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     [$\because$ 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