THE STRAIGHT LINE

General Form of a Straight Line: A first degree equation in two variable x,y represents a straight line, that is ax+by+c=0 (not both a,b=0). This is called General Form of Equation of a line.
Examples: 2x+3y+1=0; x+y-5=0
Note:
(1) Equation of X-Axis is Y=0
(2) Equation of Y-Axis is X=0
(3) Equation of a line parallel to X-Axis is Y=k ( Here k is a constant)
(4) Equation of a line parallel to Y-Axis is X=k ( Here also k is a constant)

The General Form of a equation of a line is ax+by+c=0. The line Parallel to the above line is in the form of ax+by+k=0, and the line Perpendicular to ax+by+c=0, is of the form bx-ay+k=0

Slope of a Line: The angle (theta) made by a line with X-axis in the positive direction is called the Inclination and Tan (theta) is called slope of the line.

Example: If a line is inclined at an angle 60 degrees with +ve direction of the X-axis then slope of the line tan(60 degrees)= Sqrt(3).

Formula for finding Slope of a Line:
If A(x1,y1), B(x2,y2) are two points on a line, then the slope of the line is given by m= (y2-y1)/(x2-x1) (or) (difference in y-co-ordinates)/(difference in x-co-ordinates)
Example: Find the slope of the line joining the points A(-1,2) & B(1,3)
Slope m= (3-2)/(1-(-1))= 1/2
Note:
(1) Slope of X-axis (or) any line parallel to X-axis is Zero(0)
(2) Slope of Y-axis (or) any line parallel to Y-axis is Not Defined

VARIOUS FORMS OF A LINE:
(1) General form of a line is ax+by+c=0
(2) Slope Intercept form of a line is y=mx+c, here m is the slopeof the line and c is the Y-intercept of the line
(3) Point-Slope form of a line is y-y1=m(x-x1), here (x1,y1) is a point on the line and m is the slope of the line
(4) Two-Point form of a line is ((y-y1)/(x-x1)) = ((y2-y1)/(x2-x1)), here (x1,y1) and (x2,y2) are two points on the line
(5) Intercept form of a Straight Line is ((x/a) + (y/b)) = 1, here a, b are X and Y intercepts respectively

Examples:

(1) Find the equation of a line passing through the points A(-1,0) and B( 2,3)

Solution: Two-Point form of equation of a line is given by ((y-y1)/(x-x1)) = ((y2-y1)/(x2-x1))
The required equation is ((y-0)/(x+1)) = ((3-0)/(2+1))
On simplification we get the equation is x-y+1=0

(2) Find the Equation of a line passing through the point A( 2,3) and having slope m=-1

Solution: Point Slope form of equation of a line is given by (y-y1) = m(x-x1)
The required equation is y-3 = -1(x-2)
On simplification we get the equation as x+y-5 = 0

(3) Find the equation of a line with X & Y intercepts respectively as 1/2,3

Solution: Intercept Form of a line is given by ((x/a) + (y/b)) =1
The required equation is ((x/(1/2)) + ( y/3)) = 1
On simplification we get the equation as 6x+y-3 = 0

(4) Find the equation of a line parallel to X-Axis and passing through (-1,1)

Solution: Equation of a line parallel to X-Axis is y=k
The line y=k passes through (-1,1), substituting this point in the equation we get y=1
Therefore the required equation is y=1

(5) Find the equation of a line parallel to Y-axis and passing through the point (2/3,1)

Solution: Equation of a line parallel to Y-axis is x=k,
The line x=k passes through the point (2/3,1), substituting this point in the equation we get
x= (2/3)
On simplification the required equation is 3x-2=0

(6) Find the equation of a line parallel to 2x+3y-2 = 0 and passing through the point (-2,1)

Solution: The equation of a line parallel to ax+by+c = 0 is ax+by+k = 0
The equation of a line parallel to 2x+3y-2 = 0 is 2x+3y+k =0
The above line passes through the point (-2,1), substituting this point in the equation we get
2(-2) + 3(1) + k = 0
On simplification we get k = 1,
Therefore the required equation is 2x+3y+1 = 0

(7) Find the equation of a line perpendicular to 3x+y-2 = 0 and passing through the point (4,-5)

Solution: The equation of a line perpendicular to ax+by+c = 0 is bx-ay+k = 0
The equation of a line perpendicular to 3x+y-2 = 0 is x-3y+k =0, this line passes through the point (4,5), substituting this point in the above equation we get 4-3(5)+k = 0
On simplification we get k= 11
Therefore the required equation is x-3y+11 = 0


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