Method 6

Equation of a straight line

This method is based on the sutra " Paravartaya Yojayet "which means 'transpose and apply'.

Explaining with the help of an example .

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

Answer : -

Step 1 : -

The difference of the y coordinates acts as the coefficient of x and
the difference of the x coordinates acts as the coefficient of y .
Like this

diff. of x coordinates = 3 - 4 = -1 acts as y coeficient

diff. of y coordinates = 4 - 2 = 2 acts as x coefficient

Write x first and put a - negative sign in between .

So our equation is

2.x - (-1).y
2.x + y which is the right hand part of the equation .

Step 2 : -

Substitute any given point from ( 3 , 4 ) and ( 4 , 2 ) into the above obtained equation to find the right hand side of the straight line equation .

2.(3) + 4 = 10


Hence the equation of the straight line is

2.x - y = 10


Example 2 : - Find the equation of a straight line passing through the points
( 5, 2 ) and ( 9 , 1 ) .

Answer : -

Step 1 : -

diff. of x coordinates = 5 - 9 = -4 acts as coefficient of y

diff. of y coordinates = 2 - 1 = 1 acts as coefficient of x

x - (-4).y

Hence L.H.S. of the equation is

x + 4.y

Step 2 : -

Substitute ( 5 , 2 ) or ( 9 , 1 ) in the above obtained equation

5 + 4.(2) = 13

which is the R.H.S. of the straight line equation .

Hence the equation of the straight line is

x + 4.y = 13


Example 3 : - Find the equation of the straight line passing through the points
(5 , 5) and (3, 4) .

Answer : -

Step 1 : -

diff. of x coordinates = 5 - 3 = 2 = y coefficient
diff. of y coordinates = 5 - 4 = 1 = x coefficient

Hence the L.H.S of the equation is
x - 2.y

Step 2 : -

Substitute (5 , 5) or (3 , 4) in the equation

5 - 2.5 = 5 - 10 = -5

Hence the equation of line is

x - 2.y = -5

Algebraic Proof : -

For a straight line passing through two points ( x1 , y1 ) and ( x2 , y2 )

the equation is

y - y1 = (y2 - y1 ) ( x - x1 )
(x2 - x1 )

Cross multiplying

(x2 - x1 ) (y - y1 ) = (y2 - y1 ) (x - x1 )
Solving

(x2 - x1 ) (- y1) + ( y2 - y1) (x1) = (y2 - y1) x - (x2 - x1 ) y

In the above we see that coefficient of x is difference of y coordinates and coefficient of y is difference of x coordinates .

For obtaining the constant term we know that on substituting any one point we can easily know the term .

Exercises : -

Find the equation of the straight linepassing through the points

1.) ( 9 , 0 ) and ( 2 , 1 )
2.) ( 2 , 4 ) and ( 1 , 9 )
3.) ( 5 , 3 ) and ( 4 , 2 )
4.) ( 8 , 4 ) and ( 5 , 9 )
5.) ( 6 , 3 ) and ( 4 , 9 )