Three Standard form of Equation of Straight line | kullabs.com
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## Note on Three Standard form of Equation of Straight line

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#### 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.

• 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.

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• ### 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