Three Standard form of Equation of Straight line

Slope intercept form

To find the equation of a straight line in form of y = mx + c
Let a straight line XY make an intercept c on Y-axis. then OY = c.
Let m be the slope of the line and \(\theta\) be its inclination.
Then, m = tanΘ
Let P(x, y) be any point on the XY. Draw perpendicular TR from P to X-axis.
Then, OR = x and RT = y.
Again, draw perpendicular YS from Y to the line segment TR, then
YS = OR = x
ST = RT - RS = RT - OY = y - c
Also,
∠TYS = ∠YXO = \(\theta\)
From right angled ΔYTS,
tanΘ = \(\frac{ST}{YS}\)
or, m = \(\frac{y - c}{x}\)
or, y - c = mx
or, y = mx + c, which is the equation of a straight line in the required form.

Double intercept form

To find the equation of a straight line in the form \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1
Let the straight line EF cut the axis at E and F. Let OE = a and OF = b. These are the intercept on the X-axis and the Y-axis respectively. Obviously, the coordinates of E and F are respectively (a, 0) and (0, b).
Let G(x, y) be any point on the line EF. Then,
Slope of the line segment EG = \(\frac{y - 0}{x - a}\)
Slope of the line segment GF= \(\frac{b -y}{0 - x}\)
But EG and GF are the segments of the same straight line.
So, \(\frac{y - 0}{x - a}\) = \(\frac{b - y}{0 - x}\)
or, -xy = bx - ab - xy + ay
Dividing both sides by ab we get
\(\frac{x}{a}\) + \(\frac{y}{b}\) = 1, which bis the equation of straight line in the required form.

Normal form or Perpendicular form

To find the equation of a straight line in the form x cos∝ + y sin∝ = p
Let a straight line XY cut the axes at X and Y. Then,
X-intercept = OX and Y-intercept =OY
Then, equation of the line XY is given by
\(\frac{x}{OX}\) + \(\frac{y}{OY}\) = 1 . . . . . . . . . . . . . . . . . (i)
Draw perpendicular OZ from origin to the line XY.
Let OZ = p and∠XOZ = \(\alpha\).
Then, ∠YOZ = 90° - \(\alpha\) and
∠OYZ =90° - ∠YOZ = 90° - (90° - ∝) = ∝
From right angled triangle OXZ,
sec∝ = \(\frac{OX}{OZ}\)
∴ OX = OZ sec\(\alpha\) = p sec\(\alpha\)
From right angled triangle OYZ,
cosec∝ = \(\frac{OY}{OZ}\)
∴ OY = OZ cosec\(\alpha\) = p cosec\(\alpha\)
Now,
putting the values of OA and OB in (i) we get.
\(\frac{x}{p \: secα} + \frac{y}{p \: cosecα}\) = 1
or, \(\frac{x \: cosα}{p} + \frac{y \: sinα}{p}\) = 1
or, x cosα + y sinα = p, which is the equation of a straight line in the required form.