SOLUTION: Write an equation of the line containing the given point and perpendicular to the given line. (6,-2); 7x+6y=5

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Write an equation of the line containing the given point and perpendicular to the given line. (6,-2); 7x+6y=5      Log On


   

Question 156043: Write an equation of the line containing the given point and perpendicular to the given line.
(6,-2); 7x+6y=5
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
(6,-2); 7x+6y=5
6y=-7x+5
y=-7x/6+5/6 this line has a slope=-7/6. (red line)
A perpendicular line has a slope=6/7.
-2=6*6/7+b
-2=36/7+b
b=-2-36/7
b=(-2*7-36)/7
b=(-14-36)/7
b=-50/7
Thus the equation of the perpendicular line through (6,-2)is:
y=6x/7-50/7 (green line).
+graph%28+300%2C+300%2C+-10%2C+10%2C+-10%2C+10%2C+-7x%2F6+%2B5%2F6%2C+6x%2F7+-50%2F7%29+ (graph 300x300 pixels, x from -10 to 10, y from -10 to 10, of TWO functions -7x/6 +5/6 and 6x/7 -50/7).