SOLUTION: Hi, Here is the problem I am having troubles with. Any help would be greatly appreciated Find the equation, in standard form, with all integer coefficients, of the line perp

Algebra ->  Graphs -> SOLUTION: Hi, Here is the problem I am having troubles with. Any help would be greatly appreciated Find the equation, in standard form, with all integer coefficients, of the line perp      Log On


   

Question 78146: Hi,
Here is the problem I am having troubles with. Any help would be greatly appreciated
Find the equation, in standard form, with all integer coefficients, of the line perpendicular to x + 2y = 8 and passing through (1, -6).
Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
x+2y=8
2y=-x+8
y=-x/2+8/2
y=-x/2+4 THUS THIS LINE HAS A SLOPE OF(m)=(-1/2)
THUS A PERPENDICULAR LINE HAS A SLOPE OF +2.
NOW REPLACE THE X & Y TERMS WITH 1 & -6 & SOLVE FOR THE Y INTERCEPT(b)
Y=mX+b
-6=2*1+b
-6=2+b
b=-6-2
b=-8 THE Y INTERCEPT. THUS THE EQUATION IS:
Y=2X-8
the red line is the original line & the green line is the perpendicular line.
+graph%28+300%2C+300%2C+-10%2C+10%2C+-10%2C+10%2C+y+=+-x%2F2+%2B4%2C+y+=+2x+-8%29+ (graph 300x300 pixels, x from -10 to 10, y from -10 to 10, of TWO functions y = -x/2 +4 and y = 2x -8).