SOLUTION: can you please help me find, the equation of the line: passing through (6,6) and perpendicular to 5x-3y-8=0

Question 515964: can you please help me find, the equation of the line: passing through (6,6) and perpendicular to 5x-3y-8=0
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
can you please help me find, the equation of the line: passing through (6,6) and perpendicular to 5x-3y-8=0
:
Put the equation in the slope/intercept form to find the slope
5x - 3y - 8 = 0
-3y = -5x + 8
y has to be positive, multiply both sides by -1
3y = 5x - 8
divide both sides by 3
y = x -
:
slope m1 =
the relationship of perpendicular lines: is m1*m2 = -1, to find m2
*m2 = -1
multiply both sides by 3/5
m2 = is the slope of the perpendicular line
:
Use the point/slope form to find the equation of the perpendicular line
y - y1 = m(x - x1)
x1 = 6, y1 = 6, m=-3/5
:
y - 6 = (x - 6)
y - 6 = x -
y - 6 = x + 3.6
y = x + 3.6 + 6
y = x + 9.6; the equation of the perpendicular line
:
Check this, substitute 6 for x in the equation and ensure that y = 6

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