FINDING EQUATION OF A LINE GIVEN TWO POINTS

The equation of a straight line is satisfied by the co-ordinates of every point lying on the straight line and not by any other point outside the straight line. 

Equation of a line using two points on the line :

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

Here (x1, y1) and (x2, y2) are the points on the line.

Example 1 :

Find the equation of the line which is passing through the points (2, -8)  and (-5, 2). 

Solution :

Here (x1, y1) = (2, -8) and (x2, y2) = (-5, 2).

Equation of a line :

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

(y - (-8))/(2 - (-8)) =  (x - 2)/(-5 - 2)

(y + 8)/(2 + 8) =   (x - 2)/(-7)

(y + 8)/10 =   (x - 2)/(-7)

-7(y + 8) = 10(x - 2)

-7y - 56 = 10 x - 20

10 x + 7y - 20 + 56 = 0

10 x + 7y + 36 = 0

Example 2 :

Find the equation of the line which is passing through the points (3, 0)  and (4, -1). 

Solution :

Here (x1, y1) = (3, 0) and (x2, y2) = (4, -1).

Equation of a line :

(y - 0)/(-1 - 0) =  (x - 3)/(4 - 3)

y/(-1) =   (x - 3)/1

1(y) = -1(x - 3)

y = - x + 3

x + y - 3 = 0

Example 3 :

Find the equation of the line which is passing through the points (-1, -5)  and (-3, -1).

Solution :

Here (x1, y1) = (-1, -5) and (x2, y2) = (-3, -1).

Equation of a line :

(y - (-5))/(-1 - (-5)) =  (x - (-1))/(-3 - (-1))

(y + 5)/(-6) =   (x + 1)/(-3 + 1)

(y + 5)/(-6) =   (x + 1)/(-2)

-2(y + 5) = -6(x + 1)

-2y - 10 = -6 x - 6

6 x - 2 y - 10 + 6 = 0

6 x - 2 y - 4 = 0

÷ by 2 =>   3 x - y - 2 = 0

Example 4 :

Find the equation of the line which is passing through the points (0,-5)  and (4,-6). 

Solution :

Here (x1, y1) = (0, -5) and (x2, y2) = (4, -6).

Equation of a line :

(y - (-5))/(-6 - (-5)) =  (x - 0)/(4 - 0)

                        (y + 5)/(-6 + 5) =   x/4

                        (y + 5)/(-1) =   x/4

                           4(y + 5) = -1(x)

                            4y + 20 = - 1 x

                           x + 4y + 20 = 0

Example 5 :

Find the equation of the line which is passing through the points (4, -3)  and (0, -2). 

Solution :

Here (x1, y1) = (4, -3) and (x2, y2) = (0, -2).

Equation of a line :

(y - (-3))/(-2 - (-3)) =  (x - 4)/(0-4)

(y +  3)/(-2 + 3) = (x - 4)/(-4)

(y + 3)/1 =  (x - 4)/(-4)

  -4(y + 3) = 1(x - 4)

  -4y - 12 = 1 x - 4

 x + 4y + 12 - 4 = 0

x + 4y + 8 = 0

Example 6 :

Find the equation of the line which is passing through the points (2, -1)  and (-1, -7). 

Solution :

Here (x1, y1) = (2, -1) and (x2, y2) = (-1, -7).

Equation of a line :

(y - (-1))/(-7 - (-1)) =  (x - 2)/(-1 - 2)

(y + 1)/(-7 + 1) =   (x - 2)/(-3)

(y + 1)/(-6) =   (x - 2)/(-3)

-3(y + 1) = -6(x - 2)

-3y - 3 = -6 x + 12

6 x - 3 y - 3 - 12 = 0

6 x - 3 y - 15 = 0

     ÷ by 3 =>       2 x - y - 5 = 0  

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Problems on Finding Derivative of a Function

    Mar 29, 24 12:11 AM

    Problems on Finding Derivative of a Function

    Read More

  2. How to Solve Age Problems with Ratio

    Mar 28, 24 02:01 AM

    How to Solve Age Problems with Ratio

    Read More

  3. AP Calculus BC Integration of Rational Functions by Partical Fractions

    Mar 26, 24 11:25 PM

    AP Calculus BC Integration of Rational Functions by Partical Fractions (Part - 1)

    Read More