Finding The Equation Of A Line

Finding The Equation of A Line

One of the most important topics in Algebra is finding the equation of the line. In this hub I present five problems with their solution.


Problem Number One :

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


Solution:

The point-slope form of the equation of the line through the point (X1, Y1) and slope m is :

Y – Y1=m (X - X1).

Let us first find the slopem of the line using the formula for finding the slope given two points.

m=Y2 – Y1/X2 – X1

Designate(-5, -1 ) as (X1, Y1 )and (3, 3 ) as(X2, Y2).

m = ( 3 – (-1))/ (3 – (-5))= ( 3 +1) /(3+ 5 )=4/8=½

Then you may use either of the two points to substitutein the equation :

Y – Y1= m (X – X1) .

Using the point (-5, -1 ) we got

Y – (-1)=½(X – (-5))

2[Y+ 1=½( X + 5)]

2Y + 2 = X + 5

X-2Y +3= 0is the of the line we are looking for.


Problem Number Two :

Find the equation of the line which is parallel to the line3X + 6Y = 7 and passing through(3, -5 ).


Solution :

Parallel lines have the same slope.We have to solve for the slope of the line we are looking for and substitute itin the point-slope form equation of the line. To solve for the slope we have to convert the equation 3X + 6Y= 7 into slope-intercept form of the line which isY = mX +b. m is the slope.

1/6[6Y= -3X+7

Y =-1/2X+7/6

The coefficient of X is he slopem=-1/2.-1/2 is also the slope of the line we are solving since they are parallel lines.Then substitute m = -1/2 and the point ( 3, -5 )in the point-slope form equation:

2[ Y – (-5)=- ½ (X – 3 ) ]

2Y + 10=X -3

X – 2Y -13 = 0 is the equation of the line we are looking for.



Problem Number Three :

Find an equation of the line passing through (4, 5) which is perpendicular to the line7X + 6Y = -3.


The slope of a line perpendicular to a certain line is the negative reciprocal of the slope of that certain line perpendicular to the given line . To solve for the slopeof 7X + 6Y= -3 we rearrange this equation to slope-intercept form orY = mX + b.

7X+6Y=-3

[6Y=-7X -3]

Y= -7/6 X- ½

The slope of the line we are solving for is6/7which is negative reciprocal of -7/6. Reciprocal of 7/6 and opposite in sign. Then substitutem = 6/7and ( 4, 5) to the point-slope form of the line:

7[Y – 5=6/7 (X – 4 )]

7Y – 35=6X- 24

6X – 7Y + 11= 0 is the equation of the line we are looking for.


Problem Number Four :

Find the equation of the line with the same X-interceptas the line2X – 9Y = 14 and parallel to the lineX – Y = 21,


Solution :

We have to solve for the X-interceptof the line 2X -9Y = 14. To solve for the X-intercept let Y = 0.

2X- 9(0)=14

(2X= 14) 1/2

X=7 is the X-intercept.

Since X-intercept = 7, we consider (7, 0) is appoint in the line we are solving. The line is parallel to X –Y = 21 so has the same slope as this line. Convert X-Y = 21 into slope-intercept form :

(- Y =-X+ 21) -1

Y= X + 21

m= 1

We use m = 1 and point (7,0) to solve for the line. Substituting this value into point-slope form we get :

Y=X- 7

X – Y – 7 = 0 is the equation of the line we are looking for.


Problem Number Five :

Find the equation of the line with X-intercept as 7 and Y-intercept as -3.


Solution :

(7, 0 ) and (0,-3 ) are two points of this line. WE solve first for m.

m=-3 /-7=3/7. Then we use either of the two points to substitute in the point-slope form : Let us use (7, 0):

[ y = 3/7(X – 7 ) ] 7

7Y = 3X -21

3X – 7Y -21= 0is the equation of the line we are looking for.



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Comments 4 comments

Lord De Cross profile image

Lord De Cross 5 years ago

Very instructive Cristina. Took those subjects back in my first semester of analythical Geometry. Nice for our young hubbers! Voted useful!

LORD


cristina327 profile image

cristina327 5 years ago from Manila Author

Hi lord de cross thank you for appreciating this hub. Blessings to you and to your family always. Best regards.


Dave Mathews profile image

Dave Mathews 5 years ago from NORTH YORK,ONTARIO,CANADA

Ate Cristina: Thank you. Why would this be important and to whom?


cristina327 profile image

cristina327 5 years ago from Manila Author

Hi Dave Matthews thanks for dropping by. This hub is meant for students of Algebra and Analytic Geometry. Blessings to you always and regards.

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