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).
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 ).
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.
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.
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,
We have to solve for the X-interceptof the line 2X -9Y = 14. To solve for the X-intercept let Y = 0.
(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
We use m = 1 and point (7,0) to solve for the line. Substituting this value into point-slope form we get :
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.
(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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