SOLUTION: Please help! I am confused on how to work this problem. Give the equation of a line perpendicular to the line y = -8x + 8 having the y-intercept 2 units below the y-intercept of

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: Please help! I am confused on how to work this problem. Give the equation of a line perpendicular to the line y = -8x + 8 having the y-intercept 2 units below the y-intercept of      Log On


   

Question 274169: Please help! I am confused on how to work this problem.
Give the equation of a line perpendicular to the line y = -8x + 8 having the y-intercept 2 units below the y-intercept of the given equation. At what point the two equations cross each other?

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Give the equation of a line perpendicular to the line y = -8x + 8 having the y-intercept 2 units below the y-intercept of the given equation.
Find the slope of perpendicular line (m1)
The slope of the given equation is -8
:
m1 * -8 = -1; (the slope relationship between perpendicular lines)
m1 = +1%2F8 is the slope of the perpendicular line
:
The y intercept of this line = 6; (two units below the given equation)
therefore:
y = 1%2F8x + 6; the equation of the perpendicular line
:
At what point the two equations cross each other?
When y's are equal, find x:
-8x + 8 = 1%2F8x + 6
-8x = 1%2F8x + 6 -8
-8x = 1%2F8x - 2
multiply by 8 to get rid of the denominator, results:
-64x = x - 16
-64x - x = -16
-65x = -16
x = +16%2F65
x = .246
:
Find y using the given equation
y = -8(.246) + 8
y = 6.03
:
Check in the other equation
y = 1%2F8(.246) + 6
y = 6.03
:
equations intersect at x=.246, y=6.03
:
+graph%28+300%2C+300%2C+-7%2C+7%2C+-3%2C+7%2C+-8x%2B8%2C+.125x%2B6%29+