SOLUTION: find the equation of the line passing (0,1) parallel to the line 2x-3y=6
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Question 632316: find the equation of the line passing (0,1) parallel to the line 2x-3y=6
Found 2 solutions by mananth, Maths68:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
2 x -3 y = 6
Find the slope of this line
make y the subject
-3 y = -2 x + 6
Divide by -3
y = 2/3 x -2
Compare this equation with y=mx+b
slope m = 2/3
The slope of a line parallel to the above line will be the same
The slope of the required line will be 2/3
m= 2/3 ,point ( 0 , 1 )
Find b by plugging the values of m & the point in
y=mx+b
1 = 0 + b
b= 1
m= 2/3
Plug value of the slope and b in y = mx +b
The required equation is y = 2/3 x + 1
m.ananth@hotmail.ca
Answer by Maths68(1474) (Show Source): You can put this solution on YOUR website!
Equation of the line slope-intercept form
y=mx+b
Given
Equation of the Given line
2x-3y=6
-3y=-2x+6
-3y/-3=(-2x+6)/-3
y=2x/3-2
y=(2/3)x-2
Compare with the equation of line slope-intercept form
Slope of the given = m = 2/3
Since lines are parallel their slope will be same
Slope of the required line = m = 2/3
Point (0, 1)
We have a point and slope we can get easily the require line equation.
Use Equation of the line point-slope form
m = y2-y1/x2-x1
Put the values in above equation, we have
2/3=(y-1)/(x-0)
2/3=(y-1)/x
2x=3(y-1)
2x=3y-3
-3y=-2x-3
-3y/-3=-(2x+3)/-3
y=2x/3+3/3
y=(2/3)x+1
Above equation is the equation of the required line
Graph
=====
y=(2/3)x-2 (Given line in green)
y=(2/3)x+1 (Required line in Red)
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