Straight Line Equation

The equation of straight line is generally written as    y = mx + b

Graphical Representation

where, m= slope or gradient of the straight line equation

b = the y-intercept

Suppose we want to find equation of a straight line that passes through a known point and has a known slope. Let (x ,y) represent the co-ordinates of any point on the line and let (x1 ,y1) represent the co-ordinates of other point. The slope of the straight line equation is given as,

m = (y-y_1)/(x-x_1)

After finding the slope m as we are given the co-ordinates of the point (x1 ,y1) in the equation    y= mx + b Then, the constant 'b' can also be found so that finally the straight line equation is obtained.

Other Forms of straight Line Equations

Different Forms of Straight Line Equations:

There are many other forms of Straight Line Equations.

1. Straight Line Equation through two points:

The line through two different points ( x1 ,y1) and ( x2 ,y2) is given by

y-y1  = [(y_2-y_1)/(x_2-x_1)]  . (x - x1)

2. Straight Line Equation in general form:

A straight line is defined by a linear equation as

Ax+By+C=0     where A , B are not both 0

3. Straight Line Equation in Intercept-intercept form:

Let us consider that a straight line intersects x-axis at (a , 0) and y-axis at (0 , b). Then it is defined by equation

(x/a) +(y/b) = 1

4. Straight Line Equation in Point-Slope Form:

The equation of straight line through the point (a , b) with slope m is

y = m ( x - a ) + b

Solved Examples on equations of straight line

Ex:1 Find the equation of a line passing through (2 , 3) and having a slope of 3?

sol: step 1: Compare the given point(2 , 3) with the general point (x1 , y1)

Now x1=2  and  y1=3

step 2: we have the straight line equation as  y-y1=m(x-x1)

now substitute the given point in the above equation

y-3=3(x-2)

y-3=3x-6

3x-y=6-3

3x-y-3=0

The straight line equation is 3x-y-3=0

Ex:2 Find out the straight line equation of the line passing through the points (1,2) and (2,4).

Sol: Given points are compared with (x1 , y1) and (x2 , y2) and substitute the points in the equation

y-y1  = [(y_2-y_1)/(x_2-x_1) ] . (x-x1)

y-2  =  [(4-2)/(2-1)] . (x-1)

y-2 = 2(x-1)

y-2 = 2x-2

2x - y = 0

The straight line equation is 2x-y=0.

Practice Problems on equations of straight line

Pro:1 Find the equation of the line passing through the points (-3,4) and (4,-2)

Ans: we have      

m = (y_2-y_1)/(x_2-x_1)

=(-2-4)/(4+3)

=(-6/7)

let (x,y) be compared to (-3,4)

y-4 = (-6/7) . (x-(-3))

7(y-4) = -6(x+3)

7y-28 = -6x-18

7y-28 = -6x-18

7y+6x = 10

Pro:2 Write equation of line having points and slopes as follows;

P(3,3) , m=-2

P(-2,-1) , m=1/3

P1(2,2) and P2(-4,-1)

y-intercept = 2 , m=3

Ans: The answers to the above given practice problems are

y+2x=11

3y-x=-1

2y-x=2

y-3x=2