SOLUTION: I need to write the slope-intercept form of the equation of the line that is perpendicular to 7x-8y= -2 and contains the point(6,3). I cannot get passed this step: y-3= -8/7(x-6),

Question 129653: I need to write the slope-intercept form of the equation of the line that is perpendicular to 7x-8y= -2 and contains the point(6,3). I cannot get passed this step: y-3= -8/7(x-6), how do you multiply -8/7 * -6?
Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
FIRST YOU MUST CHANGE THE FORMULA TO A LINE FORMULA OF THE FORM Y=mX+b WHERE m=SLOPE & b=Y INTERCEPT.
7X-8Y=-2
-8Y=-7X-2
Y=-7X/-8-2/-8
Y=7X/8+1/4 NOW YOU HAVE A SLOPE=7/8 & A Y INTERCEPT=1/4 (RED LINE)
FOR A LINE TO BE PERPENDICULAR TO THIS LINE IT WILL HAVE A SLOPE = THE NEGATIVE RECIPRICAL OF THIS SLOPE. THUS THE SLOPE=-8/7.
BECAUSE THIS PERPENDICULAR LINE PASSES THROUGH (6,3) YOU NOW NEED TO REPLACE THE X & Y VALUES IN THE LINE EQUATION WITH THESE VALUES.
3=-8/7*6+b
3=-56/7+B
b=3+56/7
b=(21+56)/7
b=77/7
b=11 ANSWER FOR THE Y INTERCEPT.
THUS THE EQUATION FOR THIS PERPENDICULAR LINE IS:
Y=-8X/7+11 (GREEN LINE)
(graph 300x200 pixels, x from -10 to 10, y from -10 to 10, of TWO functions 7x/8 +1/4 and -8x/7 +11).
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