Question 519008: Find the equation of the line using the given information and in the requested form.
1.Perpendicular to 7x-2y=-2 and passing through the point (7,1); in slope-intercept form.
2.Parallel to 3y-4x=3 and passing through the point (0,5); in slope-intercept form.
Answer by Maths68(1474)   (Show Source):
You can put this solution on YOUR website! 1.Perpendicular to 7x-2y=-2 and passing through the point (7,1); in slope-intercept form.
Standard Form of Equation of the line:
y=mx+b
Given
7x-2y=-2
rearrage the above equation according to the standard form
-2y=-7x-2
-2y/-2=(-7x-2)/-2
y=(7/2)x+1
Compare above equation with the standard form equation
m=7/2 and b=1
Since lines are perpendicular multiplicatin of their slope will be (-1)
So slope of the required line will be (-2/7)
Now we have a point(7,1) and slope (-2/7)of the line we can easily find required lines by putting these values in the equation of the straight line poin-slope form.
m=(y2-y1)/(x2-x1)
-2/7=(y-1))/(x-7)
-2(x-7)=7(y-1))
-2x+14=7y-7
-2x-7y=-7-14
-2x-7y=-21
Multiply by -1 both sides of the equation
2x+7y=21
7y=-2x+21
7y/7=(-2x+21)/7
y=(-2/7)x+3
Above equation is slope intercept form of the equation of the required line
2.Parallel to 3y-4x=3 and passing through the point (0,5); in slope-intercept form.
Standard Form of Equation of the line:
y=mx+b
Given
3y-4x=3
rearrage the above equation according to the standard form
3y-4x=3
3y=4x+3
3y/3=(4x+3)/3
y=4/3(x)+1
Compare above equation with the standard form equation
m=4/3 and b=1
Since lines are parallel their slopes will be same.
So slope of the required line will be (4/3)
Now we have a point(0,5) and slope (4/3)of the line we can easily find required lines by putting these values in the equation of the straight line poin-slope form.
m=(y2-y1)/(x2-x1)
4/3=(y-5))/(x-0)
4(x-0)=3(y-5)
4x=3y-15
-3y=-4x-15
-3y/-3=(-4x-15)/-3
y=(4/3)x+5
Above equation is slope intercept form of the equation of the required line
|
|
|
