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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.

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.

 

  • y = mx+ c
  • \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1
  • x cos∝ + y sin∝ = p
.

Very Short Questions

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:

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.

0%
  • The equation of a straight line having Slope = -1 and y-intercept = 5 is ______.

    x + y = 5 
    x + y = 0
    x - y + 5 = 1
    x + y - 5 = 0
  • Find the equation of straight line when Slope = (frac{3}{5}) and y-intercept = (frac{4}{7}) units.

    21x - 35y + 20 = 0
    30x + 3y + 27 = 0
    20x - 5y = 0 
    25x + y -10 = 0 
  • Find the equation of straight line if Inclination (θ) = 30° and y-intercept = 4 units.

    x - (sqrt{3})y + 4(sqrt{3}) = 0 
    (sqrt{x}) - (sqrt{3y}) = 0
    3x + y (sqrt{12}) = 0  
    7x + (sqrt{y}) - 3= 0
  • Find the equation of a straight line whose X-intercept (a) = 5 and y-intercept (b) = 6.

    3y + y = 0
    x - 2y + 9 = 0
    6x + 5y - 30 = 0 
    7x + 6yn - 10 = 0
  • Find the equation of a straight line whose X-intercept (a) = -7 and Y-intercept (b) = -2.

    2x + 7y - 14 = 0
    2x + 7y - 23 = 0
    7x + 3y = 0
    x - 6y = 0 
  • Find the equation of a straight line whose X-intercept (a) = (frac{3}{5}) and Y-intercept (b) = (frac{6}{5}).

    8x - 6y - 19 = 0  
    9x - y + 21 = 0 
    12x + 3y - 9 = 0  
     10x + 5y - 6 = 0 
  • Find the equation of straight line whose X-intercept (a) = (frac{2}{3}) and Y-intercept (b) = (frac{-4}{5}).

    6x - 2y - 12 = 0 
    5x + y + 2 = 0
    6x - 5y - 4 = 0
    5x + 7yb - 12 = 0
  • Find out the equation of a straight line whose inclination is 45° and y-intercept is 2.

    x + y - 3 = 0 
    y - x - 2 = 0
     x - 2y + 5 = 0
    2x + 2y + 2 = 0
  • find the equation of a line passing through the point (0, 5) and inclined at an angle of 60° with x-axis.

    x = 3y - 2
    x + y - 12 = 0
    y = 3x - 5
    y = (sqrt{3})x + 5
  • Find the equation of a straight line whose X-intercept and Y-intercept are 3 and -4 respectively.

    6x + y + 12 = 0 
    x + y - 2 = 0
    4x - 3y - 12 = 0
    2x + 3y - 7 = 0 
  • Find the equation of straight line if Inclination (θ) = 135° and y-intercept = (frac{3}{2}) units.

    2x + 2y - 3 = 0 
     2x + 5y - 2 = 0
    x + y + 3 = 0 
    3x - 4y - 3 = 0 
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