Three Standard form of Equation of Straight line

Subject: Optional Mathematics

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This note contents various method or way to measure the slope intercept.

Three Standard form of Equation of Straight line

Slope intercept form

Slope Intercept Form
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.

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Double intercept form

Double Intercept 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

Normal/ Perpendicular Form
Normal/ 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.

 

Things to remember
  • y = mx+ c
  • \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1
  • x cos∝ + y sin∝ = p
  • It includes every relationship which established among the people.
  • There can be more than one community in a society. Community smaller than society.
  • It is a network of social relationships which cannot see or touched.
  • common interests and common objectives are not necessary for society.
Videos for Three Standard form of Equation of Straight line
Finding the Equation of a Line in Slope-Intercept Form
How to find the equation of a straight line graph with two points
The general equation of a straight line.
Questions and Answers

Solution:

Here, 
Inclination of the line (θ) = 45° 
∴ Slope of the line (m) = tanθ 
= tan 45°  
= 1
y-intercept (c) = 2
∴ Equation of the line in slope intercept form is given by, y = mx + c
So, 
y = 1 . x + 2
or, y - x - 2 = 0 is the required equation.

Solution:

Here,
Inclination of the line (θ) = 60°
∴ Slope of the line (m) = tan 60° = \(\sqrt{3}\)
∴ Equation of the line is given by, y = mx + c
or, y = \(\sqrt{3}\) x + c . . . . . . . . . . . . (1)
Since the line passes though (0, 5) 
5 = c 
∴ c = 5
Putting the value of c in   (1) 
y = \(\sqrt{3}\) x + 5 is the required equation.

Here, 
X-intercept (a) = 3 
Y-intercept (b) = -4
Equation of a staright line is given by \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1
or, \(\frac{x}{3}\) - \(\frac{y}{4}\) = 1
or, 4x - 3y = 12 
4x - 3y - 12 = 0 is the required equation.

Solution:

Here,
The line cuts off, equal; intercepts on the axes equal in magnitude but opposite in sign. 
So, X-intercept = -Y-intercept, i.e. a = -b.
So, its equation is given by \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1
or, x - y = a . . . . . . . . . (i) 
Since it passes through the point (6, 5)
or, 6 - 5 = a
∴ a =  1
Putting the value of a in equation (i)
x - y - 1 = 0 is the required equation.

Solution:

Let AB be the required line that cuts X-axis at A and Y-axis at B such that OA = a and OB = b.
By the questions, C(-2, 3) divides the line AB in the ratio 3 : 4.
Using section formula
or, x = \(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}\), y =  \(\frac{m_1y_2 + m_2y_1}{m_1 + m_2}\)
or, -2 = \(\frac{3.0 + 4.a}{3 + 4}\), 3 = \(\frac{3.b + 4.0}{3 + 4}\)
∴ a = \(\frac{-7}{2}\), ∴ b = 7
The equation of line is \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1
\(\frac{x}{\frac{7}{-2}}\) + \(\frac{y}{7}\) = 1
∴ 2x - y + 7 = 0 is the required equation.

Solution:

Here,
perpendicular length (p) = 3
Angle made by perpendicular in the positive direction with X-axis (α) = 120°
Equation of the line is given by x cosα + y sinα = p
or, x . cos 120° + y sin 120° = 3
or, x . (\(\frac{-1}{2}\)) + y . \(\frac{\sqrt{3}}{2}\) = 3
or, -x + \(\sqrt{3}\) y = 6
∴ x - \(\sqrt{3}\) y + 6 = 0 is the required equation.

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