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 =
⇒ Cos ω =
⇒ OM =
P Cos ω
Sin ω =
⇒ Sin ω =
⇒ 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 sin
2 ω = -
x cos ω +
p cos
2 ω)
⇒
x cos ω +
y sin ω =
p(sin
2 ω + cos
2 ω)⇒
x cos ω +
y sin ω =
p [
sin
2 ω + cos
2 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
