# 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)Â on the XYÂ plane and a line L. How will we confirm whether the point is lying on line L? This is where the importance of the equation of a straight line comes into the picture in two-dimensional geometry.

The equation of a straight line contains terms in xÂ and y. If the point P(x,y)Â satisfies the equation of the line, then the point PÂ 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-axisÂ is y = a,Â where a is the y â€“ coordinate of the points on the line.

Similarly, the equation of a straight line which is vertical or parallel to the 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 the X-axis and contains the point (2,3) is y= 3.

Similarly, the equation of the line which is parallel to the Y-axis and contains the point (3,4)Â is xÂ = 3.

2. Point-slope form equation of a line

Consider a non-vertical line LÂ whose slope is m, A(x,y) be an arbitrary point on the line andÂ  P(x1, y1) be the fixed point on the same line.

Slope of the line by the definition is,

$$\begin{array}{l}m = \frac{y~-~y_1}{x~-~x_1}\end{array}$$
$$\begin{array}{l}y~-~y_1 = m(x~-~x_1)\end{array}$$

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

y – 3Â = 2(x – 2)

y= 2x-4+3

2x-y-1Â = 0

3. Two-point form equation of line

Let P(x,y)Â be the general point on the line L which passes through the points A(x1, y1) and B(x2, y2).

Since the three points are collinear,

slope of PAÂ = slope of AB

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

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-interceptÂ of the line. The point at which the line cuts the y-axis will be (0, a).

Then, equation of the line will be

y-aÂ = m(x-0)

yÂ = mx+a

Similarly, aÂ 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 LÂ having x– intercept a and y– intercept b, then the line touches X– axis at (a,0)Â and Y– axis at (0,b).

By two-point form equation,

$$\begin{array}{l}y~-~0 = \frac{b-0}{0~-~a} (x~-~a)\end{array}$$
$$\begin{array}{l}y= -\frac{b}{a} (x~-~a)\end{array}$$
$$\begin{array}{l}y = \frac{b}{a} (a~-~x)\end{array}$$
$$\begin{array}{l}\frac{x}{a} ~+ ~\frac{y}{b} = 1\end{array}$$

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

$$\begin{array}{l}\frac{x}{3} ~+ ~\frac{y}{4} = 1\end{array}$$
$$\begin{array}{l}4x~ + ~3y = 12\end{array}$$

6. Normal form

Consider a perpendicular from the origin having length lÂ to line L and it makes an angle Î²Â with the positive X-axis.

Let OP be perpendicular from the origin to line L.
Then,

$$\begin{array}{l}OQ = l~ cos \beta \end{array}$$
$$\begin{array}{l}PQ = l~ sin\beta \end{array}$$

Coordinates of the point Pare;

$$\begin{array}{l}P(l ~cos~\beta ,l ~sin~\beta )\end{array}$$

The slope of the line OP is tan Î²

Therefore,

$$\begin{array}{l}Slope~ of ~the ~line ~L = -\frac{1}{tan~\beta}=-\frac{cos~\beta}{sin~\beta }\end{array}$$

Equation of the line L having slope -cos Î² / sin Î² and passing through the point (l cos Î², l sin Î²) is,

$$\begin{array}{l}y~-~l~ sin~\beta = -\frac{cos~\beta }{sin~\beta } (x~-~l ~cos\beta )\end{array}$$
$$\begin{array}{l}y ~sin~\beta ~-~l~ sin^2~ \beta = -x ~cos~\beta ~+~l~cos^2~\beta \end{array}$$
$$\begin{array}{l}x~ cos~\beta ~ + ~y ~sin~\beta = l(sin^2~\beta ~+ ~cos^2~\beta )\end{array}$$
$$\begin{array}{l}x ~cos~\beta ~ + ~y ~sin~\beta = l\end{array}$$

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