SOLUTION: what is the equation of the line perpendicular to -8x+2y=-2 that contains point (-5,8)

Question 105078: what is the equation of the line perpendicular to -8x+2y=-2 that contains point (-5,8)
Answer by checkley75(3666) (Show Source): You can put this solution on YOUR website!
FIRST WE NEED TO FIND THE SLOPE OF THE GIVEN LINE USING ---
Y=mX+b WHERE m=SLOPE & b= THE Y INTERCEPT.
-8X+2Y=-2
2Y=8X-2
Y=8X/2-2/2
Y=4X-1 THUS THIS EQUATION HAS A SLOPE (m)=4.
A PERPENDICULAR LINE WILL HAVE A SLOPE OF THE NEGATIVE RECIPRICAL OF THIS SLOPE:
THUS THE PERPENDICULAR LINE HAS A SLOPE=-1/4.
NOW REPLACE THE X & Y TERMS BY -5 & 8 & SOLVE FOR b.
8=-1/4*-5+b
8=5/4+b
b=8-5/4
b=(32-5)/4
b=27/4 THIS IS THE Y INTERCEPT OF THE PERPENDICULAR LINE THROUGH THE POINT
(-5,8).
Y=-X/4+27/4 IS THE DESIRED EQUATION.

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