# Different Forms Of The Equation Of Line

## Trigonometry # Different Forms Of The Equation Of Line

We know that there are infinite points in the coordinate plane. Consider an arbitrary point P(x,y) on the XY plane and a line L. How will we confirm whether the point is lying on the line L? This is where the importance of equation of a straight line comes into the picture in two-dimensional geometry.

Equation of a straight line contains terms in x and y. If the point P(x,y) satisfies the equation of the line, then the point P lies on the line L.

## Different forms of equations of a straight line

1. Equations of horizontal and vertical lines

Equation of the lines which are horizontal or parallel to the X-axis is y = a, where a is the y – coordinate of the points on the line.

Similarly, equation of a straight line which is vertical or parallel to Y-axis is x = a, where a is the x-coordinate of the points on the line.

For example, the equation of the line which is parallel to X-axis and contains the point (2,3) is y= 3.

Similarly, the equation of the line which is parallel to Y-axis and contains the point (3,4) is x = 3. 2. Point-slope form equation of line

Consider a non-vertical line L whose slope is m, A(x,y) be an arbitrary point on the line and $$P(x_1,y_1)$$ be the fixed point on the same line. Slope of the line by the definition is,

$$m$$ = $$\frac{y~-~y_1}{x~-~x_1}$$

$$y~-~y_1$$ = $$m(x~-~x_1)$$

For example, equation of the straight line having a slope $$m$$ = $$2$$ and passes through the point $$(2,3)$$ is

y – 3 = 2(x – 2)

y= 2x-4+3

2x-y-1 = 0

3. Two-point form equation of line

Let P(x,y) be the general point on the line L which passes through the points $$A(x_1,y_1)$$ and $$B(x_2,y_2)$$. Since the three points are collinear,

slope of PA = slope of AB

$$\frac{y~-~y_1}{x~-~x_1}$$ = $$\frac{y_2~-~y_1}{x_2~-~x_1}$$
$$y – y_{1} = (y_{2} – y_{1}). \frac{x – x_{1} }{x_{2} – x_{1}}$$

4. Slope-intercept form equation of line

Consider a line whose slope is $$m$$ which cuts the $$Y$$-axis at a distance ‘a’ from the origin. Then the distance a is called the $$y$$– intercept of the line. The point at which the line cuts $$y$$-axis will be $$(0,a)$$. Then, equation of the line will be

y-a = m(x-0)

y = mx+a

Similarly, a straight line having slope m cuts the X-axis at a distance b from the origin will be at the point (b,0). The distance b is called x- intercept of the line.

Equation of the line will be:

y = m(x-b)

5. Intercept form

Consider a line L having x– intercept a and y– intercept b, then the line touches X– axis at (a,0) and Y– axis at (0,b). By two-point form equation,

$$y~-~0$$ = $$\frac{b-0}{0~-~a} (x~-~a)$$

$$y$$ = $$-\frac{b}{a} (x~-~a)$$

$$y$$ = $$\frac{b}{a} (a~-~x)$$

$$\frac{x}{a} ~+ ~\frac{y}{b}$$ = $$1$$

For example, equation of the line which has $$x$$– intercept $$3$$ and $$y$$– intercept $$4$$ is,

$$\frac{x}{3} ~+ ~\frac{y}{4}$$ = $$1$$

$$4x~ + ~3y$$ = $$12$$

6. Normal form

Consider a perpendicular from the origin having length l to line L and it makes an angle β with the positive X-axis. Let OP be the perpendicular from the origin to the line L.
Then,

$$OQ$$ = $$l~ cosβ$$

$$PQ$$ = $$l~ sinβ$$

Coordinates of the point $$P$$ are; $$P(l ~cos~β,l ~sin~β)$$

slope of the line $$OP$$ is $$tan~β$$

Therefore,

$$Slope~ of ~the ~line ~L$$ = $$-\frac{1}{tan~β}$$ = $$-\frac{cos~β}{sin~β}$$

Equation of the line $$L$$ having slope $$-\frac{cos~β}{sin~β}$$ and passing through the point $$(l ~cos~β,l ~sin~β)$$ is,

$$y~-~l~ sin~β$$ = $$-\frac{cos~β}{sin~β} (x~-~l ~cosβ)$$

$$y ~sin~β~-~l~ sin^2~ β$$ = $$-x ~cos~β~+~l~cos^2~β$$

$$x~ cos~β~ + ~y ~sin~β$$ = $$l(sin^2~β ~+ ~cos^2~β)$$

$$x ~cos~β~ + ~y ~sin~β$$ = $$l$$

You have learnt about the different forms of equation of a straight line. To know more about straight lines and its properties, log onto www.byjus.com.’